Chapter 7: Rational Expressions and Equations
Chapter 7: Rational Expressions and Equations 7.1 Simplifying Rational Expressions Part A: Rational Expressions Evaluate for the given set of x‐values. 1. PROBLEM: 5 ; {−1, 0, 1} x
ANSWER: Substitute the values in for x. x = −1
x=0
5 5 = = undefined x 0
5 5 = = −5 x ( −1)
2.
x =1
5 5 = =5 x 1
PROBLEM: 4x ; {−1, 0, 1} 3x 2 ANSWER: Substitute the values in for x. x = −1
4 ( −1) 4x 4 = =− 2 2 3x 3 3 ( −1) 3.
x=0
4x 4 (0) = = undefined 3x 2 3 ( 0 )
x =1
4 (1) 4 4x = = 2 2 3x 3 3 (1)
PROBLEM: 1 ; {−10, −9, 0} x+9 ANSWER: Substitute the values in for x. x = −10
1 1 = = −1 x + 9 −10 + 9
x = −9
5 5 = = undefined x 0
x=0
5 5 = =5 x 1
4.
PROBLEM: x+6 ; {−6, 0, 5} x −5 ANSWER: Substitute the values in for x. x = −6
x=0
x + 6 −6 + 6 = =0 x − 5 −6 − 5
5.
x+6 0+6 6 = =− x −5 0−5 5
x=5
x + 6 5 + 6 11 = = = undefined x−5 5−5 0
PROBLEM: 3x ( x − 2 ) ; {0, 1/2, 2} 2x −1 ANSWER: Substitute the values in for x. x=0
x=
3x ( x − 2 ) =
2x −1 3 ( 0 )( 0 − 2 ) 2 ( 0) −1
=0
x=2
1 2
⎛ 1 ⎞⎛ 1 ⎞ 3⎜ ⎟ ⎜ − 2 ⎟ 3x ( x − 2 ) 2 2 ⎠ = ⎝ ⎠⎝ 2x −1 ⎛1⎞ 2 ⎜ ⎟ −1 ⎝2⎠ 3⎛ 3⎞ ⎜− ⎟ 2⎝ 2⎠ = = undefined 0
3x ( x − 2 ) 2x −1
=
3 ( 2 )( 2 − 2 ) 2 ( 2) −1
=
0 =0 3
6.
PROBLEM: 9 x2 − 1 ; {0, 1/3, 7} x−7 ANSWER: Substitute the values in for x. x=0
9 x 2 − 1 9 ( 0 ) − 1 −1 1 = = = −7 7 x−7 0−7
7.
x=
x=7
1 3
2
⎛1⎞ 9 ⎜ ⎟ −1 2 9x −1 1−1 3 = ⎝ ⎠ = 1 20 x−7 −7 − 3 3 0 = 20 − 3 =0
9 x 2 − 1 9 ( 7 ) − 1 440 = = x−7 7−7 0 = undefined 2
PROBLEM: 5 ; {−3, 0, 3} 2 x −9 ANSWER: Substitute the values in for x. x = −3
5 5 5 = = 2 2 x − 9 ( −3) − 9 0 = undefined
x=0
5 5 5 = =− x −9 0−9 9 2
x=3
5 5 5 = = 2 x − 9 ( 3) − 9 0 2
= undefined
8.
PROBLEM: x 2 − 25 ; {−5, −4, 5} x 2 − 3x − 10 ANSWER: Substitute the values in for x. x = −5
x = −4
x=5
2 2 x 2 − 25 −5 ) − 25 −4 ) − 25 ( ( x 2 − 25 x 2 − 25 = = 2 x 2 − 3x − 10 ( −5 )2 − 3 ( −5) − 10 x 2 − 3 x − 10 ( −4 )2 − 3 ( −4 ) − 10 x − 3 x − 10 2 5 − 25 ( ) 0 1 0 −9 = = 2 = = =− ( 5) − 3 ( 5) − 10 0 25 + 15 − 10 18 2 =0 = undefined
9.
PROBLEM: Fill in the following chart:
1 x
x −3 −2 0 2 3
ANSWER:
1 x
x −3 −2 0 2 3
1 x 1 x 1 x 1 x 1 x
1 1 =− −3 3 1 1 = =− −2 2 1 = = undefined 0 1 = 2 1 = 3 =
10.
PROBLEM: Fill in the following chart:
x
1 x−2
−1 0 1 2 3 ANSWER:
x −1 0
1 2 3
1 x−2 1 1 1 = =− 3 x − 2 −1 − 2 1 1 1 = =− 2 x−2 0−2 1 1 1 = = = −1 x − 2 1 − 2 −1 1 1 1 = = = undefined x−2 2−2 0 1 1 1 = = =1 x − 2 3− 2 3− 2
11.
PROBLEM: Fill in the following chart:
x
1 x+2
−4 −3 −2 −1 0 ANSWER:
x −4 −3 −2 −1 0
1 x+2 1 1 1 1 = = =− x + 2 −4 + 2 −2 2 1 1 1 = = = −1 x + 2 −3 + 2 −1 1 1 1 = = = undefined x + 2 −2 + 2 0 1 1 1 = = =1 x + 2 −1 + 2 1 1 1 1 = = x+2 0+2 2
An objects weight depends on the height above the surface of the earth. If an object weighs 120 pounds on the surface of the Earth, then its weight W, in pounds, x miles above the surface is approximated by the formula: 120* 40002 W= (4000 + x) 2 Approximate the weight of a 120 lb object at the given height above the surface of the Earth. (1 mile = 5280 feet)
12.
PROBLEM: Fill in the following chart:
x
1 −2 x
−2 −1 0 1 2 ANSWER:
x −2 −1 0 1 2
1 −2 x 1 5 − −2 = − 2 2 1 1 − 2 = − − 2 = −3 1 x 1 − 2 = undefined 0 1 − 2 = −1 1 1 3 −2 = − 2 2
An objects weight depends on the height above the surface of the earth. If an object weighs 120 pounds on the surface of the Earth, then its weight W, in pounds, x miles above the surface is approximated by the formula:
13.
PROBLEM: 100 miles SOLUTION: 120* 40002 120* 40002 W= = = 114.2 ~ 114 lbs (4000 + x) 2 (4000 + 100) 2 ANSWER: ~ 114 lbs
14.
PROBLEM: 1000 miles SOLUTION: 120* 40002 120* 40002 W= = = 76.8 lbs (4000 + x) 2 (4000 + 1000) 2 ANSWER: 76.8 lbs
15.
PROBLEM: 44, 350 feet SOLUTION:
44,350 feet × W=
1 mile 4435 = miles 5280 feet 528
120* 40002 120* 40002 = = 119.5 lbs 2 (4000 + x) 2 ⎛ 4435 ⎞ ⎜ 4000 + ⎟ 528 ⎠ ⎝
ANSWER: 119.5 lbs
The price to earnings ratio (P/E) is a metric used to compare the valuations of similar publically traded companies. The P/E ratio is calculated using the stock price and the earnings per share (EPS) over the previous 12 month period as follows: Price per Share P/E = Earnings per Share If each share of a company stock is priced at $22.40, then calculate the P/E ratio given the following values for the earnings per share.
16.
PROBLEM: 90,000 feet SOLUTION:
90, 000 feet ×
1 mile 375 = miles 5280 feet 22
120* 40002 120* 40002 W= = = 118.9 lb ~ 119 lbs 2 (4000 + x) 2 ⎛ 375 ⎞ ⎜ 4000 + ⎟ 22 ⎠ ⎝ ANSWER: ~ 119 lbs The price to earnings ratio (P/E) is a metric used to compare the valuations of similar publically traded companies. The P/E ratio is calculated using the stock price and the earnings per share (EPS) over the previous 12 month period as follows: Price per Share P/E = Earnings per Share If each share of a company stock is priced at $22.40, then calculate the P/E ratio given the following values for the earnings per share.
17.
PROBLEM: $1.40 SOLUTION: Price per Share $22.40 P/E = = = 16 Earnings per Share $1.40 ANSWER: 16
18.
PROBLEM: $1.21 SOLUTION: Price per Share $22.40 P/E = = = 18.5 Earnings per Share $1.21 ANSWER: 18.5
19.
PROBLEM: What happens to the P/E ratio when earnings decrease? ANSWER: If the denominator increases, i.e. the earnings per share decreases the P/E ratio increases.
State the restrictions to the domain.
20.
PROBLEM: What happens to the P/E ratio when earnings increase? ANSWER: If the denominator increases, i.e. the earnings per share increases, the P/E ratio decreases.
State the restrictions to the domain.
21.
PROBLEM: 1 3x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 3x = 0 x=0
ANSWER: The domain consists of any real number x where x ≠ 0 .
22.
PROBLEM: 3x 2 7 x5 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 7 x5 = 0
x=0 ANSWER: The domain consists of any real number x where x ≠ 0 .
23.
PROBLEM: 3x ( x + 1)
x+4 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x+4=0 x + 4 − 4 = −4 x = −4 ANSWER: The domain consists of any real number x where x ≠ −4 .
24.
PROBLEM: 2 x 2 ( x − 3)
x −1 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x −1 = 0
x −1+1 = 0 +1 x =1 ANSWER: The domain consists of any real number x where x ≠ 1 .
25.
PROBLEM: 1 5x − 1 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 5x − 1 = 0 5x − 1 + 1 = 0 + 1 5x = 1 5 1 x= 5 5 1 x= 5 1 ANSWER: The domain consists of any real number x where x ≠ . 5
26.
PROBLEM: x−2 3x − 2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 3x − 2 = 0
3x − 2 + 2 = 0 + 2 3x = 2 3 2 x= 3 3 2 x= 3 ANSWER: The domain consists of any real number x where x ≠
27.
2 . 3
PROBLEM: x −9 5x ( x − 2) SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 5x ( x − 2) = 0
5x = 0 x=0
x−2=0 x−2+2 = 0+2
or
x=2 ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ 2 . 28.
PROBLEM: 1 ( x − 3)( x + 6 ) SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: ( x − 3)( x + 6) = 0
x −3 = 0
or
x − 3 + 3 = 0 + 3 or x=3 or
x+6 =0 x+6−6 = 0−6 x = −6
ANSWER: The domain consists of any real number x where x ≠ 3 and x ≠ −6 .
29.
PROBLEM: x 1 − x2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 1 − x2 = 0
(1 − x )(1 + x ) = 0 1− x = 0 1 − 1 − x = −1 x =1
or or or
1+ x = 0 1 − 1 + x = −1 x = −1
ANSWER: The domain consists of any real number x where x ≠ ±1 .
30.
PROBLEM: x2 − 9 x 2 − 36 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x 2 − 36
( x − 6)( x + 6 ) = 0 x−6 = 0 or x − 6 + 6 = 0 + 6 or x=6 or
x+6=0 x+6−6 = 0−6 x = −6
ANSWER: The domain consists of any real number x where x ≠ ±6 .
31.
PROBLEM: 1 2 x ( x + 3)( 2 x − 1) SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 x ( x + 3)( 2 x − 1) = 0
2x = 0 x=0 x=0
x + 3 = 0 or 2x −1 = 0 x + 3 − 3 = 0 − 3 or 2 x − 1 + 1 = 0 + 1 x = −3 or 2 x = 1 2 1 x = 0 or x = −3 or x = 2 2 1 x = 0 or x = −3 or x = 2 ANSWER: The domain consists of any real number x where x ≠ 0 , x ≠ −3 , and 1 x≠ . 2 32.
or or or
PROBLEM: x−3 ( 3x − 1)( 2 x + 3) SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: ( 3x − 1)( 2 x + 3) = 0
3x − 1 = 0 3x − 1 + 1 = 0 + 1 3x = 1 3 1 x= 3 3 1 x= 3
or or or or or
2x + 3 = 0 2x + 3 − 3 = 0 − 3 2 x = −3 2 3 x=− 2 2 3 x=− 2
ANSWER: The domain consists of any real number x where x ≠
1 3 and x ≠ − . 3 2
33
PROBLEM: 4 x ( 2 x + 1)
12 x 2 + x − 1 SOLUTION: 4 x(2 x + 1) 2 12 x + x − 1
4 x(2 x + 1) (4 x − 1)(3 x + 1) Therefore, 4 x − 1 = 0and 3 x + 1 = 0 1 x = and 4 1 x=− 3 =
ANSWER: The domain consists of any real number, where 34.
x≠
1 1 x≠− 4 and 3
PROBLEM: x −5 3x 2 − 15 x SOLUTION: x−5 3x 2 − 15 x x −5 = 3x( x − 5) 1 = 3x ANSWER: The domain consists of any real number where x ≠ 0
Part B: Simplifying Rational Expressions State the restrictions and then simplify.
35.
PROBLEM: 5x2 20 x3 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 20 x 3 = 0
x=0 ANSWER: The domain consists of any real number x where x ≠ 0 . 5x2 1 1 = 3− 2 = 3 20 x 4x 4x
36.
PROBLEM: 12 x 6 60 x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 60 x = 0 x=0 ANSWER: The domain consists of any real number x where x ≠ 0 . 12 x 6 1 6−1 x5 = x = 60 x 5 5
37.
PROBLEM: 3x 2 ( x − 2 ) 9 x ( x − 2)
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 9x ( x − 2) = 0
9x = 0 x=0
or or
x−2=0 x−2+2 = 0+2
x=0
or
x=2
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ 2 . 3x 2 ( x − 2 ) 3x 2 x 2−1 x = = = 9 x ( x − 2) 9x 9 9
38.
PROBLEM: 20 ( x − 3)( x − 5 ) 6 ( x − 3)( x + 1) SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 6 ( x − 3)( x + 1) = 0
( x − 3)( x + 1) = 0 x −3 = 0 x −3+3 = 0+3
or or
x +1 = 0 x +1 −1 = 0 −1
x=3
or
x = −1
ANSWER: The domain consists of any real number x where x ≠ 3 and x ≠ −1 . 20 ( x − 3)( x − 5 ) 10 ( x − 5) = 6 ( x − 3)( x + 1) 3 ( x + 1)
39.
PROBLEM: 6 x2 ( x − 8)
36 x ( x + 9 )( x − 8 )
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 36 x ( x + 9 )( x − 8 ) = 0
x=0 x=0
or or
x+9 = 0 or x + 9 − 9 = 0 − 9 or
x −8 = 0 x −8+8 = 0+8
x=0
or
x = −9
x =8
or
ANSWER: The domain consists of any real number x where x ≠ 0 , x ≠ −9 and x≠8 . 6 x2 ( x − 8) x 2−1 x = = 36 x ( x + 9 )( x − 8 ) 6 ( x + 9 ) 6 ( x + 9 )
40.
PROBLEM: 16 x 2 − 1
( 4 x + 1)
2
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 ( 4 x + 1) = 0 4x +1 = 0 4x +1−1 = 0 −1 4 x = −1 4 1 x=− 4 4 1 x=− 4
1 ANSWER: The domain consists of any real number x where x ≠ − . 4 2 16 x − 1 ( 4 x − 1)( 4 x + 1) 4 x − 1 = = 2 2 4x +1 ( 4 x + 1) ( 4 x + 1)
41.
PROBLEM: 9x2 − 6x + 1
( 3 x − 1)
2
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 ( 3x − 1) = 0 3x − 1 = 0 3x − 1 + 1 = 0 + 1 3x = 1 3 1 x= 3 3 1 x= 3
ANSWER: The domain consists of any real number x where x ≠
( 3x − 1) = 2 2 ( 3x − 1) ( 3x − 1)
9 x2 − 6 x + 1
42.
1 3
2
=1
PROBLEM: x−7 x 2 − 49 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x 2 − 49 = 0
( x + 7 )( x − 7 ) = 0 x+7 = 0
or
x+7−7 = 0−7 x = −7
or or
x−7 = 0 x−7+7 = 0+7 x=7
ANSWER: The domain consists of any real number x where x ≠ ±7 . x−7 x−7 1 = = 2 x − 49 ( x − 7 )( x + 7 ) x + 7
43.
PROBLEM: x 2 − 64 x2 + 8x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x2 + 8x = 0
x ( x + 8) = 0 x=0 x=0 x=0 x=0
x+8 = 0 x+8 = 0 x +8−8 = 0−8 x = −8
or or or or
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ −8 . x 2 − 64 ( x + 8)( x − 8 ) x − 8 = = x2 + 8x x ( x + 8) x
44.
PROBLEM: x + 10 x 2 − 100 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x 2 − 100 = 0
( x + 10 )( x − 10 ) = 0 x + 10 = 0 x + 10 − 10 = 0 − 10 x = −10
or or or
x − 10 = 0 x − 10 + 10 = 0 + 10 x = 10
ANSWER: The domain consists of any real number x where x ≠ ±10 . x + 10 x + 10 1 = = 2 x − 100 ( x + 10 )( x − 10 ) x − 10
45.
PROBLEM: 2 x3 − 12 x 2 5 x 2 − 30 x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 5 x 2 − 30 x = 0
5x ( x − 6) = 0 5x = 0 x=0 x=0
x−6 = 0 x−6+6 = 0+6 x=6
or or or
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ 6 . 2 2 x3 − 12 x 2 2 x ( x − 6 ) 2 x 2−1 2 x = = = 5 x 2 − 30 x 5 x ( x − 6 ) 5 5
46.
PROBLEM: 30 x 5 + 60 x 4 2 x3 − 8 x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 x3 − 8 x = 0
(
)
2 x x2 − 4 = 0 2 x ( x + 2 )( x − 2 ) = 0 2x = 0 x=0 x=0
x+2=0 or x + 2 − 2 = 0 − 2 or x = −2 or
or or or
x−2 = 0 x−2+2 = 0+2 x=2
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ ±2 . 4 30 x 4 ( x + 2 ) 30 x5 + 60 x 4 30 x ( x + 2 ) 15 x 4−1 15 x3 = = = = 2 x3 − 8 x 2 x ( x − 2 )( x + 2 ) ( x − 2 ) x − 2 2x x2 − 4
(
)
47.
PROBLEM: 2x −1 2 x2 + x − 6 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 x2 + x − 6 = 0 a = 2; b = 1; c = −6 2 −b ± b 2 − 4ac −1 ± 1 − 4 ( 2 )( −6 ) −1 ± 49 −1 ± 7 x= = = = 2a 2 ( 2) 2 ( 2) 4
−1 + 7 −1 − 7 or 4 4 3 = or − 2 2 =
ANSWER: The domain consists of any real number x where x ≠
The fraction cannot be simplified. 2x −1 2 x2 + x − 6
3 and x ≠ −2 . 2
48.
PROBLEM: x2 − x − 6 3x 2 − 8 x − 3 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 3x 2 − 8 x − 3 = 0
( x − 3)( 3x + 1) = 0 x −3 = 0
or
3x + 1 = 0
x −3+3 = 0+3
or
3x + 1 − 1 = 0 − 1
x=3
or
3 x = −1
x=3
or
x=3
or
3 1 x=− 3 3 1 x=− 3
ANSWER: The domain consists of any real number x where x ≠ 3 and x ≠ −
( x − 3)( x + 2 ) = x + 2 x2 − x − 6 = 2 3x − 8 x − 3 ( x − 3)( 3x + 1) 3 x + 1
1 . 3
49.
PROBLEM: 6 x 2 − 25 x + 25 3x 2 + 16 x − 35 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 3 x 2 + 16 x − 35 = 0
( 3x − 5)( x + 7 ) = 0 3x − 5 = 0
or
x+7 =0
3x − 5 + 5 = 0 + 5
or
x+7−7 = 0−7
3x = 5
or
x = −7
or
x = −7
or
x = −7
3 5 x= 3 3 5 x= 3
ANSWER: The domain consists of any real number x where x ≠
6 x 2 − 25 x + 25 ( 2 x − 5)( 3x − 5 ) 2 x − 5 = = 3x 2 + 16 x − 35 ( 3x − 5)( x + 7 ) x+7 50.
5 and x ≠ −7 . 3
PROBLEM: 3x 2 + 4 x − 15 x2 − 9 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x2 − 9 = 0
( x + 3)( x − 3) = 0 x+3= 0 x +3−3 = 0−3 x = −3
or or or
x −3 = 0 x −3+3 = 0+3 x=3
ANSWER: The domain consists of any real number x where x ≠ ±3 . 3x 2 + 4 x − 15 ( 3x − 5)( x + 3) 3x − 5 = = x2 − 9 ( x − 3)( x + 3) x − 3
51.
PROBLEM: x 2 − 10 x + 21 x 2 − 4 x − 21 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x 2 − 4 x − 21 = 0
x 2 − 7 x + 3 x − 21 = 0 x ( x − 7) + 3( x − 7) = 0
( x + 3)( x − 7 ) = 0 x+3= 0 x +3−3 = 0−3 x = −3
or or or
x−7 = 0 x−7+7 = 0+7 x=7
ANSWER: The domain consists of any real number x where x ≠ −3 and x ≠ 7 . x 2 − 10 x + 21 ( x − 3)( x − 7 ) x − 3 = = x 2 − 4 x − 21 ( x + 3)( x − 7 ) x + 3
52.
PROBLEM: x3 − 1 x2 − 1 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x2 − 1 = 0
( x − 1)( x + 1) = 0 x −1 x −1 + 1 = 0 + 1 x =1
or or or
x +1 = 0 x +1−1 = 0 −1 x = −1
ANSWER: The domain consists of any real number x where x ≠ ±1 . 2 x3 − 1 ( x − 1) x + x + 1 x 2 + x + 1 = = x2 − 1 x +1 ( x − 1)( x + 1)
(
)
53.
PROBLEM: x3 + 8 x2 − 4 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x2 − 4 = 0
( x + 2 )( x − 2 ) = 0 x+2=0 x+2−2 = 0−2 x = −2
x−2=0 x−2+2 = 0+2 x=2
or or or
ANSWER: The domain consists of any real number x where x ≠ ±2 . x3 + 8 ( x + 2 )( x − 2 x + 4 ) x − 2 x + 4 = = x2 − 4 x−2 ( x + 2 )( x − 2 )
54.
PROBLEM: x 4 − 16 x2 − 4 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x2 − 4 = 0
( x + 2 )( x − 2 ) = 0 x+2=0 x+2−2 = 0−2 x = −2
x−2=0 x−2+2 = 0+2 x=2
or or or
ANSWER: The domain consists of any real number x where x ≠ ±2 . x2 − 4 x2 + 4 x 4 − 16 = = x2 + 4 2 2 x −4 x −4
(
)(
)
Part C: Simplifying Rational Expressions State the restrictions and then simplify.
55.
PROBLEM: x −9 9− x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 9− x = 0 9 − 9 x = −9 − x = −9 x=9
ANSWER: The domain consists of any real number x where x ≠ 9 . x −9 x −9 = = −1 9 − x − ( x − 9)
56.
PROBLEM: 3x − 2 2 − 3x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 − 3x = 0 2 − 2 − 3x = 0 − 2 − 3 x = −2 −3 −2 x= −3 −3 2 x= 3 2 ANSWER: The domain consists of any real number x where x ≠ . 3 3x − 2 3x − 2 = = −1 2 − 3x − ( 3x − 2 )
57.
PROBLEM: x+6 6+ x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 6+ x = 0 6−6+ x = 0−6 x = −6 ANSWER: The domain consists of any real number x where x ≠ −6 . x+6 x+6 = =1 6+ x x+6
58.
PROBLEM: 3x + 1 1 + 3x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 1 + 3x = 0 1 − 1 + 3x = 0 − 1 3x = −1 3 1 x=− 3 3 1 x=− 3 1 ANSWER: The domain consists of any real number x where x ≠ − . 3 3x + 1 3x + 1 = =1 1 + 3x 3x + 1
59.
PROBLEM: ( 2 x − 5)( x − 7 )
( 7 − x )( 2 x − 1)
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: ( 7 − x )( 2 x − 1) = 0
7−x =0 7−7− x = 0−7 x = −7
or or or
x=7
or
x=7
or
2x −1 = 0 2x −1+1 = 0 +1 2x = 1 2 1 x= 2 2 1 x= 2
ANSWER: The domain consists of any real number x where x ≠
( 2 x − 5)( x − 7 ) = ( 2 x − 5)( x − 7 ) = − 2 x − 5 ( 7 − x )( 2 x − 1) − ( x − 7 )( 2 x − 1) 2 x − 1 60.
1 and x ≠ 7 . 2
PROBLEM: ( 3x + 2 )( x + 5 )
( x − 5 )( 2 + 3x )
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: ( x − 5)( 2 + 3x ) = 0
x −5 = 0 x −5+5 = 0+5 x=5
or or or
x=5
or
x=5
or
2 + 3x = 0 2 − 2 + 3x = 0 − 2 3 x = −2 3 2 x=− 3 3 2 x=− 3
2 ANSWER: The domain consists of any real number x where x ≠ − and x ≠ 5 . 3 3 x + 2 x + 5 3 x + 2 x + 5 ( )( ) = ( )( ) = x + 5 ( x − 5)( 2 + 3x ) ( x − 5)( 3x + 2 ) x − 5
61.
PROBLEM: x2 − 4
(2 − x)
2
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve:
(2 − x)
2
=0
2− x = 0 2−2− x = 0−2 x=2
ANSWER: The domain consists of any real number x where x ≠ 2 . x 2 − 4 ( x − 2 )( x + 2 ) x + 2 = = 2 2 x−2 (2 − x) ( x − 2)
62.
PROBLEM: 16 − 9 x 2
( 3x + 4 )
2
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve:
( 3x + 4 )
2
=0
3x + 4 = 0 3x + 4 − 4 = 0 − 4 3 4 x=− 3 3 4 x=− 3 4 ANSWER: The domain consists of any real number x where x ≠ − . 3 2 ( 3x + 4 )( 4 − 3x ) = 4 − 3x 16 − 9 x = 2 2 3x + 4 ( 3x + 4 ) ( 3x + 4 )
63.
PROBLEM: 4 x 2 (10 − x ) 3 x 3 − 300 x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 3 x 3 − 300 x = 0
(
)
3 x x 2 − 100 = 0 3x = 0
or
x 2 − 100 = 0
x=0
or
( x + 10 )( x − 10 ) = 0
x=0 x=0
or or
x + 10 = 0 x + 10 − 10 = 0 − 10
or x − 10 = 0 or x − 10 + 10 = 0 + 10
x=0
or
x = −10
or
x = 10
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ ±10 . 4 x 2 (10 − x ) 4 x 2 (10 − x ) 4x = =− 3 3x − 300 x −3 x ( x + 10 )(10 − x ) 3 ( x + 10 )
64.
PROBLEM: −2 x + 14 x3 − 49 x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x3 − 49 x = 0
(
)
x x 2 − 49 = 0 x ( x − 7 )( x + 7 ) = 0 x=0 x=0 x=0
or or or
x−7 = 0 or x − 7 + 7 = 0 + 7 or x=7 or
x+7 =0 x+7−7 = 0−7 x = −7
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ ±7 . −2 ( x − 7 ) −2 x + 14 2 = =− 3 x − 49 x x ( x − 7 )( x + 7 ) x ( x + 7)
65.
PROBLEM: 2x2 − 7 x − 4 1 − 4x2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 1 − 4x2 = 0
(1 + 2 x )(1 − 2 x ) = 0 1 + 2x = 0
or
1− 2x = 0
1 −1+ 2x = 0 −1
or
1 − 1 − 2 x = −1
2 x = −1
or
− 2 x = −1
2 1 x=− 2 2 1 x=− 2
−2 −1 x= −2 −2 1 x= 2
or or
ANSWER: The domain consists of any real number x where x ≠ ±
2 x 2 − 7 x − 4 ( 2 x + 1)( x − 4 ) x−4 = = 2 1 − 4x ( 2 x + 1)(1 − 2 x ) 1 − 2 x 66.
1 . 2
PROBLEM: 9 x2 − 4 4 x − 6x2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 4 x − 6 x2 = 0 2 x ( 2 − 3x ) = 0 2x = 0
or
2 − 3x = 0
x=0
or
2 − 2 − 3x = 0 − 2
x=0
or
x=0
or
x=0
or
− 3 x = −2 −3 −2 x= −3 −3 2 x= 3
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠
9 x 2 − 4 ( 3x − 2 )( 3x + 2 ) 3x + 2 = =− 2 −2 x ( 3x − 2 ) 4x − 6x 2x
2 . 3
67.
PROBLEM: x 2 − 5 x − 14 7 − 15 x + 2 x 2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 7 − 15 x + 2 x 2 = 0
( 2 x − 1)( x − 7 ) = 0 2x −1 = 0
or
x−7 = 0
2x −1+1 = 0 +1
or
x−7+7 = 0+7
2x = 1
or
x=7
or
x=7
or
x=7
2 1 x= 2 2 1 x= 2
ANSWER: The domain consists of any real number x where x ≠
( x + 2 )( x − 7 ) = x + 2 x 2 − 5 x − 14 = 2 7 − 15 x + 2 x ( 2 x − 1)( x − 7 ) 2 x − 1 68.
1 and x ≠ 7 . 2
PROBLEM: 2 x3 + x 2 − 2 x − 1 1 + x − 2x2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 1 + x − 2x2 = 0
( 2 x + 1)( − x + 1) = 0 2x +1 = 0 2x +1−1 = 0 −1 2 x = −1 1 x=− 2
or or or or
− x +1 = 0 − x + 1 −1 = 0 −1 − x = −1 x =1
1 ANSWER: The domain consists of any real number x where x ≠ − and x ≠ 1 . 2 3 2 ( x + 1)( 2 x + 1)(1 − x ) = − x − 1 2x + x − 2x −1 =− 2 1+ x − 2x ( 2 x + 1)(1 − x )
69.
PROBLEM: x3 + 2 x − 3x 2 − 6 2 + x2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 + x2 = 0
x2 + 2 − 2 = 0 − 2 x 2 = −2 x = ± −2 ANSWER: The roots of x are complex numbers. Hence, there are no restrictions to the domain. The fraction cannot be simplified further. 2 + x 2 ( x − 3) x3 + 2 x − 3x 2 − 6 = = x −3 2 + x2 2 + x2
(
70.
(
)
)
PROBLEM: 27 + x3 x2 + 6 x + 9 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x2 + 6 x + 9 = 0
( x + 3)
2
=0
x+3= 0 x +3−3 = 0−3 x = −3 ANSWER: The domain consists of any real number x where x ≠ −3 . ( x + 3) x 2 − 3 x + 9 x 2 − 3 x + 9 27 + x3 = = 2 x2 + 6 x + 9 x+3 ( x + 3)
(
)
71.
PROBLEM: 64 − x3 x 2 − 8 x + 16 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x 2 − 8 x + 16 = 0
( x − 4)
2
=0
x−4 = 0 x−4+4 = 0+4 x=4 ANSWER: The domain consists of any real number x where x ≠ 4 . − ( x − 4 ) x 2 + 4 x + 16 x 2 + 4 x + 16 64 − x3 = = − 2 x 2 − 8 x + 16 x−4 ( x − 4)
(
72.
)
PROBLEM: x2 + 4 4 − x2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 4 − x2 = 0 2− x = 0 or 2+ x = 0 2−2− x = 0−2 or 2−2+ x = 0−2 − x = −2 or x = −2 or x = −2 x=2 ANSWER: The domain consists of any real number x where x ≠ ±2 . The fraction cannot be simplified. x2 + 4 4 − x2
Simplify. (Assume all denominators are nonzero.)
73.
PROBLEM: −15 x 3 y 2 5 xy 2 ( x + y ) SOLUTION: 3x 2 −15 x 3 y 2 = − 5 xy 2 ( x + y ) x+ y ANSWER: −
74.
3x 2 x+ y
PROBLEM: 4 14 x 7 y 2 ( x − 2 y ) 7 x8 y ( x − 2 y )
2
SOLUTION: 4 14 x 7 y 2 ( x − 2 y ) 7 x8 y ( x − 2 y )
ANSWER:
75.
2
=
2 y 2−1 ( x − 2 y )
2y ( x − 2y)
x8−7 2
x
PROBLEM: y+x x2 − y 2 SOLUTION: ( x + y) = 1 y+x = x 2 − y 2 ( x + y )( x − y ) x − y ANSWER:
1 x− y
4−2
=
2y ( x − 2y) x
2
76.
PROBLEM: y−x x2 − y2 SOLUTION: −( x − y) y−x 1 = =− 2 2 x −y ( x − y )( x + y ) x + y ANSWER: −
77.
1 x+ y
PROBLEM: x2 − y 2
( x − y)
2
SOLUTION: x 2 − y 2 ( x − y )( x + y ) x + y = = 2 2 x− y ( x − y) ( x − y)
ANSWER:
78.
x+ y x− y
PROBLEM: a 2 − ab − 6b 2 a 2 − 6ab + 9b 2 SOLUTION: a 2 − ab − 6b 2 ( a + 2b )( a − 3b ) a + 2b = = 2 a 2 − 6ab + 9b 2 a − 3b ( a − 3b ) ANSWER:
a + 2b a − 3b
79.
PROBLEM: 2a 2 − 11a + 12 −32 + 2a 2 SOLUTION: 2a 2 − 11a + 12 ( a − 4 )( 2a − 3) 2a − 3 = = 2 −32 + 2a 2 ( a − 4 )( a + 4 ) 2 ( a + 4 ) ANSWER:
80.
2a − 3 2 ( a + 4)
PROBLEM: a 2b − 3a 2 3a 2 − 3ab SOLUTION: 2 a 2b − 3a 2 a ( b − 3) a ( b − 3) = = 3a 2 − 3ab 3a ( a − b ) 3 ( a − b ) ANSWER:
81.
a ( b − 3)
3( a − b)
PROBLEM: xy 2 − x + y 3 − y x − xy 2 SOLUTION:
(
)
2 xy 2 − x + y 3 − y ( x + y ) y − 1 x+ y = =− 2 2 x − xy x −x y −1
(
ANSWER: −
x+ y x
)
82.
PROBLEM: x3 − xy 2 − x 2 y + y 3 x 2 − 2 xy + y 2 SOLUTION: x3 − xy 2 − x 2 y + y 3 ( x − y )( x + y )( x − y ) = = x+ y 2 x 2 − 2 xy + y 2 ( x − y) ANSWER: x + y
83.
PROBLEM: x3 − 27 x 2 + 3x + 9 SOLUTION:
(
)
( x − 3) x 2 + 3 x + 9 x3 − 27 = = x −3 x 2 + 3x + 9 x 2 + 3x + 9
ANSWER: x − 3
84.
PROBLEM: x2 − x + 1 x3 + 1 SOLUTION:
(
)
x3 + 1 x2 − x + 1 1 = = 3 3 x +1 ( x + 1) x + 1 x + 1
ANSWER:
1 x +1
(
)
Part D: Rational Functions Calculate the following.
85.
PROBLEM: 5x f ( x) = ; f ( 0) , f ( 2) , f ( 4) x −3 ANSWER: Substitute the values in for x. f ( x ) = f (0)
f ( x) = 86.
5 ( 0) 0−3
f ( x ) = f ( 2)
f ( x) =
=0
5 ( 2) 2−3
= −10
f ( x ) = f ( 4)
f ( x) =
5 ( 4) 4−3
= 20
PROBLEM: x+7 f ( x) = 2 ; f ( −1) , f ( 0 ) , f (1) x +1 ANSWER: Substitute the values in for x. f ( x ) = f ( −1)
f ( x) = 87.
( −1) + 7 ( −1)
2
+1
=
f ( x ) = f (0)
f ( x) =
6 =3 2
0+7 =7 0 +1
f ( x ) = f (1)
f ( x) =
1+ 7 =4 12 + 1
PROBLEM: x3 ; g ( 0 ) , g ( 2 ) , g ( −2 ) g ( x) = 2 ( x − 2) ANSWER: Substitute the values in for x. g ( x ) = g (0)
g ( x) =
0
( 0 − 2)
2
g ( x ) = g ( 2)
=0
g ( x) =
23
( 2 − 2)
2
= undefined
g ( x ) = g ( −2 )
g ( x) =
( −2 )
3
( −2 − 2 )
2
=
−8 1 =− 16 2
88.
PROBLEM: x2 − 9 g ( x) = ; g ( −2 ) , g ( 0 ) , g ( 2 ) 9 − x2 ANSWER: Substitute the values in for x. g ( x ) = g ( −2 )
g ( x) = 89.
( −2 ) − 9 = −1 2 9 − ( −2 ) 2
g ( x ) = g (0)
g ( x) =
0−9 = −1 9−0
g ( x ) = g ( 2)
g ( x) =
22 − 9 = −1 9 − 22
PROBLEM: x3 g ( x) = 2 ; g ( −1) , g ( 0 ) , g (1) x +1 ANSWER: Substitute the values in for x. g ( x ) = g ( −1)
g ( x) = 90.
( −1) = − 1 2 ( −1) + 1 2 3
g ( x ) = g (0)
g ( x) =
0 =0 0 +1
g ( x ) = g (1)
g ( x) =
13 1 = 2 1 +1 2
PROBLEM: 5x + 1 g ( x) = 2 ; g ( −1/ 5 ) , g ( −1) , g ( −5 ) x − 25 ANSWER: Substitute the values in for x. ⎛ 1⎞ g ( x) = g ⎜ − ⎟ ⎝ 5⎠
⎛ 1⎞ 5⎜ − ⎟ +1 5 g ( x) = ⎝ 2 ⎠ ⎛ 1⎞ ⎜ − ⎟ − 25 ⎝ 5⎠ 0 = =0 2 ⎛ 1⎞ ⎜ − ⎟ − 25 ⎝ 5⎠
g ( x ) = g ( −1)
g ( x) =
5 ( −1) + 1
( −1)
2
− 25
=
g ( x ) = g ( −5 )
−4 1 = −24 6
g ( x) =
5 ( −5 ) + 1
( −5 )
= undefined
2
− 25
=
−24 0
State the restrictions to the domain and then simplify.
91.
PROBLEM: −3x 2 − 6 x f ( x) = 2 x + 4x + 4 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x2 + 4 x + 4 = 0
( x + 2)
2
=0
x+2=0 x + 2 − 2 = −2 x = −2 ANSWER: The domain consists of any real number x where x ≠ −2 . −3x 2 − 6 x −3x ( x + 2 ) 3x = =− 2 2 x + 4x + 4 x+2 ( x + 2)
f ( x) = − 92.
3x x+2
PROBLEM: x2 + 6 x + 9 f ( x) = 2 2 x + 5x − 3 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 x2 + 5x − 3 = 0
( 2 x − 1)( x + 3) = 0 2x −1 = 0 2x −1+1 = 0 +1 x=
1 2
or or
x+3= 0 x + 3 − 3 = −3
or
x = −3
ANSWER: The domain consists of any real number x where x ≠
( x + 3) x2 + 6x + 9 x+3 = = 2 2 x + 5 x − 3 ( 2 x − 1)( x + 3) 2 x − 1 2
f ( x) =
x+3 2x −1
1 and x = −3 . 2
93.
PROBLEM: 9− x g ( x) = 2 x − 81 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x 2 − 81 = 0
( x + 9 )( x − 9 ) = 0 x+9 = 0
or
x+9−9 = 0−9 x = −9
or or
x −9 = 0 x −9+9 = 0+9 x=9
ANSWER: The domain consists of any real number x where x ≠ ±9 . − ( x − 9) 9− x 1 = =− 2 x − 81 ( x + 9 )( x − 9 ) x−9
g ( x) = − 94.
1 x −9
PROBLEM: x3 − 27 g ( x) = 3− x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 3− x = 0 3−3− x = 0−3 − x = −3 x=3
ANSWER: The domain consists of any real number x where x ≠ 3 . 2 x3 − 27 ( x − 3) x + 3x + 9 = = − x 2 + 3x + 9 3− x − ( x − 3)
(
(
g ( x ) = − x 2 + 3x + 9
)
)
(
)
95.
PROBLEM: 3x − 15 g ( x) = 10 − 2 x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 10 − 2 x = 0 10 + 10 − 2 x = 0 − 10 −2 x = −10 −2 −10 x= −2 −2 x=5 ANSWER: The domain consists of any real number x where x ≠ 5 . 3x − 15 3 ( x − 5) 3 = =− 10 − 2 x −2 ( x − 5) 2
96.
PROBLEM: 25 − 5 x g ( x) = 4 x − 20 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 4 x − 20 = 0 4 x − 20 + 20 = 0 + 20 4 x = 20 4 20 x= 4 4 x=5 ANSWER: The domain consists of any real number x where x = 5 . 25 − 5 x −5 ( x − 5) 5 = =− 4 x − 20 4 ( x − 5) 4
g ( x) = −
5 4
97.
PROBLEM: The cost, in dollars, of producing coffee mugs with a company logo is given by C ( x ) = x + 40 where x represents the number of mugs produced. Calculate the
average cost of producing 100 mugs and the average cost of producing 500 mugs. SOLUTION: C ( x ) = x + 40
C (100 ) = 100 + 40 = 140 C ( 500 ) = 500 + 40 = 540 Total cost Average cost = Number of mugs 140 = $1.40 . For 100 mugs, average cost = 100 540 = $1.08 . For 500 mugs, average cost = 500 ANSWER: The average cost of producing 100 mugs is $1.40 per mug. The average cost of producing 500 mugs is $1.08 per mug.
98.
PROBLEM: The cost, in dollars, of renting a moving truck for the day is given by C ( x ) = 0.45 x + 90 where x represents the number of miles driven. Calculate the
average cost per mile if the truck is driven 250 miles in one day. SOLUTION: C ( x ) = 0.45 x + 90
C ( 250 ) = 0.45 ( 250 ) + 90 = 202.5 Average cost =
Total cost Number of miles
ANSWER: For 250 miles, average cost =
202.5 = $0.81 . 250
99.
PROBLEM: The cost, in dollars, of producing sweatshirts with a custom design on the back is given by C ( x) = 1200 + (12 − 0.05 x) x where x represents the number of sweatshirts produced. Calculate the average cost of producing 150 custom sweatshirts. SOLUTION: C ( x) = 1200 + (12 − 0.05 x) x
C (150) = 1200 + (12 − 0.05 (150 ) )150 = 1875 Average cost =
Total cost Number of sweatshirts
ANSWER: For 150 sweat shirts, average cost = 100.
1875 = $12.5 150
PROBLEM: The cost, in dollars, of producing a custom injected molded part is given by C ( x) = 500 + (3 − 0.001x) x where x represents the number of parts produced. Calculate the average cost of producing 1000 custom parts. SOLUTION: C ( x) = 500 + (3 − 0.001x) x C (1000) = 500 + ( 3 − 0.001(1000 ) )1000 = 2500
Average cost =
Total cost Number of parts
ANSWER: For 1000 parts, average cost =
2500 = $2.5 1000
Part E: Discussion Board
101.
PROBLEM:
Explain why
b−a = −1 and illustrate this fact by substituting in some numbers. a −b
ANSWER: b−a b−a = = −1 a − b − (b − a )
(Students answers may vary.) Put a = 1; b = 2 b − a 2 −1 1 = = = −1 a − b 1 − 2 −1
102.
PROBLEM:
Explain why
b+a = 1 and illustrate this fact by substituting in some numbers. a+b
ANSWER: b+a a+b = =1 a+b a+b (Students answers may vary.) Put a = 1; b = 2 b + a 2 +1 3 = = =1 a + b 1+ 2 3
103.
PROBLEM:
Explain why we cannot cancel x in the expression
x . x +1
ANSWER: The denominator cannot be factored into a multiple of x. Hence, we cannot cancel x in the expression. 7.2 Multiplying and Dividing Rational Expressions Part A: Multiplying Rational Expressions Multiply. (Assume all denominators are nonzero)
1.
PROBLEM: 2x 9 ⋅ 3 4x2 SOLUTION: 2x 9 3 3 ⋅ 2 = 2−1 = 3 4x 2x 2x ANSWER:
3 2x
2.
PROBLEM: −5 x 3 y 2 ⋅ y 25 x SOLUTION: −5 x3 y 2 x 3−1 y 2−1 x2 y ⋅ =− =− y 25 x 5 5 ANSWER: −
3.
x2 y 5
PROBLEM: 5x2 4 y 2 ⋅ 2 y 15 x3 SOLUTION: 5 x 2 4 y 2 2 y 2−1 2 y ⋅ = = 2 y 15 x3 3x3−2 3x
4.
ANSWER:
2y 3x
PROBLEM: 16a 4 49b ⋅ 7b 2 32a3 SOLUTION: 16a 4 49b 7a 4−3 7a ⋅ = = 7b2 32a3 2b 2−1 2b ANSWER:
7a 2b
5.
PROBLEM: x − 6 24 x 2 ⋅ 12 x 3 x − 6 SOLUTION: x − 6 24 x 2 2 2 ⋅ = 3− 2 = 3 12 x x − 6 x x ANSWER:
6.
2 x
PROBLEM: x + 10 x − 2 ⋅ 2 x − 1 x + 10 SOLUTION: x + 10 x − 2 x−2 ⋅ = 2 x − 1 x + 10 2 x − 1 ANSWER:
7.
x−2 2x −1
PROBLEM:
( y − 1) y +1
2
⋅
1 y −1
SOLUTION: 2 ( y − 1) ⋅ 1 = y − 1 y +1 y −1 y +1 ANSWER:
y −1 y +1
8.
PROBLEM: y2 − 9 2 y − 3 ⋅ y +3 y −3 SOLUTION: y 2 − 9 2 y − 3 ( y − 3)( y + 3)( 2 y − 3) ⋅ = = 2y −3 y +3 y −3 ( y + 3)( y − 3) ANSWER: 2 y − 3
9.
PROBLEM: 2a − 5 2a + 5 ⋅ a − 5 4a 2 − 25 SOLUTION: ( 2a − 5 ) ⋅ ( 2 a + 5 ) = 1 2a − 5 2 a + 5 ⋅ 2 = a − 5 4a − 25 ( a − 5 ) ( 2a + 5 )( 2a − 5 ) a − 5 ANSWER:
10.
1 a −5
PROBLEM: 2a 2 − 9a + 4 2 ⋅ ( a + 4a ) a 2 − 16 SOLUTION: ( 2a − 1)( a − 4 ) a ( a + 4 ) = a 2a − 1 2a 2 − 9 a + 4 2 ⋅ ( a + 4a ) = ( ) 2 a − 16 ( a + 4 )( a − 4 ) ANSWER: a ( 2a − 1)
11.
PROBLEM: 2 x 2 + 3x − 2 2 x ⋅ 2 ( 2 x − 1) x + 2 SOLUTION: ( 2 x − 1)( x + 2 ) 2 x = 2 x 2 x 2 + 3x − 2 2 x ⋅ = 2 2 ( 2 x − 1) x + 2 ( 2 x − 1) ( x + 2 ) 2 x − 1 ANSWER:
12.
2x 2x −1
PROBLEM: 9 x 2 + 19 x + 2 x 2 − 4 x + 4 ⋅ 2 4 − x2 9x − 8x −1 SOLUTION:
( x + 2 )( 9 x + 1)( x − 2 ) 9 x 2 + 19 x + 2 x 2 − 4 x + 4 x−2 ⋅ 2 = =− 2 4− x 9 x − 8 x − 1 − ( x − 2 )( x + 2 )( x − 1)( 9 x + 1) x −1 x−2 ANSWER: − x −1 2
13.
PROBLEM: x 2 + 8 x + 16 x 2 − 3x − 4 ⋅ 2 16 − x 2 x + 5x + 4 SOLUTION:
( x + 4 ) ( x − 4 )( x + 1) = −1 x 2 + 8 x + 16 x 2 − 3x − 4 ⋅ = 16 − x 2 x 2 + 5 x + 4 − ( x − 4 )( x + 4 )( x + 1)( x + 4 ) 2
ANSWER: −1
14.
PROBLEM: x 2 − x − 2 x 2 + 2 x − 15 ⋅ x2 + 8x + 7 x2 − 5x + 6 SOLUTION: x 2 − x − 2 x 2 + 2 x − 15 ( x + 1)( x − 2 )( x − 3)( x + 5 ) x + 5 ⋅ = = x 2 + 8 x + 7 x 2 − 5 x + 6 ( x + 1)( x + 7 )( x − 2 )( x − 3) x + 7 ANSWER:
15.
x+5 x+7
PROBLEM: x +1 3 − x ⋅ x −3 x +5 SOLUTION: x + 1 3 − x x + 1 − ( x − 3) x +1 ⋅ = ⋅ =− x −3 x +5 x −3 x +5 x+5 ANSWER: −
16.
x +1 x+5
PROBLEM: 2x −1 x + 6 ⋅ x −1 1 − 2x SOLUTION: 2x −1 x + 6 2x −1 x + 6 x+6 ⋅ = ⋅ =− x − 1 1 − 2 x x − 1 − ( 2 x − 1) x −1 ANSWER: −
x+6 x −1
17.
PROBLEM: 9+ x 3 ⋅ 3x + 1 x + 9 SOLUTION: 9+ x 3 x+9 3 3 ⋅ = ⋅ = 3x + 1 x + 9 3x + 1 x + 9 3x + 1 ANSWER:
18.
3 3x + 1
PROBLEM: 1 5x + 2 ⋅ 2 + 5x 5x SOLUTION: 1 5x + 2 1 2 + 5x 1 ⋅ = ⋅ = 2 + 5x 5x 2 + 5x 5x 5x ANSWER:
19.
1 5x
PROBLEM: 100 − y 2 25 y 2 ⋅ y − 10 y + 10 SOLUTION: 2 100 − y 2 25 y 2 − ( y + 10 )( y − 10 ) 25 y ⋅ = = −25 y 2 y − 10 y + 10 ( y + 10 )( y − 10 ) ANSWER: −25y 2
20.
PROBLEM: 3 y 3 36 y 2 − 25 ⋅ 6y −5 5+ 6y SOLUTION: 3 3 y 3 36 y 2 − 25 3 y ( 6 y − 5 )( 6 y + 5 ) ⋅ = = 3 y3 6y − 5 5 + 6y ( 6 y − 5)( 6 y + 5) ANSWER: 3y 3
21.
PROBLEM: 3a 2 + 14a − 5 3a + 1 ⋅ a2 + 1 1 − 9a 2 SOLUTION: ( 3a − 1)( a + 5)( 3a + 1) = − a + 5 3a 2 + 14a − 5 3a + 1 ⋅ = 2 2 a +1 1 − 9a a2 + 1 − ( a 2 + 1) ( 3a + 1)( 3a − 1) ANSWER: −
22.
a+5 a2 + 1
PROBLEM: 4a 2 − 16a 1 − 16a 2 ⋅ 2 4a − 1 4a − 15a − 4 SOLUTION: −4a ( a − 4 )( 4a + 1)( 4a − 1) 4a 2 − 16a 1 − 16a 2 ⋅ 2 = 4a − 1 4a − 15a − 4 ( 4a − 1)( 4a + 1)( a − 4 )
= −4a
ANSWER: −4a
23.
PROBLEM: x+9 ⋅ x 2 − 81 2 − x + 14 x − 45
(
)
SOLUTION:
( x + 9 )( x + 9 )( x − 9 ) = − ( x + 9 ) x+9 ⋅ ( x 2 − 81) = 2 − x + 14 x − 45 − ( x − 5)( x − 9 ) x −5 ( x + 9) ANSWER: − x−5
2
2
24.
PROBLEM: 1 ⋅ 25 x 2 + 20 x + 4 2 + 5x
(
)
SOLUTION:
( 2 + 5x ) = 5x + 2 1 ⋅ ( 25 x 2 + 20 x + 4 ) = 2 + 5x ( 2 + 5x ) 2
ANSWER: 5 x + 2
25.
PROBLEM: x2 + x − 6 2x2 − 8 ⋅ 3x 2 + 15 x + 18 x 2 − 4 x + 4 SOLUTION: ( x + 3)( x − 2 ) 2 ( x − 2 )( x + 2 ) = 2 x2 + x − 6 2x2 − 8 ⋅ = 2 2 2 3x + 15 x + 18 x − 4 x + 4 3 3 ( x + 2 )( x + 3)( x − 2 ) ANSWER:
26.
2 3
PROBLEM: 5 x 2 − 4 x − 1 25 x 2 − 10 x + 1 ⋅ 5x2 − 6 x + 1 3 − 75 x 2 SOLUTION:
( 5x + 1)( x − 1)( 5x − 1) 5 x 2 − 4 x − 1 25 x 2 − 10 x + 1 1 ⋅ = =− 2 2 5x − 6x + 1 3 − 75 x − ( x − 1)( 5 x − 1) 3 ( 5 x + 1)( 5 x − 1) 3 2
ANSWER: −
1 3
Part B: Dividing Rational Expressions Divide. (Assume all denominators are nonzero)
27.
PROBLEM: 5 x 15 x 2 ÷ 8 4 SOLUTION: 5 x 15 x 2 5 x 4 1 ÷ = ⋅ = 2 8 4 8 15 x 6x ANSWER:
28.
1 6x
PROBLEM: 3 −15 ÷ 8 y 2 y2 SOLUTION: 3 −15 3 ⎛ 2 y 2 ⎞ y ÷ 2 = ⋅⎜ − ⎟=− 8y 2y 8 y ⎝ 15 ⎠ 20 ANSWER: −
29.
y 20
PROBLEM: 5 x9 3 y3 25 x10 9 y5 SOLUTION: 5 x9 3 y2 5 x9 9 y 5 3 y3 = ⋅ = 25 x10 3 y 3 25 x10 5 x 9 y5 ANSWER:
3y2 5x
30.
PROBLEM: 12 x 4 y 2 21z 5 6 x3 y 2 7 z3 SOLUTION: 12 x 4 y 2 4 2 3 21z 5 = 12 x y ⋅ 7 z = 2 x 6 x3 y 2 21z 5 6 x3 y 2 3z 2 7 z3 ANSWER:
31.
2x 3z 2
PROBLEM: 2 ( x − 4) ÷ x − 4 30 x 4 15 x SOLUTION: 2 2 ( x − 4 ) ÷ x − 4 = ( x − 4 ) ⋅ 15 x = x − 4 30 x 4 15 x 30 x 4 x − 4 2 x3 ANSWER:
32.
x−4 2 x3
PROBLEM: 5 y4 10 y 5 ÷ 2 3 10 ( 3 y − 5) 2 ( 3 y − 5 ) SOLUTION:
5 y4 10 ( 3 y − 5)
2
ANSWER:
÷
10 y 5 2 ( 3 y − 5)
3y − 5 10 y
3
=
5 y4 10 ( 3 y − 5)
2
⋅
2 ( 3 y − 5) 10 y
5
3
=
3y − 5 10 y
33.
PROBLEM: x2 − 9 ÷ ( x − 3) 5x SOLUTION: ( x + 3)( x − 3) = x + 3 x2 − 9 x2 − 9 1 ÷ ( x − 3) = ⋅ = 5x 5 x ( x − 3) 5 x ( x − 3) 5x ANSWER:
34.
x+3 5x
PROBLEM: y 2 − 64 ÷ (8 + y ) 8y SOLUTION: ( y + 8)( y − 8) = y − 8 y 2 − 64 y 2 − 64 1 ÷ (8 + y ) = ⋅ = 8y 8y 8+ y 8 y ( y + 8) 8y ANSWER:
35.
y −8 8y
PROBLEM: 2 ( a − 8) ÷ a − 8 2a 2 + 10a a SOLUTION: 2 2 2 ( a − 8) ÷ a − 8 = ( a − 8) ⋅ a = ( a − 8) ⋅ a = a − 8 2a 2 + 10a a 2a 2 + 10a a − 8 2a ( a + 5) a − 8 2 ( a + 5 ) ANSWER:
a −8 2 ( a + 5)
36.
PROBLEM:
24a 2b3 (a − 2b) ÷ SOLUTION:
24a 2b3 (a − 2b) ÷
12ab ( a − 2b ) 5 12ab ( a − 2b ) 5
= 24a 2b3 (a − 2b) ⋅
5 = 10ab 2 12ab ( a − 2b )
ANSWER: 10ab 2
37.
PROBLEM: x 2 + 7 x + 10 1 ÷ 2 2 x + 4x + 4 x − 4 SOLUTION: ( x + 2 )( x + 5)( x + 2 )( x − 2 ) = x + 5 x − 2 x 2 + 7 x + 10 1 x 2 + 7 x + 10 2 ÷ = ⋅x −4= ( )( ) 2 2 2 2 x + 4x + 4 x − 4 x + 4x + 4 ( x + 2) ANSWER: ( x + 5 )( x − 2 )
38.
PROBLEM: 2x2 − x −1 1 ÷ 2 2 2 x − 3x + 1 4 x − 1 SOLUTION: ( x − 1)( 2 x + 1)( 2 x + 1)( 2 x − 1) 2x2 − x −1 1 2 x2 − x −1 ÷ = ⋅ 4 x2 − 1 = 2 2 2 2 x − 3x + 1 4 x − 1 2 x − 3x + 1 ( 2 x − 1)( x − 1)
= ( 2 x + 1) ANSWER: ( 2 x + 1)
2
2
39.
PROBLEM: y +1 y2 −1 ÷ y2 − 3y y2 − 6 y + 9 SOLUTION:
( y + 1)( y − 3) y +1 y2 −1 y +1 y2 − 6 y + 9 ÷ = ⋅ = 2 2 2 2 y − 3y y − 6 y + 9 y − 3y y −1 y ( y − 3)( y + 1)( y − 1) 2
= ANSWER:
40.
y −3 y ( y − 1)
y −3 y ( y − 1)
PROBLEM: 9 − a2 2a 2 − 10a ÷ a 2 − 8a + 15 a 2 − 10a + 25 SOLUTION:
a 2 − 10a + 25 − ( a − 3)( a + 3)( a − 5 ) 9 − a2 2a 2 − 10a 9 − a2 ÷ = ⋅ = a 2 − 8a + 15 a 2 − 10a + 25 a 2 − 8a + 15 2a 2 − 10a ( a − 3)( a − 5) 2a ( a − 5) 2
=−
ANSWER: −
41.
a+3 2a
a+3 2a
PROBLEM: a 2 − 3a − 18 a 2 + a − 6 ÷ 2a 2 − 11a − 6 2a 2 − a − 1 SOLUTION: a 2 − 3a − 18 a 2 + a − 6 a 2 − 3a − 18 2a 2 − a − 1 ÷ = ⋅ 2a 2 − 11a − 6 2a 2 − a − 1 2a 2 − 11a − 6 a 2 + a − 6 ( a + 3)( a − 6 )( 2a + 1)( a − 1) = ( 2a + 1)( a − 6 )( a − 2 )( a + 3)
= ANSWER:
a −1 a−2
a −1 a−2
42.
PROBLEM: y 2 − 7 y + 10 y 2 + 5 y − 14 2 y2 − 9 y − 5 y 2 + 14 y + 49 SOLUTION: y 2 − 7 y + 10 2 y − 2 )( y − 5)( y + 7 ) ( y 2 − 7 y + 10 y 2 + 14 y + 49 y 2 + 5 y − 14 = 2 ⋅ = 2 y2 − 9 y − 5 y + 5 y − 14 2 y 2 − 9 y − 5 ( y − 2 )( y + 7 )( 2 y + 1)( y − 5) y 2 + 14 y + 49 y+7 = 2 y +1 y+7 ANSWER: 2 y +1
43.
PROBLEM: 6 y2 + y −1 4 y2 + 4 y +1 3 y2 + 2 y −1 2 y2 − 7 y − 4 SOLUTION: 6 y2 + y −1 6 y2 + y −1 2 y2 − 7 y − 4 4 y2 + 4 y +1 = ⋅ 3 y2 + 2 y −1 4 y2 + 4 y +1 3 y2 + 2 y −1 2 y2 − 7 y − 4 =
( 3 y − 1)( 2 y + 1)( 2 y + 1)( y − 4 ) 2 ( 2 y + 1) ( 3 y − 1)( y + 1)
y−4 y +1 y−4 ANSWER: y +1 =
44.
PROBLEM: x 2 − 7 x − 18 x 2 − 81 ÷ x 2 + 8 x + 12 x 2 + 12 x + 36 SOLUTION: x 2 − 7 x − 18 x 2 − 81 x 2 − 7 x − 18 x 2 + 12 x + 36 ÷ = ⋅ x 2 + 8 x + 12 x 2 + 12 x + 36 x 2 + 8 x + 12 x 2 − 81
( x − 9 )( x + 2 )( x + 6 ) = ( x + 2 )( x + 6 )( x − 9 )( x + 9 ) 2
= ANSWER:
45.
x+6 x+9
x+6 x+9
PROBLEM: 4a 2 − b 2 2 ÷ ( b − 2a ) b + 2a SOLUTION: − ( b + 2a )( b − 2a ) 4a 2 − b 2 4a 2 − b 2 1 2 ÷ ( b − 2a ) = ⋅ = b + 2a b + 2a ( b − 2a )2 ( b + 2a )( b − 2a )2
=− ANSWER:
46.
1 1 = b − 2a 2a − b
1 2a − b
PROBLEM: x2 − y2 2 ÷ ( y − x) y+x SOLUTION: ( x + y )( x − y ) x2 − y2 x2 − y2 1 2 ÷ ( y − x) = ⋅ = 2 2 y+x y + x ( y − x) ( x + y )( x − y ) =
ANSWER:
1 x− y
1 x− y
47.
PROBLEM: 5 y 2 ( y − 3) 25 y ( 3 − y ) ÷ 4 x3 2 x2 SOLUTION: 5 y 2 ( y − 3) 25 y ( 3 − y ) 5 y 2 ( y − 3) 5 y 2 ( y − 3) ⋅ 2 x 2 2 x2 ÷ = ⋅ = 4 x3 2x2 4 x3 25 y ( 3 − y ) −4 x 3 ⋅ 25 y ( y − 3) =−
ANSWER: −
48.
y 10 x
y 10 x
PROBLEM: 15 x 3 25 x 6 ÷ 3 ( y + 7 ) 9 ( 7 + y )2 SOLUTION:
15 x 3 25 x 6 15 x3 9 ( 7 + y ) 15 x 3 9 ( y + 7 ) ÷ = ⋅ = ⋅ 3 ( y + 7 ) 9 ( 7 + y )2 3 ( y + 7 ) 25 x 6 3( y + 7) 25 x 6 2
=
ANSWER:
49.
9 ( y + 7)
9 ( y + 7) 5 x3
5 x3
PROBLEM: 3x + 4 7 x ÷ x −8 8− x SOLUTION: 3x + 4 7 x 3x + 4 8 − x 3x + 4 x − 8 ÷ = ⋅ =− ⋅ x − 8 8 − x x − 8 7x x − 8 7x 3x + 4 =− 7x ANSWER: −
3x + 4 7x
2
50.
PROBLEM: 3x − 2 2 − 3x ÷ 2x +1 3x SOLUTION: 3x − 2 2 − 3x 3x − 2 3x 3x − 2 3x ÷ = ⋅ = ⋅ 2x +1 3x 2 x + 1 2 − 3x 2 x + 1 − ( 3x − 2 )
=− ANSWER: −
51.
3x 2x + 1
3x 2x +1
PROBLEM:
( 7 x − 1) 4x +1
2
÷
28 x 2 − 11x + 1 1 − 4x
SOLUTION: 2 2 ( 7 x − 1) ÷ 28 x 2 − 11x + 1 = ( 7 x − 1) ⋅ 1 − 4 x 4x +1 1− 4x 4 x + 1 28 x 2 − 11x + 1
− ( 7 x − 1) ( 4 x − 1) 2
=
( 4 x + 1)( 4 x − 1)( 7 x − 1)
=− ANSWER: −
52.
7 x −1 4x +1
7x −1 4x +1
PROBLEM: 4x 2− x ÷ 2 2 ( x + 2) x − 4 SOLUTION: 4x 2− x 4x x 2 − 4 4 x ( x − 2 )( x + 2 ) ÷ = ⋅ = 2 2 2 2 ( x + 2) x − 4 ( x + 2) 2 − x − ( x + 2) ( x − 2) =−
ANSWER: −
4x x+2
4x x+2
53.
PROBLEM: a 2 − b2 2 ÷ (b − a ) a SOLUTION: ( a + b )( a − b ) a 2 − b2 a 2 − b2 1 2 ÷ (b − a ) = ⋅ = 2 2 a a a ( a − b) (b − a )
= ANSWER:
54.
a+b a ( a − b)
a+b a (a − b)
PROBLEM: (a − 2b) 2 ÷ 2b 2 + ab − a 2 2b
(
)
SOLUTION: (a − 2b) 2 (a − 2b) 2 (a − 2b) 2 1 1 ÷ 2b 2 + ab − a 2 = ⋅ = ⋅ 2 2 2b 2b 2b ( 2b − a )( a + b ) 2b + ab − a
(
)
(
( 2b − a ) = 2b
=
ANSWER:
2b − a 2b ( a + b )
2
⋅
2b − a 2b ( a + b )
)
1 ( 2b − a )( a + b )
55.
PROBLEM: x2 − 6 x + 9 9 − x2 ÷ x 2 + 7 x + 12 x 2 + 8 x + 16 SOLUTION: x2 − 6 x + 9 9 − x2 x 2 − 6 x + 9 x 2 + 8 x + 16 ÷ = ⋅ x 2 + 7 x + 12 x 2 + 8 x + 16 x 2 + 7 x + 12 9 − x2
( x − 3) ( x + 4 ) = − ( x + 4 )( x + 3)( x + 3)( x − 3) ( x − 3)( x + 4 ) =− 2 ( x + 3) ( x − 3)( x + 4 ) ANSWER: − 2 ( x + 3) 2
56.
2
PROBLEM: 2x2 − 9x − 5 1 − 4x + 4x2 ÷ 25 − x 2 −2 x 2 − 9 x + 5 SOLUTION: 2 x2 − 9 x − 5 1 − 4 x + 4 x2 2 x 2 − 9 x − 5 −2 x 2 − 9 x + 5 ÷ = ⋅ 25 − x 2 25 − x 2 −2 x 2 − 9 x + 5 1 − 4x + 4x2 − ( 2 x + 1)( x − 5 )( 2 x − 1)( x + 5 ) = 2 − ( x − 5 )( x + 5 )( 2 x − 1)
= ANSWER:
2x +1 2x −1
2x +1 2x −1
57.
PROBLEM: 3 x 2 − 16 x + 5 100 − 4 x 2 9x2 − 6x + 1 3 x 2 + 14 x − 5 SOLUTION: 3 x 2 − 16 x + 5 3x 2 − 16 x + 5 3x 2 + 14 x − 5 100 − 4 x 2 = ⋅ 2 9x2 − 6x + 1 100 − 4 x 2 9x − 6x +1 2 3 x + 14 x − 5 ( 3x − 1)( x − 5)( 3x − 1)( x + 5) = 2 −4 ( x + 5 )( x − 5 )( 3x − 1)
=− ANSWER: −
58.
1 4
1 4
PROBLEM: 10 x 2 − 25 x − 15 x2 − 6x + 9 9 − x2 x2 + 6 x + 9 SOLUTION: 10 x 2 − 25 x − 15 2 2 2 x 2 − 6 x + 9 = 10 x − 25 x − 15 ⋅ x + 6 x + 9 = 5 ( 2 x + 1)( x − 3)( x + 3) 2 9 − x2 x2 − 6x + 9 9 − x2 − ( x − 3) ( x − 3)( x + 3) x2 + 6 x + 9 5 ( 2 x + 1)( x + 3) =− 2 ( x − 3) ANSWER: −
5 ( 2 x + 1)( x + 3)
( x − 3)
2
Recall that multiplication and division is to be performed in the order as they appear from left to right. Simplify the following:
59.
PROBLEM: 1 x −1 x −1 ⋅ ÷ x 2 x + 3 x3 SOLUTION: 1 x − 1 x − 1 1 x − 1 x3 x ⋅ ÷ 3 = 2⋅ ⋅ = 2 x x+3 x x x + 3 x − 1 ( x + 3) ANSWER:
60.
x ( x + 3)
PROBLEM: x−7 1 x−7 ⋅ ÷ x + 9 x3 x SOLUTION: x−7 1 x−7 x−7 1 x 1 ⋅ 3÷ = ⋅ 3⋅ = 2 x+9 x x x + 9 x x − 7 x ( x + 9) ANSWER:
61.
1 x ( x + 9) 2
PROBLEM: x +1 x x2 ÷ ⋅ x − 2 x − 5 x +1 SOLUTION: x ( x − 5) x +1 x x2 x + 1 x − 5 x2 ÷ ⋅ = ⋅ ⋅ = x − 2 x − 5 x +1 x − 2 x x +1 x−2 ANSWER:
x ( x − 5) x−2
62.
PROBLEM: x+4 x −3 x + 4 ÷ ⋅ 2x + 5 2x + 5 x − 3 SOLUTION:
x+4 x − 3 x + 4 x + 4 2x + 5 x + 4 ⎛ x + 4 ⎞ ÷ ⋅ = ⋅ ⋅ =⎜ ⎟ 2x + 5 2x + 5 x − 3 2x + 5 x − 3 x − 3 ⎝ x − 3 ⎠ ⎛ x+4⎞ ANSWER: ⎜ ⎟ ⎝ x−3 ⎠ 63.
2
2
PROBLEM: 2x −1 x − 4 x − 4 ÷ ⋅ x + 1 x2 + 1 2x −1 SOLUTION: 2 x − 1 x − 4 x − 4 2 x − 1 x2 + 1 x − 4 x2 + 1 ÷ ⋅ = ⋅ ⋅ = x + 1 x2 + 1 2x −1 x + 1 x − 4 2x −1 x + 1 ANSWER:
64.
x2 + 1 x +1
PROBLEM: 4 x 2 − 1 2 x − 1 3x + 2 ÷ ⋅ 3x + 2 x + 5 2 x + 1 SOLUTION: 4 x 2 − 1 2 x − 1 3x + 2 4 x 2 − 1 x + 5 3x + 2 ÷ ⋅ = ⋅ ⋅ 3x + 2 x + 5 2 x + 1 3x + 2 2 x − 1 2 x + 1 ( 2 x − 1)( 2 x + 1)( x + 5)( 3x + 2 ) = ( 3x + 2 )( 2 x − 1)( 2 x + 1) ANSWER: x + 5
= x+5
Part C: Multiplying and Dividing Rational Functions Calculate ( f ⋅ g ) ( x) and determine the restrictions to the domain.
65.
PROBLEM: 1 1 f ( x) = and g ( x) = x x −1 SOLUTION: 1 1 f ( x) = and g ( x) = x x −1 1 1 1 = ( f ⋅ g ) ( x) = ⋅ x x − 1 x ( x − 1) In this case, the domain of f ( x) consists of all real numbers except 0, and the domain of g ( x) consists all real numbers except 1. 1 x ( x − 1) The domain consists of any real number x where x ≠ 0 and x ≠ 1 .
ANSWER: ( f ⋅ g ) ( x) =
66.
PROBLEM: x +1 f ( x) = and g ( x) = x 2 − 1 x −1 SOLUTION: x +1 f ( x) = and g ( x) = x 2 − 1 x −1 ( x + 1)( x + 1)( x − 1) = x + 1 2 x +1 2 ⋅ x −1 = ( f ⋅ g ) ( x) = ( ) x −1 ( x − 1) In this case, the domain of f ( x) consists of all real numbers except 1, and the domain of g ( x) consists all real numbers. ANSWER: ( f ⋅ g ) ( x) = ( x + 1)
2
The domain consists of any real number x where x ≠ 1 .
67.
PROBLEM: 3x + 2 x2 − 4 f ( x) = and g ( x) = 2 x+2 ( 3x + 2 ) SOLUTION: 3x + 2 x2 − 4 f ( x) = and g ( x) = 2 x+2 ( 3x + 2 )
( f ⋅ g ) ( x) =
( 3x + 2 )( x − 2 )( x + 2 ) = x − 2 3x + 2 x 2 − 4 ⋅ = 2 2 x + 2 ( 3x + 2 ) 3x + 2 ( x + 2 )( 3x + 2 )
In this case, the domain of f ( x) consists of all real numbers except –2, and the 2 domain of g ( x) consists all real numbers except − . 3 ANSWER: ( f ⋅ g ) ( x) =
x−2 3x + 2
2 The domain consists of any real number x where x ≠ −2 and x ≠ − . 3
68.
PROBLEM: 2 2 1 − 3x ) x − 6) ( ( f ( x) = and g ( x) = 9x2 −1 x−6 SOLUTION:
(1 − 3x ) f ( x) =
2
( x − 6) g ( x) =
2
and 9x2 −1 x−6 2 2 2 2 1 − 3x ) ( x − 6 ) 3x − 1) ( x − 6 ) ( ( 3x − 1)( x − 6 ) ( ⋅ 2 = = ( f ⋅ g ) ( x) = x−6 9 x − 1 ( x − 6 )( 3 x − 1)( 3 x + 1) ( 3x + 1) In this case, the domain of f ( x) consists of all real numbers except 6, and the 1 domain of g ( x) consists all real numbers except ± . 3 3 x − 1 x − 6 ( )( ) ANSWER: ( f ⋅ g ) ( x) = ( 3x + 1) 1 The domain consists of any real number x where x ≠ 6 and x ≠ ± . 3
69.
PROBLEM: 25 x 2 − 1 x2 − 9 f ( x) = 2 and g ( x) = x + 6x + 9 5x + 1 SOLUTION: 25 x 2 − 1 x2 − 9 f ( x) = 2 and g ( x ) = x + 6x + 9 5x + 1 2 2 25 x − 1 x − 9 ( 5 x + 1)( 5 x − 1)( x − 3)( x + 3) ( 5 x − 1)( x − 3) ⋅ = = ( f ⋅ g ) ( x) = 2 2 x + 6 x + 9 5x + 1 ( x + 3) ( x + 3) ( 5 x + 1)
In this case, the domain of f ( x) consists of all real numbers except –3, and the 1 domain of g ( x) consists all real numbers except − . 5 ANSWER: ( f ⋅ g ) ( x) =
( 5 x − 1)( x − 3) ( x + 3)
1 The domain consists of any real number x where x ≠ −3 and x ≠ − . 5
70.
PROBLEM: x 2 − 49 4 x2 − 4 x + 1 f ( x) = 2 and g ( x) = 2 x + 13x − 7 7−x SOLUTION: x 2 − 49 4 x2 − 4 x + 1 f ( x) = 2 and g ( x) = 7−x 2 x + 13x − 7 2 x 2 − 49 4 x 2 − 4 x + 1 ( x + 7 )( x − 7 )( 2 x − 1) ⋅ = = 1− 2x ( f ⋅ g ) ( x) = 2 2 x + 13x − 7 7−x − ( 2 x − 1)( x + 7 )( x − 7 )
In this case, the domain of f ( x) consists of all real numbers except –7 and and the domain of g ( x) consists all real numbers except 7 . ANSWER: ( f ⋅ g ) ( x) = 1 − 2x
The domain consists of any real number x where x ≠ ±7 and x ≠
1 . 2
1 , 2
Calculate ( f / g ) ( x) and state the restrictions.
71.
PROBLEM: 1 x−2 f ( x) = and g ( x) = x x −1 SOLUTION: 1 x−2 f ( x) = and g ( x ) = x x −1 1 1 x −1 x −1 = ( f / g ) ( x) = x −x 2 = ⋅ x x − 2 x ( x − 2) x −1 In this case, the domain of f ( x) consists of all real numbers except 0, and the domain of g ( x) consists all real numbers except 1. In addition, the reciprocal of g ( x) has a restriction of 2. x −1 x ( x − 2) The domain consists of any real number x where x ≠ 0 , x ≠ 1 , and x ≠ 2 .
ANSWER: ( f / g ) ( x) =
72.
PROBLEM: 2 5 x + 3) ( 5x + 3 f ( x) = and g ( x) = 2 x 6− x SOLUTION: 2 5 x + 3) ( 5x + 3 f ( x) = and g ( x ) = 2 x 6− x
( 5 x + 3)
2
2 ( 5 x + 3) 6 − x ( 5 x + 3)( 6 − x ) ⋅ = ( f / g ) ( x) = 5 xx+ 3 = 5x + 3 x2 x2 6− x In this case, the domain of f ( x) consists of all real numbers except 0, and the domain of g ( x) consists all real numbers except 6. In addition, the reciprocal of 3 g ( x) has a restriction of − . 5 ( 5 x + 3)( 6 − x ) ANSWER: ( f / g ) ( x) = x2 3 The domain consists of any real number x where x ≠ 0 , x ≠ 6 , and x ≠ − . 5 2
73.
PROBLEM: x 2 − 25 5− x g ( x ) and = f ( x) = 2 x −8 ( x − 8) SOLUTION: 5− x x 2 − 25 f ( x) = and = g ( x ) 2 x −8 ( x − 8)
5− x
(f
/ g)
( x − 8) ( x) =
2
x − 25 x −8 2
=
5− x
( x − 8)
2
⋅
− ( x − 5 )( x − 8 ) 1 x −8 = =− 2 2 x − 25 ( x − 8 ) ( x − 5 )( x + 5 ) ( x + 5)( x − 8)
In this case, the domain of f ( x) consists of all real numbers except 8, and the domain of g ( x) consists all real numbers except 8. In addition, the reciprocal of g ( x) has a restriction of ±5 . ANSWER: ( f / g ) ( x) = −
1
( x + 5 )( x − 8)
The domain consists of any real number x where x ≠ 8 and x ≠ ±5 .
74.
PROBLEM: x 2 − 2 x − 15 2 x2 − 5x − 3 f ( x) = 2 and g ( x) = 2 x − 3 x − 10 x − 7 x + 12 SOLUTION: x 2 − 2 x − 15 2 x2 − 5x − 3 f ( x) = 2 and g ( x ) = 2 x − 3 x − 10 x − 7 x + 12 2 x − 2 x − 15 2 x 2 − 2 x − 15 x 2 − 7 x + 12 ⋅ ( f / g ) ( x) = x 2− 3x − 10 = 2 2 x − 5 x − 3 x − 3 x − 10 2 x 2 − 5 x − 3 x 2 − 7 x + 12 ( x + 3)( x − 5)( x − 3)( x + 4 ) = ( x − 5 )( x + 2 )( 2 x + 1)( x − 3) =
( x + 3)( x + 4 ) ( x + 2 )( 2 x + 1)
In this case, the domain of f ( x) consists of all real numbers except 5 and –2 , and 1 the domain of g ( x) consists all real numbers except 3 and − . In addition, the 2 reciprocal of g ( x) has a restriction of 3 and 4 . ANSWER: ( f / g ) ( x) =
( x + 3)( x + 4 ) ( x + 2 )( 2 x + 1)
The domain consists of any real number x where x ≠ 5 , x ≠ −2 , x ≠ 3 , and x ≠ − .
1 2
75.
PROBLEM: 3x 2 + 11x − 4 x2 − 2 x + 1 f ( x) = and g ( x ) = 9x2 − 6x + 1 3x 2 − 4 x + 1 SOLUTION: 3x 2 + 11x − 4 x2 − 2 x + 1 f ( x) = and g ( x) = 2 9 x2 − 6 x + 1 3x − 4 x + 1 2 3x + 11x − 4 2 3x 2 + 11x − 4 3 x 2 − 4 x + 1 ⋅ ( f / g ) ( x) = 9 x2 − 6 x + 1 = 2 x − 2x +1 9x − 6x + 1 x2 − 2x + 1 3x 2 − 4 x + 1 ( 3x − 1)( x + 4 )( 3x − 1)( x − 1) = 2 2 ( 3x − 1) ( x − 1)
=
x+4 x −1
1 , and the 3 domain of g ( x) consists all real numbers except 1. In addition, the reciprocal of 1 g ( x) has a restriction of and 1 . 3
In this case, the domain of f ( x) consists of all real numbers except
ANSWER: ( f / g ) ( x) =
x+4 x −1
1 The domain consists of any real number x where x ≠ 1 and x ≠ . 3
76.
PROBLEM: 36 − x 2 x 2 − 12 x + 36 f ( x) = 2 and g ( x) = 2 x + 12 x + 36 x + 4 x − 12 SOLUTION: 36 − x 2 x 2 − 12 x + 36 f ( x) = 2 and g ( x ) = 2 x + 12 x + 36 x + 4 x − 12 2 36 − x 2 36 − x 2 x 2 + 4 x − 12 ⋅ 2 ( f / g ) ( x) = x 2 + 12 x + 36 = 2 x − 12 x + 36 x + 12 x + 36 x − 12 x + 36 x 2 + 4 x − 12 − ( x − 6 )( x + 6 )( x + 6 )( x − 2 ) = 2 2 ( x + 6) ( x − 6)
x−2 x−6 In this case, the domain of f ( x) consists of all real numbers except 6, and the domain of g ( x) consists all real numbers except –6. In addition, the reciprocal of g ( x) has a restriction of –6 and 2 . =−
x−2 x−6 The domain consists of any real number x where x ≠ ±6 and x ≠ 2 . ANSWER: ( f / g ) ( x) = −
Part D: Discussion Board Topics
77.
PROBLEM: History of fractions, who is credited for the first use of the fraction bar? ANSWER: The earliest known use of fractions is ca. 2800 BC as Ancient Indus Valley units of measurement. The Egyptians used Egyptian fractions ca. 1000 BC. The Greeks used unit fractions and later continued fractions and followers of the Greek philosopher Pythagoras, ca. 530 BC, discovered that the square root of two cannot be expressed as a fraction. In 150 BC Jain mathematicians in India wrote the “Sthananga Sutra”, which contains work on the theory of numbers, arithmetical operations, operations with fractions. The fraction bar is called the vinculum, which was developed by in the 12th century by the Moroccan mathematician Abu Bakr al-Hassar. (Source: http://en.wikipedia.org/wiki/Fraction_%28mathematics%29)
78.
PROBLEM: How did the ancient Egyptians use fractions? ANSWER: An Egyptian fraction is the sum of distinct unit fractions. 1 1 1 For e.g.: + + 2 3 16 Each fraction in the expression has a numerator equal to 1 and a denominator that is a positive integer, and all the denominators differ from each other. The sum of an expression of this type is a positive rational number a/b; for instance the Egyptian fraction above sums to 43/48. Every positive rational number can be represented by an Egyptian fraction. Sums of this type, and similar sums also including 2/3 and 3/4 as summands, were used as a serious notation for rational numbers by the ancient Egyptians, and continued to be used by other civilizations into medieval times. In modern mathematical notation, Egyptian fractions have been superseded by vulgar fractions and decimal notation. (Source: http://en.wikipedia.org/wiki/Egyptian_fraction)
79.
PROBLEM:
Explain why x = 7 is a restriction to
1 x−7 ÷ . x x−2
ANSWER: 1 x−7 1 x−2 x−2 ÷ = ⋅ = x x − 2 x x − 7 x ( x − 7)
We can find the restrictions but equating the denominator to zero. x ( x − 7) = 0
x = 0 or x = 7
7.3 Adding and Subtracting Rational Expressions
Part A: Adding and Subtracting with Common Denominators Simplify. (Assume all denominators are nonzero.)
1.
PROBLEM: 3 7 + x x SOLUTION: 3 7 3 + 7 10 + = = x x x x ANSWER:
2.
10 x
PROBLEM: 9 10 − x x SOLUTION: 9 10 9 − 10 1 − = =− x x x x ANSWER: −
3.
1 x
PROBLEM: x 3 − y y SOLUTION: x 3 x −3 − = y y y ANSWER:
x−3 y
4.
PROBLEM: 4 6 + x −3 x −3 SOLUTION: 4 6 4+6 10 + = = x −3 x −3 x −3 x −3 ANSWER:
5.
10 x−3
PROBLEM: 7 x − 2x −1 2x −1 SOLUTION: 7 x 7−x − = 2x −1 2x −1 2x −1 ANSWER:
6.
7−x 2x −1
PROBLEM: 8 3x − 3x − 8 3x − 8 SOLUTION: 8 3x 8 − 3x − ( 3x − 8 ) − = = = −1 3x − 8 3x − 8 3x − 8 3x − 8 ANSWER: −1
7.
PROBLEM: 2 x − 11 + x −9 x −9 SOLUTION: 2 x − 11 2 + x − 11 x − 9 + = = =1 x −9 x −9 x −9 x −9 ANSWER: 1
8.
PROBLEM: y+2 y +3 − 2y + 3 2y + 3 SOLUTION: y+2 y +3 y + 2− y −3 1 − = =− 2y + 3 2y + 3 2y + 3 2y + 3 ANSWER: −
9.
1 2y + 3
PROBLEM: 2x − 3 x − 4 − 4x −1 4x −1 SOLUTION: 2x − 3 x − 4 2x − 3 − x + 4 x +1 − = = 4x −1 4x −1 4x −1 4x −1 ANSWER:
10.
x +1 4x −1
PROBLEM: 2 x 3x + 4 x − 2 − + x −1 x −1 x −1 SOLUTION: 2 x 3x + 4 x − 2 2 x − 3x − 4 + x − 2 6 − + = =− x −1 x −1 x −1 x −1 x −1 ANSWER: −
6 x −1
11.
PROBLEM: 1 2 y − 9 13 − 5 y − − 3y 3y 3y SOLUTION: 1 2 y − 9 13 − 5 y 1 − 2 y + 9 − 13 + 5 y 3 y − 3 − − = = 3y 3y 3y 3y 3y
=
ANSWER:
12.
3 ( y − 1) 3y
=
y −1 y
y −1 y
PROBLEM: −3 y + 2 y+7 3y + 4 + − 5 y − 10 5 y − 10 5 y − 10 SOLUTION: −3 y + 2 y+7 3 y + 4 −3 y + 2 + y + 7 − 3 y − 4 −5 y + 5 + − = = 5 y − 10 5 y − 10 5 y − 10 5 y − 10 5 y − 10
=
ANSWER:
13.
5 (1 − y )
5 ( y − 2)
=
1− y y−2
1− y y−2
PROBLEM: x 3 − ( x + 1)( x − 3) ( x + 1)( x − 3) SOLUTION: x 3 x −3 1 − = = ( x + 1)( x − 3) ( x + 1)( x − 3) ( x + 1)( x − 3) x + 1 ANSWER:
1 x +1
14.
PROBLEM: 3x + 5 x+6 − ( 2 x − 1)( x − 6 ) ( 2 x − 1)( x − 6 ) SOLUTION: 3x + 5 x+6 3x + 5 − x − 6 2x −1 − = = ( 2 x − 1)( x − 6 ) ( 2 x − 1)( x − 6 ) ( 2 x − 1)( x − 6 ) ( 2 x − 1)( x − 6 ) =
ANSWER:
15.
1 x−6
1 x−6
PROBLEM: x 6 + 2 2 x − 36 x − 36 SOLUTION: ( x + 6) = 1 x x+6 6 + = = x 2 − 36 x 2 − 36 x 2 − 36 ( x + 6 )( x − 6 ) x − 6 ANSWER:
16.
1 x−6
PROBLEM: x 9 − 2 2 x − 81 x − 81 SOLUTION: x 9 x −9 x−9 1 − 2 = 2 = = 2 x − 81 x − 81 x − 81 ( x − 9 )( x + 9 ) x + 9 ANSWER:
1 x+9
17.
PROBLEM: x2 + 2 x − 22 + 2 2 x + 3 x − 28 x + 3 x − 28 SOLUTION: x2 + 2 x − 22 x 2 + x − 20 ( x + 5 )( x − 4 ) + = = x 2 + 3 x − 28 x 2 + 3x − 28 x 2 + 3 x − 28 ( x − 4 )( x + 7 )
=
ANSWER:
18.
x+5 x+7
x+5 x+7
PROBLEM: x 3− x − 2 2 2x − x − 3 2x − x − 3 SOLUTION: x 3− x x −3+ x 2x − 3 1 − 2 = 2 = = 2 2 x − x − 3 2 x − x − 3 2 x − x − 3 ( 2 x − 3)( x + 1) x + 1
ANSWER:
1 x +1
Part B: Adding and Subtracting with Unlike Denominators Simplify. (Assume all denominators are nonzero.)
19.
PROBLEM: 1 1 + 2 3x SOLUTION: LCD = 6 x To obtain equivalent terms with this common denominator, we will multiply the 3x 2 first term by and the second term by . 3x 2 1 1 1 3x 1 2 3x 2 3x + 2 + = ⋅ + ⋅ = + = 2 3x 2 3x 3x 2 6 x 6 x 6x ANSWER:
3x + 2 6x
20.
PROBLEM: 1 1 − 5x 2 x SOLUTION: LCD = 5 x 2 To obtain equivalent terms with this common denominator, we will multiply the 1 5x2 first term by and the second term by 2 . 1 5x 2 1 1 1 1 1 5x 1 5x 1 − 5x − = 2⋅ − ⋅ 2 = 2− 2 = 2 x 5x 1 x 5x 5x 5x 5x 5x2 ANSWER:
21.
1 − 5x 5x2
PROBLEM: 1 3 + 2 12 y 10 y 3 SOLUTION: LCD = 60 y 3 To obtain equivalent terms with this common denominator, we will multiply the 5y 6 first term by and the second term by . 5y 6 1 3 1 5y 3 6 5y 18 5 y + 18 + = ⋅ + ⋅ = + = 2 3 2 3 3 3 12 y 10 y 12 y 5 y 10 y 6 60 y 60 y 60 y 3 ANSWER:
22.
5 y + 18 60 y 3
PROBLEM: 1 1 − x 2y SOLUTION: LCD = 2 xy To obtain equivalent terms with this common denominator, we will multiply the 2y x first term by and the second term by . 2y x 1 1 1 2y 1 x 2y x 2y − x − = ⋅ − ⋅ = − = x 2 y x 2 y 2 y x 2 xy 2 xy 2 xy ANSWER:
2y − x 2 xy
23.
PROBLEM: 1 −2 y SOLUTION: LCD = y To obtain equivalent terms with this common denominator, we will multiply the 1 y first term by and the second term by . y 1 1 1 1 y 1 2 y 1− 2 y − 2 = ⋅ − 2⋅ = − = y y 1 y y y y ANSWER:
24.
1− 2 y y
PROBLEM: 3 −4 y+2 SOLUTION: LCD = y + 2 To obtain equivalent terms with this common denominator, we will multiply the 1 y+2 first term by and the second term by . y+2 1 3 3 1 3 4y +8 y+2 −4 = ⋅ − 4⋅ = − y+2 y+2 1 y+2 y+2 y+2 3− 4y −8 4y + 5 = =− y+2 y+2 ANSWER: −
4y + 5 y+2
25.
PROBLEM: 2 +2 x+4 SOLUTION: LCD = x + 4 To obtain equivalent terms with this common denominator, we will multiply the 1 x+4 first term by and the second term by . 1 x+4 2 2 1 x + 4 2 + 2x + 8 = +2= ⋅ + 2⋅ x+4 1 x+4 x+4 x+4 2 x + 10 = x+4 2 ( x + 5) = x+4 ANSWER:
26.
2 ( x + 5) x+4
PROBLEM: 2 1 − y y2 SOLUTION: LCD = y 2 To obtain equivalent terms with this common denominator, we will multiply the 1 y first term by and the second term by . y 1 2 1 2 y 1 1 2y 1 − = ⋅ − ⋅ = − y y2 y y y2 1 y2 y2 2 y −1 = y2 ANSWER:
2 y −1 y2
27.
PROBLEM: 3 1 + x +1 x SOLUTION: LCD = x ( x + 1)
To obtain equivalent terms with this common denominator, we will multiply the x x +1 first term by and the second term by . x x +1 3 1 3 x 1 x +1 3x x +1 + = ⋅ + ⋅ = + x + 1 x x + 1 x x x + 1 x ( x + 1) x ( x + 1)
3x + x + 1 4x +1 = x ( x + 1) x ( x + 1)
=
ANSWER:
28.
4x +1 x ( x + 1)
PROBLEM: 1 2 − x −1 x SOLUTION: LCD = x ( x − 1)
To obtain equivalent terms with this common denominator, we will multiply the x x −1 first term by and the second term by . x x −1 1 2 1 x 2 x −1 x 2x − 2 = − − = ⋅ − ⋅ x − 1 x x − 1 x x x − 1 x ( x − 1) x ( x − 1)
=
x − 2x + 2 2− x = x ( x − 1) x ( x − 1)
ANSWER:
2− x x ( x − 1)
29.
PROBLEM: 1 1 + x −3 x +5 SOLUTION: LCD = ( x − 3)( x + 5)
To obtain equivalent terms with this common denominator, we will multiply the x+5 x−3 first term by . and the second term by x+5 x−3 1 1 1 x+5 1 x −3 x+5 x −3 + = + + = ⋅ ⋅ x − 3 x + 5 x − 3 x + 5 x + 5 x − 3 ( x − 3)( x + 5) ( x − 3)( x + 5 )
= =
ANSWER:
30.
x +5+ x −3 2x + 2 = ( x − 3)( x + 5) ( x − 3)( x + 5) 2 ( x + 1)
( x − 3)( x + 5) 2 ( x + 1)
( x − 3)( x + 5)
PROBLEM: 1 1 − x + 2 x −3 SOLUTION: LCD = ( x + 2 )( x − 3)
To obtain equivalent terms with this common denominator, we will multiply the x−3 x+2 first term by . and the second term by x−3 x+2 1 1 1 x −3 1 x+2 x −3 x+2 = − − = ⋅ − ⋅ x + 2 x − 3 x + 2 x − 3 x − 3 x + 2 ( x + 2 )( x − 3) ( x + 2 )( x − 3)
=
ANSWER: −
5 x −3− x − 2 =− ( x + 2 )( x − 3) ( x + 2 )( x − 3) 5
( x + 2 )( x − 3)
31.
PROBLEM: x 2 − x +1 x − 2 SOLUTION: LCD = ( x + 1)( x − 2 )
To obtain equivalent terms with this common denominator, we will multiply the x−2 x +1 first term by . and the second term by x +1 x−2 x 2 x x−2 2 x +1 x2 − 2 x 2x + 2 − = ⋅ − ⋅ = − x + 1 x − 2 x + 1 x − 2 x − 2 x + 1 ( x + 1)( x − 2 ) ( x + 1)( x − 2 )
=
ANSWER:
32.
x2 − 2 x − 2 x − 2 x2 − 4x − 2 = ( x + 1)( x − 2 ) ( x + 1)( x − 2 )
x2 − 4x − 2 ( x + 1)( x − 2 )
PROBLEM: 2x − 3 x − x +5 x −3 SOLUTION: LCD = ( x + 5 )( x − 3)
To obtain equivalent terms with this common denominator, we will multiply the x−3 x+5 first term by . and the second term by x−3 x+5 2x − 3 x 2x − 3 x − 3 x x + 5 2x2 − 9 x + 9 x2 + 5x = − − = ⋅ − ⋅ x + 5 x − 3 x + 5 x − 3 x − 3 x + 5 ( x + 5)( x − 3) ( x + 5)( x − 3)
2 x2 − 9 x + 9 − x2 − 5x x 2 − 14 x + 9 = = ( x + 5)( x − 3) ( x + 5)( x − 3) ANSWER:
x 2 − 14 x + 9 ( x + 5)( x − 3)
33.
PROBLEM: y +1 y −1 + y −1 y +1 SOLUTION: LCD = ( y + 1)( y − 1)
To obtain equivalent terms with this common denominator, we will multiply the y +1 y −1 first term by . and the second term by y +1 y −1 y + 1 y −1 y +1 y + 1 y −1 y −1 y2 + 2 y +1 y2 − 2 y +1 + = + + = ⋅ ⋅ y − 1 y + 1 y − 1 y + 1 y + 1 y − 1 ( y − 1)( y + 1) ( y − 1)( y + 1) = =
ANSWER:
34.
y2 + 2 y +1+ y2 − 2 y +1 2 y2 + 2 = ( y − 1)( y + 1) ( y − 1)( y + 1)
(
)
2 y2 +1
( y − 1)( y + 1)
(
)
2 y2 +1
( y − 1)( y + 1)
PROBLEM: 3y −1 y + 4 − 3y y−2 SOLUTION: LCD = ( 3 y )( y − 2 )
To obtain equivalent terms with this common denominator, we will multiply the y−2 3y first term by and the second term by . y−2 3y 3 y − 1 y + 4 3 y − 1 y − 2 y + 4 3 y 3 y 2 − 7 y + 2 3 y 2 + 12 y − = ⋅ − ⋅ = − 3y y−2 3y y − 2 y − 2 3y 3 y ( y − 2) 3 y ( y − 2) =
ANSWER:
3 y 2 − 7 y + 2 − 3 y 2 − 12 y 2 − 19 y = 3 y ( y − 2) 3 y ( y − 2)
2 − 19 y 3 y ( y − 2)
35.
PROBLEM: 2x − 5 2x + 5 − 2x + 5 2x − 5 SOLUTION: LCD = ( 2 x + 5 )( 2 x − 5 )
To obtain equivalent terms with this common denominator, we will multiply the 2x − 5 2x + 5 first term by . and the second term by 2x − 5 2x + 5 2x − 5 2x + 5 2x − 5 2x − 5 2x + 5 2x + 5 − = ⋅ − ⋅ 2x + 5 2x − 5 2x + 5 2x − 5 2x − 5 2x + 5 4 x 2 − 20 x + 25 4 x 2 + 20 x + 25 = − ( 2 x + 5)( 2 x − 5) ( 2 x + 5)( 2 x − 5)
4 x 2 − 20 x + 25 − 4 x 2 − 20 x − 25 = ( 2 x + 5)( 2 x − 5) =−
ANSWER: −
40 x ( 2 x + 5) ( 2 x − 5)
40 x ( 2 x + 5)( 2 x − 5)
36.
PROBLEM: 2 2x +1 − 2x −1 1− 2x SOLUTION: LCD = ( 2 x − 1)(1 − 2 x )
To obtain equivalent terms with this common denominator, we will multiply the 1− 2x 2x −1 first term by . and the second term by 1− 2x 2x −1 2 2x +1 2 1 − 2x 2x +1 2x −1 − = ⋅ − ⋅ 2x −1 1 − 2x 2x −1 1 − 2x 1 − 2x 2x −1 2 − 4x 4 x2 − 1 = − 2 2 − ( 2 x − 1) − ( 2 x − 1)
= = =
2 − 4x − 4x2 + 1 − ( 2 x − 1)
2
(
− 4 x2 + 4 x − 3 − ( 2 x − 1)
)
2
4 x2 + 4 x − 3
( 2 x − 1) ( 2 x − 1)( 2 x + 3) = 2 ( 2 x − 1) =
ANSWER:
2
2x + 3 2x −1
2x + 3 2x −1
37.
PROBLEM: 3x + 4 2 − x −8 8− x SOLUTION: LCD = ( x − 8 )( 8 − x )
To obtain equivalent terms with this common denominator, we will multiply the 8− x x −8 first term by . and the second term by 8− x x −8 3x + 4 2 3x + 4 8 − x 2 x −8 − = ⋅ − ⋅ x −8 8− x x −8 8− x 8− x x −8 −3x 2 + 20 x + 32 2 x − 16 = − 2 2 − ( x − 8) − ( x − 8)
= = = =
ANSWER:
−3x 2 + 20 x + 32 − 2 x + 16 − ( x − 8)
2
−3x 2 + 18 x + 48 − ( x − 8)
2
−3 ( x + 2 )( x − 8) − ( x − 8) 3( x + 2)
( x − 8)
3( x + 2)
( x − 8)
2
38.
PROBLEM: 1 1 + y −1 1 − y SOLUTION: LCD = ( y − 1)(1 − y )
To obtain equivalent terms with this common denominator, we will multiply the 1− y y −1 first term by . and the second term by y −1 1− y 1 1 1 1− y 1 y −1 1− y y −1 + = ⋅ + ⋅ = + 2 y − 1 1 − y y − 1 1 − y 1 − y y − 1 − ( y − 1) − ( y − 1)2 =
1 − y + y −1 − ( y − 1)
=0 ANSWER: 0
2
=
0 − ( y − 1)
2
39.
PROBLEM: x + 15 2 x2 + 2 x − 9 9 − x2
(
)(
SOLUTION: LCD = x 2 − 9 9 − x 2
)
To obtain equivalent terms with this common denominator, we will multiply the 9 − x2 x2 − 9 first term by . and the second term by 9 − x2 x2 − 9 2 x2 x + 15 2 x 2 9 − x 2 x + 15 x 2 − 9 + = ⋅ + ⋅ x2 − 9 9 − x2 x2 − 9 9 − x2 9 − x2 x2 − 9 9 − x 2 2 x 2 − x − 15 = 2 − x2 − 9
(
)( ) ( ) ( 9 − x ) ( 2 x + 5)( x − 3) = − ( x − 9) 2
2
=
ANSWER:
( 2 x + 5)( x − 3)
(x
2
−9
)
=
( 2 x + 5)( x − 3) ( x − 3)( x + 3)
=
2x + 5 x+3
2x + 5 x+3
2
40.
PROBLEM: x 1 15 − x + − x + 3 x − 3 ( x + 3)( x − 3) SOLUTION: LCD = ( x + 3)( x − 3)
To obtain equivalent terms with this common denominator, we will multiply the x−3 x+3 1 and the third term by . , the second term by first term by x−3 x+3 1 x 1 15 − x x x−3 1 x+3 15 − x 1 + − = ⋅ + ⋅ − ⋅ x + 3 x − 3 ( x + 3)( x − 3) x + 3 x − 3 x − 3 x + 3 ( x + 3)( x − 3) 1
x 2 − 3x x+3 15 − x = + − ( x + 3)( x − 3) ( x + 3)( x − 3) ( x + 3)( x − 3) =
x 2 − 3 x + x + 3 − 15 + x ( x + 3) ( x − 3)
x 2 − x − 12 = ( x + 3)( x − 3)
ANSWER:
x−4 x −3
=
( x − 4 )( x + 3) ( x + 3)( x − 3)
=
x−4 x−3
41.
PROBLEM: 2x 1 2( x − 1) − + 3 x − 1 3 x + 1 ( 3 x − 1)( 3 x + 1) SOLUTION: LCD = ( 3x − 1)( 3x + 1)
To obtain equivalent terms with this common denominator, we will multiply the 3x + 1 3x − 1 1 , and the third term by . , the second term by first term by 3x + 1 3x − 1 1 2x 1 2( x − 1) 2 x 3x + 1 1 3x − 1 2( x − 1) 1 − + = ⋅ − ⋅ + ⋅ 3x − 1 3 x + 1 ( 3x − 1)( 3x + 1) 3 x − 1 3x + 1 3x + 1 3x − 1 ( 3 x − 1)( 3 x + 1) 1 6 x2 + 2 x 3x − 1 2x − 2 = − + ( 3x − 1)( 3x + 1) ( 3x − 1)( 3x + 1) ( 3x − 1)( 3x + 1) =
6 x 2 + 2 x − 3x + 1 + 2 x − 2 ( 3x − 1)( 3x + 1)
6 x2 + x − 1 = ( 3x − 1)( 3x + 1)
ANSWER:
2x +1 3x + 1
=
( 3x − 1)( 2 x + 1) ( 3x − 1)( 3x + 1)
=
2x +1 3x + 1
42.
PROBLEM: 4x x 16 x − 3 − + 2 x + 1 x − 5 ( 2 x + 1)( x − 5 ) SOLUTION: LCD = ( 2 x + 1)( x − 5 )
To obtain equivalent terms with this common denominator, we will multiply the x−5 2x +1 1 and the third term by . , the second term by first term by x−5 2x +1 1 4x x 16 x − 3 4x x − 5 x 2x +1 16 x − 3 1 − + = ⋅ − ⋅ + ⋅ 2 x + 1 x − 5 ( 2 x + 1)( x − 5 ) 2 x + 1 x − 5 x − 5 2 x + 1 ( 2 x + 1)( x − 5 ) 1
ANSWER:
x−3 x−5
=
4 x 2 − 20 x 2x2 + x 16 x − 3 − + ( 2 x + 1)( x − 5) ( 2 x + 1)( x − 5) ( 2 x + 1)( x − 5)
=
4 x 2 − 20 x − 2 x 2 − x + 16 x − 3 ( 2 x + 1)( x − 5 )
=
2x2 − 5x − 3 ( 2 x + 1)( x − 5 )
=
( 2 x + 1)( x − 3) ( 2 x + 1)( x − 5 )
=
x−3 x −5
43.
PROBLEM: x 2 4 + + 3x x − 2 3x ( x − 2 ) SOLUTION: LCD = 3x ( x − 2 )
To obtain equivalent terms with this common denominator, we will multiply the x−2 3x 1 , the second term by , and the third term by . first term by x−2 3x 1 x 2 4 x x−2 2 3x 4 1 + + = ⋅ + ⋅ + ⋅ 3x x − 2 3x ( x − 2 ) 3x x − 2 x − 2 3x 3x ( x − 2 ) 1 =
x2 − 2x 6x 4 + + 3x ( x − 2 ) 3x ( x − 2 ) 3x ( x − 2 )
=
x2 − 2 x + 6 x + 4 3x ( x − 2 )
=
x2 + 4 x + 4 3x ( x − 2 )
x2 + 4x + 4 ANSWER: 3x ( x − 2 )
44.
PROBLEM: −2 x 3x 18( x − 2) − − x + 6 6 − x ( x + 6 )( x − 6 ) SOLUTION: LCD = ( x + 6 )( x − 6 )
To obtain equivalent terms with this common denominator, we will multiply the x−6 x+6 1 and the third term by . , the second term by first term by x−6 x+6 1 −2 x 3x 18( x − 2) −2 x x − 6 3x x + 6 18( x − 2) 1 − − = ⋅ − ⋅ − ⋅ x + 6 6 − x ( x + 6 )( x − 6 ) x + 6 x − 6 6 − x x + 6 ( x + 6 )( x − 6 ) 1 =
−2 x 2 + 12 x 3 x 2 + 18 x 18 x − 36 + − ( x + 6 )( x − 6 ) ( x + 6 )( x − 6 ) ( x + 6 )( x − 6 )
=
−2 x 2 + 12 x + 3x 2 + 18 x − 18 x + 36 ( x + 6 )( x − 6 )
=
x 2 + 12 x + 36 ( x + 6 )( x − 6 )
( x + 6) = ( x + 6 )( x − 6 ) 2
= ANSWER:
x+6 x−6
x+6 x−6
45.
PROBLEM: x 1 25 − 7 x − − x + 5 x − 7 ( x + 5 )( x − 7 ) SOLUTION: LCD = ( x + 5 )( x − 7 )
To obtain equivalent terms with this common denominator, we will multiply the x−7 x+5 1 , and the third term by . , the second term by first term by x−7 x+5 1 x 1 25 − 7 x x x−7 1 x+5 25 − 7 x 1 − − = ⋅ − ⋅ − ⋅ x + 5 x − 7 ( x + 5)( x − 7 ) x + 5 x − 7 x − 7 x + 5 ( x + 5)( x − 7 ) 1 x2 − 7 x 25 − 7 x x+5 = − − ( x + 5)( x − 7 ) ( x + 5)( x − 7 ) ( x + 5)( x − 7 ) =
x 2 − 7 x − x − 5 − 25 + 7 x ( x + 5)( x − 7 )
x 2 − x − 30 = ( x + 5)( x − 7 )
ANSWER:
x−6 x−7
=
( x + 5)( x − 6 ) ( x + 5)( x − 7 )
=
x−6 x−7
46.
PROBLEM: x 2 + 2 x − 2x − 3 x − 3 SOLUTION: x 2 x 2 + = + 2 x − 2 x − 3 x − 3 ( x − 3)( x + 1) x − 3
LCD = ( x − 3)( x + 1) To obtain equivalent terms with this common denominator, we will multiply the 1 x +1 . first term by and the second term by 1 x +1 2 1 2 x +1 x x + = ⋅ + ⋅ ( x − 3)( x + 1) x − 3 ( x − 3)( x + 1) 1 x − 3 x + 1
=
ANSWER:
x
( x − 3)( x + 1)
=
x + 2x + 2 ( x − 3)( x + 1)
=
3x + 2 ( x − 3)( x + 1)
3x + 2 ( x − 3)( x + 1)
+
2x + 2 ( x − 3)( x + 1)
47.
PROBLEM: 1 x2 − 2 x + 5 x − 25 SOLUTION: 1 x2 1 x2 − 2 = − x + 5 x − 25 x + 5 ( x + 5)( x − 5)
LCD = ( x + 5)( x − 5) To obtain equivalent terms with this common denominator, we will multiply the x−5 1 and the second term by . first term by x−5 1 2 1 1 x−5 1 x x2 − = ⋅ − ⋅ x + 5 ( x + 5 )( x − 5 ) x + 5 x − 5 ( x + 5 )( x − 5 ) 1 =
x −5 x2 − ( x + 5)( x − 5) ( x + 5)( x − 5)
=
− x2 + x − 5 ( x + 5)( x − 5)
− x2 + x − 5 ANSWER: ( x + 5)( x − 5)
48.
PROBLEM: 5x − 2 2 − 2 x −4 x−2 SOLUTION: 5x − 2 2 5x − 2 2 − = − 2 x − 4 x − 2 ( x + 2 )( x − 2 ) x − 2
LCD = ( x − 3)( x + 1) To obtain equivalent terms with this common denominator, we will multiply the 1 x+2 . first term by and the second term by 1 x+2 5x − 2 2 5x − 2 1 2 x+2 − = ⋅ − ⋅ ( x + 2 )( x − 2 ) x − 2 ( x + 2 )( x − 2 ) 1 x − 2 x + 2 =
5x − 2 2x + 4 − ( x + 2 )( x − 2 ) ( x + 2 )( x − 2 )
=
5x − 2 − 2 x − 4 ( x + 2 )( x − 2 )
=
3x − 6 ( x + 2 )( x − 2 )
= = ANSWER:
3 x+2
3 ( x − 2)
( x + 2 )( x − 2 ) 3 x+2
49.
PROBLEM: 1 6x − 3 − 2 x +1 x − 7x − 8 SOLUTION: 1 6x − 3 1 6x − 3 − 2 = − x + 1 x − 7 x − 8 x + 1 ( x − 8 )( x + 1)
LCD = ( x − 8)( x + 1) To obtain equivalent terms with this common denominator, we will multiply the x −8 1 and the second term by . first term by x −8 1 1 6x − 3 1 x −8 6x − 3 1 − = ⋅ − ⋅ x + 1 ( x − 8 )( x + 1) x + 1 x − 8 ( x − 8 )( x + 1) 1 =
x −8 6x − 3 − ( x − 8)( x + 1) ( x − 8)( x + 1)
=
x − 8 − 6x + 3 ( x − 8)( x + 1)
=
−5 x − 5 ( x − 8)( x + 1)
=
( x − 8)( x + 1)
=−
ANSWER: −
5 x −8
−5 ( x + 1)
5 x −8
50.
PROBLEM: 3x 1 − 2 9 x − 16 3x + 4 SOLUTION: 3x 1 3x 1 − = − 2 9 x − 16 3x + 4 ( 3x + 4 )( 3x − 4 ) 3x + 4
LCD = ( 3x + 4 )( 3x − 4 ) To obtain equivalent terms with this common denominator, we will multiply the 1 3x − 4 . first term by and the second term by 1 3x − 4 3x 1 3x 1 1 3x − 4 − = ⋅ − ⋅ ( 3x + 4 )( 3x − 4 ) 3x + 4 ( 3x + 4 )( 3x − 4 ) 1 3x + 4 3x − 4 =
3x 3x − 4 − ( 3x + 4 )( 3x − 4 ) ( 3x + 4 )( 3x − 4 )
=
3x − 3x + 4 ( 3x + 4 )( 3x − 4 )
= ANSWER:
4
4
( 3x + 4 )( 3x − 4 )
( 3x + 4 )( 3x − 4 )
51.
PROBLEM: 2x 1 + 2 2 x −1 x + x SOLUTION: 2x 1 2x 1 + 2 = + 2 x − 1 x + x ( x − 1)( x + 1) x ( x + 1)
LCD = x ( x + 1)( x − 1) To obtain equivalent terms with this common denominator, we will multiply the x x −1 . first term by and the second term by x x −1 2x 1 2x x 1 x −1 + = ⋅ + ⋅ ( x − 1)( x + 1) x ( x + 1) ( x − 1)( x + 1) x x ( x + 1) x − 1 =
x −1 2 x2 + x ( x − 1)( x + 1) x ( x − 1)( x + 1)
2x2 + x −1 = x ( x − 1)( x + 1)
ANSWER:
2x −1 x ( x − 1)
=
( 2 x − 1)( x + 1) x ( x − 1) ( x + 1)
=
2x −1 x ( x − 1)
52.
PROBLEM: x ( 4 x − 1) x − 2 2x + 7 x − 4 4 + x SOLUTION: x ( 4 x − 1) x ( 4 x − 1) x x − = − 2 2 x + 7 x − 4 4 + x ( 2 x − 1)( x + 4 ) x + 4
LCD = ( 2 x − 1)( x + 4 ) To obtain equivalent terms with this common denominator, we will multiply the 1 2x −1 . first term by and the second term by 1 2x −1 x ( 4 x − 1) x ( 4 x − 1) 1 x x 2x −1 − = ⋅ − ⋅ ( 2 x − 1)( x + 4 ) x + 4 ( 2 x − 1)( x + 4 ) 1 x + 4 2 x − 1
ANSWER:
=
4 x2 − x 2 x2 − x − ( 2 x − 1)( x + 4 ) ( 2 x − 1)( x + 4 )
=
4 x2 − x − 2 x2 + x ( 2 x − 1)( x + 4 )
=
2 x2 ( 2 x − 1)( x + 4 )
2 x2 ( 2 x − 1)( x + 4 )
53.
PROBLEM: 3x 2 2x − 2 3x + 5 x − 2 3x − 1 SOLUTION: 3x 2 2x 3x 2 2x − = − 2 3x + 5 x − 2 3x − 1 ( x + 2 )( 3x − 1) 3 x − 1
LCD = ( x + 2 )( 3x − 1) To obtain equivalent terms with this common denominator, we will multiply the 1 x+2 . first term by and the second term by 1 x+2 3x 2 1 2x x + 2 3x 2 2 x2 + 4 x ⋅ − ⋅ = − ( x + 2 )( 3x − 1) 1 3x − 1 x + 2 ( x + 2 )( 3x − 1) ( x + 2 )( 3x − 1) 3x 2 − 2 x 2 − 4 x = ( x + 2 )( 3x − 1) = =
ANSWER:
x ( x − 4)
( x + 2 )( 3x − 1)
x2 − 4 x ( x + 2 )( 3x − 1) x ( x − 4)
( x + 2 )( 3x − 1)
54.
PROBLEM: 2x 11x + 4 − 2 x − 4 x − 2x − 8 SOLUTION: 2x 11x + 4 2x 11x + 4 − 2 = − x − 4 x − 2 x − 8 x − 4 ( x + 2 )( x − 4 )
LCD = ( x + 2 )( x − 4 ) To obtain equivalent terms with this common denominator, we will multiply the x+2 1 and the second term by . first term by x+2 1 2x 11x + 4 2x x + 2 11x + 4 1 − = ⋅ − ⋅ x − 4 ( x + 2 )( x − 4 ) x − 4 x + 2 ( x + 2 )( x − 4 ) 1 =
2x2 + 4x 11x + 4 − ( x + 2 )( x − 4 ) ( x + 2 )( x − 4 )
2 x 2 + 4 x − 11x + 4 = ( x + 2 )( x − 4 )
ANSWER:
2x +1 x+2
=
2 x2 − 7 x + 4 ( x + 2 )( x − 4 )
=
( 2 x + 1)( x − 4 ) ( x + 2 )( x − 4 )
=
2x +1 x+2
55.
PROBLEM: x 6 x − 24 + 2 2x +1 2x − 7x − 4 SOLUTION: x 6 x − 24 x 6 x − 24 + 2 = + 2 x + 1 2 x − 7 x − 4 2 x + 1 ( 2 x + 1)( x − 4 )
LCD = ( 2 x + 1)( x − 4 ) To obtain equivalent terms with this common denominator, we will multiply the x−4 1 and the second term by . first term by x−4 1 6 x − 24 6 x − 24 1 x x x−4 + = ⋅ + ⋅ 2 x + 1 ( 2 x + 1)( x − 4 ) 2 x + 1 x − 4 ( 2 x + 1)( x − 4 ) 1
ANSWER:
x+6 2x +1
=
6 x − 24 x2 − 4 x + ( 2 x + 1)( x − 4 ) ( 2 x + 1)( x − 4 )
=
x 2 − 4 x + 6 x − 24 ( 2 x + 1)( x − 4 )
=
( x − 4 )( x + 6 ) ( 2 x + 1)( x − 4 )
=
x+6 2x +1
56.
PROBLEM: 1 1 + 2 2 x − x − 6 x − 3x − 10 SOLUTION: 1 1 1 1 + 2 = + 2 x − x − 6 x − 3x − 10 ( x − 3)( x + 2 ) ( x − 5)( x + 2 )
LCD = ( x − 5 )( x − 3)( x + 2 ) To obtain equivalent terms with this common denominator, we will multiply the x−5 x−3 . first term by and the second term by x−5 x−3 x−5 x−3 1 1 1 1 + = ⋅ + ⋅ ( x − 3)( x + 2 ) ( x − 5)( x + 2 ) ( x − 3)( x + 2 ) x − 5 ( x − 5)( x + 2 ) x − 3 =
x −5 x −3 + ( x − 5 )( x − 3)( x + 2 ) ( x − 5)( x − 3)( x + 2 )
=
x −5+ x −3 ( x − 5 )( x − 3)( x + 2 )
=
2x − 8 ( x − 5 )( x − 3)( x + 2 )
= ANSWER:
2 ( x − 4)
2 ( x − 4)
( x − 5 )( x − 3)( x + 2 )
( x − 5)( x − 3)( x + 2 )
57.
PROBLEM: x 3 − 2 2 x + 4x + 3 x − 4x − 5 SOLUTION: x 3 x 3 − 2 = − 2 x + 4 x + 3 x − 4 x − 5 ( x + 3)( x + 1) ( x − 5 )( x + 1)
LCD = ( x + 3)( x + 1)( x − 5 ) To obtain equivalent terms with this common denominator, we will multiply the x−5 x+3 . and the second term by first term by x−5 x+3 3 3 x x x −5 x+3 − = ⋅ − ⋅ ( x + 3)( x + 1) ( x − 5)( x + 1) ( x + 3)( x + 1) x − 5 ( x − 5)( x + 1) x + 3
ANSWER:
x −9 ( x + 3)( x − 5)
=
3x + 9 x2 − 5x − ( x + 3)( x + 1)( x − 5) ( x + 3)( x + 1)( x − 5 )
=
x 2 − 5 x − 3x + 9 ( x + 3)( x + 1)( x − 5)
=
x2 − 8x + 9 ( x + 3)( x + 1)( x − 5)
=
( x − 9 )( x + 1) ( x + 3)( x + 1)( x − 5)
=
x−9 ( x + 3)( x − 5)
58.
PROBLEM: y +1 y − 2 2 2 y + 5 y − 3 4 y −1 SOLUTION: y +1 y y +1 y − 2 = − 2 2 y + 5 y − 3 4 y − 1 ( 2 y − 1)( y + 3) ( 2 y − 1)( 2 y + 1)
LCD = ( 2 y − 1)( 2 y + 1)( y + 3) To obtain equivalent terms with this common denominator, we will multiply the 2 y +1 y+3 . first term by and the second term by 2 y +1 y+3 y +1 y y +1 2 y +1 y y +3 − = ⋅ − ⋅ ( 2 y − 1)( y + 3) ( 2 y − 1)( 2 y + 1) ( 2 y − 1)( y + 3) 2 y + 1 ( 2 y − 1)( 2 y + 1) y + 3
ANSWER:
=
2 y2 + 3 y +1 y2 + 3 y − ( 2 y − 1)( 2 y + 1)( y + 3) ( 2 y − 1)( 2 y + 1)( y + 3)
=
2 y2 + 3y +1− y2 − 3y ( 2 y − 1)( 2 y + 1)( y + 3)
=
y2 +1 ( 2 y − 1)( 2 y + 1)( y + 3)
y2 +1 ( 2 y − 1)( 2 y + 1)( y + 3)
59.
PROBLEM: y −1 2 − 2 2 y − 25 y − 10 y + 25 SOLUTION: y −1 y −1 2 2 − 2 = − 2 y − 25 y − 10 y + 25 ( y − 5 )( y + 5) ( y − 5)2
LCD = ( y − 5 ) ( y + 5 ) To obtain equivalent terms with this common denominator, we will multiply the y −5 y+5 . and the second term by first term by y −5 y+5 2 2 y −1 y −1 y −5 y+5 − = ⋅ − ⋅ 2 2 ( y − 5 )( y + 5) ( y − 5) ( y − 5)( y + 5 ) y − 5 ( y − 5) y + 5 2
= = =
ANSWER:
y2 − 6 y + 5
y 2 − 6 y + 5 − 2 y − 10
( y − 5) ( y + 5) 2
y2 − 8 y − 5
( y − 5) ( y + 5)
y2 − 8 y − 5
( y − 5) ( y + 5) 2
−
2 y + 10
( y − 5) ( y + 5) ( y − 5) ( y + 5) 2
2
2
60.
PROBLEM: 3x 2 + 24 12 − 2 x − 2x − 8 x − 4 SOLUTION: 3x 2 + 24 12 3x 2 + 24 12 − = − 2 x − 2 x − 8 x − 4 ( x − 4 )( x + 2 ) x − 4
LCD = ( x − 4 )( x + 2 ) To obtain equivalent terms with this common denominator, we will multiply the 1 x+2 . first term by and the second term by 1 x+2 3x 2 + 24 12 3x 2 + 24 1 12 x + 2 − = ⋅ − ⋅ ( x − 4 )( x + 2 ) x − 4 ( x − 4 )( x + 2 ) 1 x − 4 x + 2 3 x 2 + 24 12 x + 24 = − ( x − 4 )( x + 2 ) ( x − 4 )( x + 2 ) =
3x 2 + 24 − 12 x − 24 ( x − 4 )( x + 2 )
3x 2 − 12 x = ( x − 4 )( x + 2 ) = = ANSWER:
3x x+2
3x ( x − 4 )
( x − 4 )( x + 2 ) 3x x+2
61.
PROBLEM: 4 x 2 + 28 28 − 2 x − 6x − 7 x − 7 SOLUTION: 4 x 2 + 28 28 4 x 2 + 28 28 − = − 2 x − 6 x − 7 x − 7 ( x − 7 )( x + 1) x − 7
LCD = ( x − 7 )( x + 1) To obtain equivalent terms with this common denominator, we will multiply the 1 x +1 . first term by and the second term by x +1 1 4 x 2 + 28 28 4 x 2 + 28 1 28 x + 1 − = ⋅ − ⋅ ( x − 7 )( x + 1) x − 7 ( x − 7 )( x + 1) 1 x − 7 x + 1 4 x 2 + 28 28 x + 28 = − ( x − 7 )( x + 1) ( x − 7 )( x + 1) = = =
ANSWER:
4x x +1
4 x 2 − 28 x ( x − 7 )( x + 1) 4x ( x − 7)
( x − 7 )( x + 1) 4x x +1
62.
PROBLEM: a a 2 − 9a + 18 + 2 4 − a a − 13a + 36 SOLUTION: a a 2 − 9a + 18 a a 2 − 9a + 18 + 2 = + 4 − a a − 13a + 36 4 − a ( a − 4 )( a − 9 )
LCD = ( a − 4 )( a − 9 ) To obtain equivalent terms with this common denominator, we will multiply the a −9 1 first term by and the second term by . a −9 1 2 2 a a − 9a + 18 a a − 9 a − 9a + 18 1 + = ⋅ + ⋅ 4 − a ( a − 4 )( a − 9 ) 4 − a a − 9 ( a − 4 )( a − 9 ) 1
ANSWER:
=
a 2 − 9a + 18 − a 2 + 9a + ( a − 4 )( a − 9 ) ( a − 4 )( a − 9 )
=
− a 2 + 9a + a 2 − 9a + 18 ( a − 4 )( a − 9 )
=
18 ( a − 4 )( a − 9 )
=
18 ( a − 4 )( a − 9 )
18 ( a − 4 )( a − 9 )
63.
PROBLEM: 3a − 12 a+2 − 2 a − 8a + 16 4 − a SOLUTION: 3a − 12 a + 2 3a − 12 a + 2 − = − 2 a − 8a + 16 4 − a ( 4 − a )2 4 − a
LCD = ( 4 − a ) To obtain equivalent terms with this common denominator, we will multiply the 1 4−a . first term by and the second term by 1 4−a 3a − 12 a + 2 3a − 12 1 a + 2 4 − a − = ⋅ − ⋅ 2 2 (4 − a) 4 − a (4 − a) 1 4 − a 4 − a 2
= = =
3a − 12
(4 − a)
2
−
− a 2 + 2a + 8
(4 − a)
3a − 12 + a 2 − 2a − 8
(4 − a)
2
a 2 + a − 20
(4 − a) ( a + 5) ( a − 4 ) = 2 ( a − 4) =
ANSWER:
a+5 a−4
2
2
a+5 a−4
64.
PROBLEM: a 2 − 14 5 − 2 2a − 7 a − 4 1 + 2a SOLUTION: a 2 − 14 5 a 2 − 14 5 − = − 2 2a − 7a − 4 1 + 2a ( 2a + 1)( a − 4 ) 2a + 1
LCD = ( 2a + 1)( a − 4 ) To obtain equivalent terms with this common denominator, we will multiply the 1 a−4 . first term by and the second term by 1 a−4 5 1 5 a 2 − 14 a 2 − 14 a−4 − = ⋅ − ⋅ ( 2a + 1)( a − 4 ) 2a + 1 ( 2a + 1)( a − 4 ) 1 2a + 1 a − 4 =
5a − 20 a 2 − 14 − ( 2a + 1)( a − 4 ) ( 2a + 1)( a − 4 )
=
a 2 − 14 − 5a + 20 ( 2a + 1)( a − 4 )
a 2 − 5a + 6 = ( 2a + 1)( a − 4 ) =
ANSWER:
( a − 2 )( a − 3) ( 2a + 1)( a − 4 )
( a − 2 )( a − 3) ( 2a + 1)( a − 4 )
65.
PROBLEM: 1 x 3 − 2 + 2 x + 3 x − 6x + 9 x − 9 SOLUTION: 1 3 1 3 x x − 2 + 2 = − + 2 x + 3 x − 6 x + 9 x − 9 x + 3 ( x − 3) ( x + 3)( x − 3)
LCD = ( x − 3) ( x + 3) To obtain equivalent terms with this common denominator, we will multiply the 2 x − 3) ( x+3 x−3 , the second term by first term by , and the third term by . 2 x+3 x−3 ( x − 3) 2
1 x 3 1 ( x − 3) x x+3 3 x −3 − + = ⋅ − ⋅ + ⋅ 2 2 2 x + 3 ( x − 3) ( x + 3)( x − 3) x + 3 ( x − 3) ( x − 3) x + 3 ( x + 3)( x − 3) x − 3 2
= =
x2 − 6x + 9
6x
( x − 3) ( x + 3) 2
x 2 + 3x
+
3x − 9
( x − 3) ( x + 3) ( x − 3) ( x + 3) ( x − 3) ( x + 3) 2
2
x 2 − 6 x + 9 − x 2 − 3x + 3x − 9
=−
ANSWER: −
−
( x − 3) ( x + 3) 2
6x
( x − 3) ( x + 3) 2
2
66.
PROBLEM: 3x 2x 23x − 10 − + 2 x + 7 x − 2 x + 5 x − 14 SOLUTION: 3x 2x 23x − 10 3x 2x 23x − 10 − + 2 = − + x + 7 x − 2 x + 5 x − 14 x + 7 x − 2 ( x + 7 )( x − 2 )
LCD = ( x + 7 )( x − 2 ) To obtain equivalent terms with this common denominator, we will multiply the x−2 x+7 1 first term by , the second term by and the third term by . x−2 x+7 1 1 3x 2x 23 x − 10 3x x − 2 2 x x + 7 23 x − 10 − + = ⋅ − ⋅ + ⋅ x + 7 x − 2 ( x + 7 )( x − 2 ) x + 7 x − 2 x − 2 x + 7 ( x + 7 )( x − 2 ) 1
ANSWER:
x+5 x+7
=
3x 2 − 6 x 2 x 2 + 14 x 23 x − 10 − + ( x + 7 )( x − 2 ) ( x + 7 )( x − 2 ) ( x + 7 )( x − 2 )
=
3x 2 − 6 x − 2 x 2 − 14 x + 23 x − 10 ( x + 7 )( x − 2 )
=
x 2 + 3 x − 10 ( x + 7 )( x − 2 )
=
( x + 5)( x − 2 ) ( x + 7 )( x − 2 )
=
x+5 x+7
67.
PROBLEM: x + 3 x − 1 x ( x + 11) + − x − 1 x + 2 x2 + x − 2 SOLUTION: x ( x + 11) x + 3 x − 1 x ( x + 11) x + 3 x − 1 + − 2 = + − x − 1 x + 2 x + x − 2 x − 1 x + 2 ( x − 1)( x + 2 )
LCD = ( x − 1)( x + 2 ) To obtain equivalent terms with this common denominator, we will multiply the x+2 x −1 1 , the second term by first term by , and the third term by . x+2 1 x −1 x ( x + 11) x ( x + 11) 1 x + 3 x −1 x + 3 x + 2 x −1 x −1 + − = ⋅ + ⋅ − ⋅ x − 1 x + 2 ( x − 1)( x + 2 ) x − 1 x + 2 x + 2 x − 1 ( x − 1)( x + 2 ) 1 =
ANSWER:
x−7 x+2
x2 + 5x + 6 x2 − 2x + 1 x 2 + 11x + − ( x − 1)( x + 2 ) ( x − 1)( x + 2 ) ( x − 1)( x + 2 )
=
x 2 + 5 x + 6 + x 2 − 2 x + 1 − x 2 − 11x ( x − 1)( x + 2 )
=
x2 − 8x + 7 ( x − 1)( x + 2 )
=
( x − 7 )( x − 1) ( x − 1)( x + 2 )
=
x−7 x+2
68.
PROBLEM: 4 ( x + 5) −2 x 4 − + 2 3x + 1 x − 2 3x − 5 x − 2 SOLUTION: 4 ( x + 5) 4 ( x + 5) −2 x 4 −2 x 4 − + 2 = − + 3x + 1 x − 2 3x − 5 x − 2 3x + 1 x − 2 ( 3 x + 1)( x − 2 )
LCD = ( 3x + 1)( x − 2 ) To obtain equivalent terms with this common denominator, we will multiply the x−2 3x + 1 1 first term by , the second term by and the third term by . x−2 3x + 1 1 4 ( x + 5) 4 ( x + 5) −2 x −2 x x − 2 1 4 4 3x + 1 − + = ⋅ − ⋅ + ⋅ 3 x + 1 x − 2 ( 3 x + 1)( x − 2 ) 3 x + 1 x − 2 x − 2 3x + 1 ( 3 x + 1)( x − 2 ) 1 =
−2 x 2 + 4 x 12 x + 4 4 x + 20 − + ( 3x + 1)( x − 2 ) ( 3x + 1)( x − 2 ) ( 3x + 1)( x − 2 )
=
−2 x 2 + 4 x − 12 x − 4 + 4 x + 20 ( 3x + 1)( x − 2 )
−2 x 2 − 4 x + 16 = ( 3x + 1)( x − 2 ) =
ANSWER: −
2 ( x + 4)
( 3x + 1)
−2 ( x + 4 )( x − 2 )
( 3x + 1)( x − 2 ) 2 ( x + 4) =− ( 3x + 1)
69.
PROBLEM: 3 ( x + 5) x −1 x + 3 − − 2 4 x − 1 2 x + 3 8 x + 10 x − 3 SOLUTION: 3 ( x + 5) 3 ( x + 5) x −1 x + 3 x −1 x + 3 − − 2 = − − 4 x − 1 2 x + 3 8 x + 10 x − 3 4 x − 1 2 x + 3 ( 4 x − 1)( 2 x + 3)
LCD = ( 4 x − 1)( 2 x + 3) To obtain equivalent terms with this common denominator, we will multiply the 2x + 3 4x −1 1 , the second term by first term by , and the third term by . 2x + 3 4x −1 1 3 ( x + 5) 3 ( x + 5) x −1 x + 3 x −1 2x + 3 x + 3 4x −1 1 − − = ⋅ − ⋅ − ⋅ 4 x − 1 2 x + 3 ( 4 x − 1)( 2 x + 3) 4 x − 1 2 x + 3 2 x + 3 4 x − 1 ( 4 x − 1)( 2 x + 3) 1 =
2x2 + x − 3 4 x 2 + 11x − 3 3 x + 15 − − ( 4 x − 1)( 2 x + 3) ( 4 x − 1)( 2 x + 3) ( 4 x − 1)( 2 x + 3)
=
2 x 2 + x − 3 − 4 x 2 − 11x + 3 − 3 x − 15 ( 4 x − 1)( 2 x + 3)
=
−2 x 2 − 13x − 15 ( 4 x − 1)( 2 x + 3)
=− =−
ANSWER: −
x+5 4x −1
( 2 x + 3)( x + 5 ) ( 4 x − 1) ( 2 x + 3)
x+5 4x −1
70.
PROBLEM: 3x 2 6 x2 − 5x − 9 − − 2x − 3 2x + 3 4x2 − 9 SOLUTION: 3x 2 6 x2 − 5x − 9 3x 2 6 x2 − 5x − 9 − − = − − 2x − 3 2x + 3 4x2 − 9 2 x − 3 2 x + 3 ( 2 x − 3)( 2 x + 3)
LCD = ( 2 x − 3)( 2 x + 3) To obtain equivalent terms with this common denominator, we will multiply the 2x + 3 2x − 3 1 first term by , the second term by and the third term by . 2x + 3 2x − 3 1 2 3x 2 6 x − 5x − 9 3x 2 x + 3 2 2x − 3 6 x2 − 5x − 9 1 − − = ⋅ − ⋅ − ⋅ 2 x − 3 2 x + 3 ( 2 x − 3)( 2 x + 3) 2 x − 3 2 x + 3 2 x + 3 2 x − 3 ( 2 x − 3)( 2 x + 3) 1 6x2 + 9x 4x − 6 6 x2 − 5x − 9 = − − ( 2 x − 3)( 2 x + 3) ( 2 x − 3)( 2 x + 3) ( 2 x − 3)( 2 x + 3) =
6 x2 + 9 x − 4 x + 6 − 6 x2 + 5x + 9 ( 2 x − 3)( 2 x + 3)
=
10 x + 15 ( 2 x − 3)( 2 x + 3)
= =
ANSWER:
5 2x − 3
5 ( 2 x + 3)
( 2 x − 3)( 2 x + 3) 5 2x − 3
71.
PROBLEM: 1 1 2 + + 2 y +1 y y −1 SOLUTION: 1 1 2 1 1 2 + + 2 = + + y + 1 y y − 1 y + 1 y ( y + 1)( y − 1)
LCD = y ( y + 1)( y − 1) To obtain equivalent terms with this common denominator, we will multiply the y ( y − 1) ( y − 1)( y + 1) , and the third term by y . , the second term by first term by y y ( y − 1) ( y − 1)( y + 1) 1 1 2 1 y ( y − 1) 1 ( y − 1)( y + 1) 2 y + + = ⋅ + ⋅ + ⋅ y + 1 y ( y + 1)( y − 1) y + 1 y ( y − 1) y ( y − 1)( y + 1) ( y + 1)( y − 1) y
ANSWER:
2 y −1 y ( y − 1)
=
y2 − y + y2 −1+ 2 y y ( y + 1)( y − 1)
=
2 y2 + y −1 y ( y + 1)( y − 1)
=
( 2 y − 1)( y + 1) y ( y + 1)( y − 1)
=
2 y −1 y ( y − 1)
72.
PROBLEM: 1 1 1 − + y y +1 y −1 SOLUTION: LCD = y ( y − 1)( y + 1)
To obtain equivalent terms with this common denominator, we will multiply the ( y − 1)( y + 1) , the second term by y ( y − 1) and the third term by first term by y ( y − 1) ( y − 1)( y + 1) y ( y + 1) y ( y + 1)
.
1 1 1 1 ( y − 1)( y + 1) 1 y ( y − 1) 1 y ( y + 1) − + = ⋅ − ⋅ + ⋅ y y + 1 y − 1 y ( y − 1)( y + 1) y + 1 y ( y − 1) y − 1 y ( y + 1) =
y2 −1 y2 − y y2 + y − + y ( y − 1)( y + 1) y ( y − 1)( y + 1) y ( y − 1)( y + 1)
=
y2 −1 − y2 + y + y2 + y y ( y − 1)( y + 1)
=
y2 + 2 y −1 y ( y − 1)( y + 1)
=
y2 + 2 y −1 y ( y − 1)( y + 1)
ANSWER:
y2 + 2 y −1 y ( y − 1)( y + 1)
73.
PROBLEM: 5−2 + 2−1 SOLUTION: 1 1 1 1 5−2 + 2−1 = 2 + = + 5 2 25 2 LCD = 50 To obtain equivalent terms with this common denominator, we will multiply the 2 25 first term by and the second term by . 2 25 1 1 1 2 1 25 2 25 2 + 25 27 + = ⋅ + ⋅ = + = = 25 2 25 2 2 25 50 50 50 50 ANSWER:
74.
27 50
PROBLEM: 6−1 + 4−2 SOLUTION: 1 1 1 1 6−1 + 4−2 = + 2 = + 6 4 6 16 LCD = 48 To obtain equivalent terms with this common denominator, we will multiply the 8 3 first term by and the second term by . 8 3 1 1 1 8 1 3 8 3 8 + 3 11 + = ⋅ + ⋅ = + = = 6 16 6 8 16 3 48 48 48 48 11 ANSWER: 48
75.
PROBLEM: x −1 + y −1 SOLUTION: 1 1 x −1 + y −1 = + x y LCD = xy To obtain equivalent terms with this common denominator, we will multiply the y x first term by and the second term by . y x 1 1 1 y 1 x y x x+ y + = ⋅ + ⋅ = + = x y x y y x xy xy xy ANSWER:
76.
x+ y xy
PROBLEM: x −2 − y −1 SOLUTION: 1 1 x −2 − y −1 = 2 − x y 2 LCD = x y To obtain equivalent terms with this common denominator, we will multiply the y x2 first term by and the second term by 2 . y x 1 1 1 y 1 x2 y x2 y − x2 − = ⋅ − ⋅ = − = x2 y x2 y y x2 x2 y x2 y x2 y y − x2 ANSWER: 2 x y
77.
PROBLEM: −1 ( 2 x − 1) − x −2 SOLUTION:
( 2 x − 1)
−1
− x −2 =
1 1 − 2 2x −1 x
LCD = x 2 ( 2 x − 1) To obtain equivalent terms with this common denominator, we will multiply the x2 2x −1 . first term by 2 and the second term by 2x −1 x x2 1 2 x −1 2 1 1 − 2 = ⋅ 2− 2⋅ x 2x −1 2x −1 x 2x −1 x 2 x 2x −1 = 2 − 2 x ( 2 x − 1) x ( 2 x − 1) =
x2 − 2 x + 1 x 2 ( 2 x − 1)
( x − 1) = 2 x ( 2 x − 1) 2 x − 1) ( ANSWER: 2 x ( 2 x − 1) 2
78.
PROBLEM: −1 ( x − 4 ) − ( x + 1)−1 SOLUTION:
( x − 4)
−1
1 1 − x − 4 x +1
− ( x + 1) −1 =
LCD = ( x − 4 )( x + 1) To obtain equivalent terms with this common denominator, we will multiply the x +1 x−4 and the second term by first term by . x−4 x +1 1 1 x +1 x−4 − = − x − 4 x + 1 ( x − 4 )( x + 1) ( x − 4 )( x + 1) = = ANSWER:
79.
x +1− x + 4 ( x − 4 )( x + 1) 5
( x − 4 )( x + 1) 5
( x − 4 )( x + 1)
PROBLEM: 3x 2 ( x − 1) −1 − 2 x SOLUTION:
3x 2 3x ( x − 1) − 2 x = − 2x x −1 LCD = x − 1 To obtain equivalent terms with this common denominator, we will multiply the 1 x −1 first term by and the second term by . 1 x −1 x − 1 3x 2 2 x 2 − 2 x 3x 2 3x 2 1 − 2x = ⋅ − 2x ⋅ = − x −1 x −1 x −1 x −1 1 x −1 2 2 2 3x − 2 x + 2 x x + 2 x = = x −1 x −1 x ( x + 2) = x −1 2
−1
ANSWER:
x ( x + 2) x −1
80.
PROBLEM: −2 −1 2 ( y − 1) − ( y − 1) SOLUTION: 2 ( y − 1) − ( y − 1) = −2
2
−1
( y − 1)
2
−
1 y −1
LCD = ( y − 1) To obtain equivalent terms with this common denominator, we will multiply the 1 y −1 first term by and the second term by . 1 y −1 1 2 1 2 1 y −1 − = ⋅ − ⋅ 2 2 ( y − 1) y − 1 ( y − 1) 1 y − 1 y − 1 2
= = =
ANSWER:
2
( y − 1)
2
−
2 − y +1
( y − 1)
2
3− y
( y − 1)
3− y
( y − 1)
2
2
y −1
( y − 1)
2
Part C: Adding and Subtracting Rational Functions Calculate ( f + g ) ( x) and ( f − g ) ( x) and state the restrictions to the domain.
81.
PROBLEM: 1 1 f ( x) = and g ( x ) = 3x x−2 SOLUTION:
1 1 + 3x x − 2 LCD = 3x ( x − 2 ) To obtain equivalent terms with this common denominator, we will multiply the x−2 3x and the second term by . first term by x−2 3x 1 1 1 x−2 1 3x 3x x−2 + = ⋅ + ⋅ = + 3x x − 2 3x x − 2 x − 2 3x 3x ( x − 2 ) 3x ( x − 2 )
( f + g ) ( x) =
=
x − 2 + 3x 3x ( x − 2 )
=
4x − 2 3x ( x − 2 )
=
2 ( 2 x − 1)
3x ( x − 2 )
2 ( 2 x − 1)
( f + g ) ( x) =
3x ( x − 2 )
1 1 − 3x x − 2 1 1 1 x−2 1 3x 3x x−2 − = ⋅ − ⋅ = − 3x x − 2 3x x − 2 x − 2 3x 3x ( x − 2 ) 3x ( x − 2 )
( f − g ) ( x) =
=
x − 2 − 3x 3x ( x − 2 )
=
−4 x − 2 3x ( x − 2 )
=−
( f − g ) ( x)
2 ( 2 x + 1)
3x ( x − 2 )
=−
2 ( 2 x + 1)
3x ( x − 2 )
In this case, the domain of f ( x) consists of all real numbers except 0, and the domain of g ( x) consists all real numbers except 2. ANSWER: ( f + g ) ( x) =
2 ( 2 x − 1)
, ( f − g ) ( x) = −
2 ( 2 x + 1)
3x ( x − 2 ) 3x ( x − 2 ) The domain consists of any real number x where x ≠ 0 and x ≠ 2 . 82.
PROBLEM: 1 1 f ( x) = and g ( x ) = x −1 x+5 SOLUTION:
1 1 + x −1 x + 5 LCD = ( x − 1)( x + 5) To obtain equivalent terms with this common denominator, we will multiply the x+5 x −1 and the second term by first term by . x+5 x −1 1 1 1 x+5 1 x −1 x+5 x −1 + = ⋅ + ⋅ = + x − 1 x + 5 x − 1 x + 5 x + 5 x − 1 ( x − 1)( x + 5 ) ( x − 1)( x + 5 )
( f + g ) ( x) =
=
x + 5 + x −1 ( x − 1)( x + 5)
=
2x + 4 ( x − 1)( x + 5)
=
2 ( x + 2)
( x − 1)( x + 5) 2 ( x + 2) ( f + g ) ( x) = ( x − 1)( x + 5)
( f − g ) ( x) =
1 1 − x −1 x + 5
x+5 x −1 1 1 1 x+5 1 x −1 − = ⋅ − ⋅ = − x − 1 x + 5 x − 1 x + 5 x + 5 x − 1 ( x − 1)( x + 5 ) ( x − 1)( x + 5 ) = =
( f − g ) ( x) =
x + 5 − x +1 ( x − 1)( x + 5 ) 6
( x − 1)( x + 5 ) 6
( x − 1)( x + 5)
In this case, the domain of f ( x) consists of all real numbers except –1, and the domain of g ( x) consists all real numbers except –5. ANSWER: ( f + g ) ( x) =
( f − g ) ( x) =
2 ( x + 2)
( x − 1)( x + 5)
6
( x − 1)( x + 5)
The domain consists of any real number x where x ≠ −1 and x ≠ −5 .
83.
PROBLEM: x 1 f ( x) = and g ( x ) = 4− x x−4 SOLUTION:
x 1 + x−4 4− x x 1 x 1 x −1 + = − = x−4 4− x x−4 x−4 x−4 x −1 ( f + g ) ( x) = x−4
( f + g ) ( x) =
x 1 − x−4 4− x x 1 x 1 x +1 − = + = x−4 4− x x−4 x−4 x−4 x +1 ( f − g ) ( x) = x−4 In this case, the domain of f ( x) consists of all real numbers except 4, and the domain of g ( x) consists all real numbers except 4.
( f − g ) ( x) =
x −1 x +1 , ( f − g ) ( x) = x−4 x−4 The domain consists of any real number x where x ≠ 4 . ANSWER: ( f + g ) ( x) =
84.
PROBLEM: x 1 f ( x) = and g ( x ) = 2x − 3 x −5 SOLUTION:
x 1 + x − 5 2x − 3 LCD = ( x + 5)( 2 x − 3) To obtain equivalent terms with this common denominator, we will multiply the 2x − 3 x−5 and the second term by first term by . 2x − 3 x−5 1 1 x x 2x − 3 x − 5 2 x 2 − 3x + x − 5 + = ⋅ + ⋅ = x − 5 2 x − 3 x − 5 2 x − 3 2 x − 3 x − 5 ( x + 5 )( 2 x − 3)
( f + g ) ( x) =
=
( f + g ) ( x) =
2 x2 − 2x − 5 ( x + 5)( 2 x − 3)
2 x2 − 2 x − 5 ( x + 5)( 2 x − 3)
x 1 − x − 5 2x − 3 1 1 x x 2x − 3 x − 5 2 x 2 − 3x − x + 5 − = ⋅ − ⋅ = x − 5 2 x − 3 x − 5 2 x − 3 2 x − 3 x − 5 ( x + 5 )( 2 x − 3)
( f − g ) ( x) =
=
2x2 − 4x + 5 ( x + 5)( 2 x − 3)
2 x2 − 4 x + 5 ( x + 5 )( 2 x − 3) In this case, the domain of f ( x) consists of all real numbers except 5, and the 3 domain of g ( x) consists all real numbers except . 2 2 2x − 2x − 5 ANSWER: ( f + g ) ( x) = ( x + 5)( 2 x − 3)
( f − g ) ( x) =
( f − g ) ( x) =
2 x2 − 4 x + 5 ( x + 5 )( 2 x − 3)
The domain consists of any real number x where x ≠ 5 and x ≠
3 . 2
85.
PROBLEM: x −1 4 f ( x) = 2 and g ( x ) = 2 x −4 x − 6 x − 16 SOLUTION:
x −1 4 + 2 2 x − 4 x − 6 x − 16 x −1 4 x −1 4 + 2 = + 2 x − 4 x − 6 x − 16 ( x − 2 )( x + 2 ) ( x − 8 )( x + 2 )
( f + g ) ( x) =
LCD = ( x − 2 )( x + 2 )( x − 8) To obtain equivalent terms with this common denominator, we will multiply the x −8 x−2 and the second term by first term by . x −8 x−2 x −1 4 x −1 x −8 4 x−2 + = ⋅ + ⋅ ( x − 2 )( x + 2 ) ( x − 8)( x + 2 ) ( x − 2 )( x + 2 ) x − 8 ( x − 8)( x + 2 ) x − 2 =
4x − 8 x2 − 9 x + 8 + ( x − 2 )( x + 2 )( x − 8) ( x − 2 )( x + 2 )( x − 8)
x2 − 9x + 8 + 4x − 8 = ( x − 2 )( x + 2 )( x − 8) = =
x2 − 5x ( x − 2 )( x + 2 )( x − 8) x ( x − 5)
( x − 2 )( x + 2 )( x − 8)
( f + g ) ( x) =
x( x − 5) ( x − 2 )( x + 2 )( x − 8)
( f − g ) ( x) =
x −1 4 − 2 2 x − 4 x − 6 x − 16
x −1 4 x −1 x −8 4 x−2 − = ⋅ − ⋅ ( x − 2 )( x + 2 ) ( x − 8 )( x + 2 ) ( x − 2 )( x + 2 ) x − 8 ( x − 8 )( x + 2 ) x − 2 =
x2 − 9 x + 8 4x − 8 − ( x − 2 )( x + 2 )( x − 8) ( x − 2 )( x + 2 )( x − 8)
=
x2 − 9x + 8 − 4x + 8 ( x − 2 )( x + 2 )( x − 8)
x 2 − 13 x + 16 = ( x − 2 )( x + 2 )( x − 8)
( f − g ) ( x) =
x 2 − 13 x + 16 ( x − 2 )( x + 2 )( x − 8 )
In this case, the domain of f ( x) consists of all real numbers except 2 and –2, and the domain of g ( x) consists all real numbers except –2 and 8. ANSWER: ( f + g ) ( x) =
x( x − 5) , ( x − 2 )( x + 2 )( x − 8 )
x 2 − 13 x + 16 ( x − 2 )( x + 2 )( x − 8 ) The domain consists of any real number x where x ≠ ±2 and x ≠ 8 .
( f − g ) ( x) =
86.
PROBLEM: 5 f ( x) = and g ( x ) = 3x + 4 x+2 SOLUTION:
( f + g ) ( x) =
5 + 3x + 4 x+2
LCD = x + 2 To obtain equivalent terms with this common denominator, we will multiply the 1 x+2 first term by and the second term by . 1 x+2 5 5 1 x+2 + 3x + 4 = ⋅ + 3x + 4 ⋅ x+2 x+2 1 x+2 2 5 3 x + 10 x + 8 = + x+2 x+2 2 5 + 3 x + 10 x + 8 = x+2 2 3 x + 10 x + 13 = x+2 2 3 x + 10 x + 13 ( f + g ) ( x) = x+2
5 − ( 3x + 4 ) x+2 5 5 1 x+2 − ( 3x + 4 ) = ⋅ − ( 3x + 4 ) ⋅ x+2 x+2 1 x+2 2 5 3 x + 10 x + 8 = − x+2 x+2 2 5 − 3 x − 10 x − 8 = x+2 −3 x 2 − 10 x − 3 = x+2 2 3 x + 10 x + 3 ( f − g ) ( x) = − x+2 In this case, the domain of f ( x) consists of all real numbers except –2, and the domain of g ( x) consists all real numbers.
( f − g ) ( x) =
ANSWER: ( f + g ) ( x) =
3 x 2 + 10 x + 13 x+2
3 x 2 + 10 x + 3 x+2 The domain consists of any real number x where x ≠ −2 .
( f − g ) ( x) = −
Calculate ( f + f ) ( x) and state the restrictions to the domain.
87.
PROBLEM: 1 f ( x) = x SOLUTION:
( f + f )( x ) =
1 1 + x x
1 1 2 + = x x x 2 x In this case, the domain of f ( x) consists of all real numbers except 0.
( f + f )( x ) =
2 x The domain consists of any real number x where x ≠ 0 . ANSWER: ( f + f )( x ) =
88.
PROBLEM: 1 f ( x) = 2x SOLUTION:
1 1 + 2x 2x 1 1 2 1 + = = 2x 2x 2x x 1 ( f + f )( x ) = x In this case, the domain of f ( x) consists of all real numbers except 0.
( f + f )( x ) =
1 x The domain consists of any real number x where x ≠ 0 . ANSWER: ( f + f )( x ) =
89.
PROBLEM: x f ( x) = 2x −1 SOLUTION:
x x + 2x −1 2x −1 x x 2x + = 2x −1 2x −1 2x −1 2x ( f + f )( x ) = 2x −1
( f + f )( x ) =
In this case, the domain of f ( x) consists of all real numbers except ANSWER: ( f + f )( x ) =
2x 2x −1
The domain consists of any real number x where x ≠ 90.
1 . 2
1 . 2
PROBLEM: 1 f ( x) = x+2 SOLUTION:
1 1 + x+2 x+2 1 1 2 + = x+2 x+2 x+2 2 ( f + f )( x ) = x+2 In this case, the domain of f ( x) consists of all real numbers except –2.
( f + f )( x ) =
2 x+2 The domain consists of any real number x where x ≠ −2 . ANSWER: ( f + f )( x ) =
Part D: Discussion Board
91.
PROBLEM:
Explain to a classmate why this is incorrect:
1 2 3 + 2 = 2. 2 2x x x
ANSWER: 1 2 3 + 2 = 2 2 x x x The sum of the two rational expressions on the right hand side of the equation is 3 3 not . 2 2x 2 x
92.
PROBLEM: Explain to a classmate how to find the common denominator when adding algebraic expressions. Give an example. ANSWER: Adding and subtracting rational expressions is similar to adding and subtracting fractions. Recall that if the denominators are the same, we can add or subtract the numerators and write the result over the common denominator. 3 7 3+ 7 + = 13 13 13 10 = 13 x−5 1 x − 5 −1 − = Simplify the numerator. 2x −1 2x −1 2x −1 x−6 = 2x −1
To add rational expressions with unlike denominators, we need to first find equivalent expressions with common denominators. We do this just as we have with fractions. If the denominators of fractions are relatively prime, then the LCD is their product. For example, 1 1 + ⇒ LCD = 3*5 = 15 3 5 1 1 1 y 1 x + = ⋅ + ⋅ x y x y y x y x = + Equivalent terms with xy xy a common denominator. =
y+x xy
Add the numerators and place the result over the common denominator xy.
7.4 Complex Rational Expressions
Part A: Complex Rational Expressions Simplify. (Assume all denominators are nonzero.)
1.
PROBLEM: 1 2 5 4 SOLUTION: 1 2 = 1⋅4 = 2 5 2 5 5 4 ANSWER:
2.
2 5
PROBLEM: 7 8 5 4 SOLUTION: 7 8 = 7⋅4 = 7 5 8 5 10 4 7 ANSWER: 10
3.
PROBLEM: 10 3 20 9 SOLUTION: 10 3 = 10 ⋅ 9 = 3 20 3 20 2 9 ANSWER:
4.
3 2
PROBLEM: 4 − 21 8 7 SOLUTION: 4 − 21 = − 4 ⋅ 7 = − 1 8 21 8 6 7 1 ANSWER: − 6
5.
PROBLEM: 2 3 5 6 SOLUTION: 2 3 = 2⋅6 = 4 5 3 5 5 6 ANSWER:
4 5
6.
PROBLEM: 7 4 14 3 SOLUTION: 7 4 = 7⋅ 3 = 3 14 4 14 8 3 3 ANSWER: 8
7.
PROBLEM: 3 1− 2 5 1 − 4 3 SOLUTION: 3 1 1− − 2 = 2 = − 1 ⋅ 12 = − 6 5 1 11 2 11 11 − 4 3 12 ANSWER: −
8.
6 11
PROBLEM: 1 −5 2 1 1 + 2 3 SOLUTION: 1 9 −5 − 9 6 27 2 = 2 =− ⋅ =− 1 1 5 2 5 5 + 2 3 6 ANSWER: −
27 5
9.
PROBLEM: 3 1+ 2 1 1− 4 SOLUTION: 3 5 1+ 2 = 2 = 5 ⋅ 4 = 10 1 3 2 3 3 1− 4 4 ANSWER:
10.
10 3
PROBLEM: 1 2− 2 3 1+ 4 SOLUTION: 1 3 2− 2 = 2 = 3⋅4 = 6 3 7 2 7 7 1+ 4 4 ANSWER:
11.
6 7
PROBLEM: 5x2 x +1 25 x x +1 SOLUTION: 5x2 2 x + 1 = 5x ⋅ x + 1 = x 25 x x + 1 25 x 5 x +1 ANSWER:
x 5
12.
PROBLEM: 7+ x 7x x+7 14 x 2 SOLUTION: 7+ x 2 7 x = x + 7 ⋅ 14 x = 2 x x+7 7x x + 7 2 14 x ANSWER: 2x
13.
PROBLEM: 3y x y2 x −1 SOLUTION: 3y x = 3 y ⋅ x − 1 = 3 ( x − 1) y2 x y2 xy x −1 ANSWER:
14.
3 ( x − 1) xy
PROBLEM: 5a 2 b −1 15a 3
( b − 1)
2
SOLUTION: 5a 2 2 2 b − 1 = 5a ⋅ ( b − 1) = b − 1 15a 3 b − 1 15a 3 3a 2 ( b − 1) b −1 ANSWER: 3a
15.
PROBLEM: 1 1+ x 1 2− x SOLUTION: 1 x 1 x 1 x +1 1+ 1⋅ + + x = x x = x x = x 1 x 1 2x 1 2x −1 2− 2⋅ − − x x x x x x x +1 x = ⋅ x 2x −1 x +1 = 2x −1 ANSWER:
16.
x +1 2x −1
PROBLEM: 2 +1 x 1 3− x SOLUTION: 2 2 x 2 x 2+ x +1 + 1⋅ + x x = x x = x = x 1 x 1 3x 1 3x − 1 3− 3⋅ − − x x x x x x x 2+ x = ⋅ x 3x − 1 x+2 = 3x − 1 x+2 ANSWER: 3x − 1
17.
PROBLEM: 2 −4 3y 1 6− y SOLUTION: 2 2 3y 2 12 y 2 − 12 y −4 − 4⋅ − 3y 3y 3y 3y 3y 3y = = = y 1 1 6y 1 6 y −1 6− 6⋅ − − y y y y y y −2 ( 6 y − 1) −2 ( 6 y − 1) y 3y = = ⋅ 6 y −1 3y 6 y −1 y 2 =− 3
ANSWER: −
18.
2 3
PROBLEM: 5 1 − y 2 10 − y y2 SOLUTION: y 5 1 5 2 1 y 10 10 − y − ⋅ − ⋅ − y 2 y 2 2 y 2y 2y 2y = = = 10 − y 10 − y 10 − y 10 − y 2 2 2 y y y y2 10 − y y 2 ⋅ 2 y 10 − y y = 2 y ANSWER: 2 =
19.
PROBLEM: 1 1 − 5 x 1 1 − 25 x 2 SOLUTION: 1 1 1 x 1 5 x 5 x−5 − ⋅ − ⋅ − 5 x = 5 x x 5 = 5x 5x = 5x 1 1 x2 x 2 − 25 1 x 2 1 25 25 − 2 ⋅ 2− 2⋅ − 25 x 25 x x 25 25 x 2 25 x 2 25 x 2 x −5 5x = ( x − 5)( x + 5) 25 x 2 x−5 25 x 2 = ⋅ 5 x ( x − 5 )( x + 5 ) =
5x x+5
ANSWER:
5x x+5
20.
PROBLEM: 1 1 + x 5 1 1 − 25 x 2 SOLUTION: 1 1 1 x 1 5 x 5 x+5 + ⋅ + ⋅ + 5 x = 5 x x 5 = 5x 5x = 5x 1 1 x2 x 2 − 25 1 x 2 1 25 25 − 2 ⋅ 2− 2⋅ − 25 x 25 x x 25 25 x 2 25 x 2 25 x 2 x+5 5x = ( x − 5)( x + 5) 25 x 2 x+5 25 x 2 = ⋅ 5 x ( x − 5 )( x + 5 ) 5x x −5 5x ANSWER: x−5 =
21.
PROBLEM: 1 1 − x 3 1 1 − 9 x2 SOLUTION: 1 1 1 3 1 x 3 x 3− x − ⋅ − ⋅ − x 3 = x 3 3 x = 3x 3x = 3x 2 1 1 x2 9 x2 − 9 − 2 1 ⋅ x 2 − 12 ⋅ 9 − 9 x 9 x x 9 9 x2 9 x2 9 x2 3− x x−3 − x −3 9x2 3 x 3 x = 2 = =− ⋅ x − 9 ( x − 3)( x + 3) 3x ( x − 3)( x + 3) 9x2 9x2 3x =− x+3 ANSWER: −
3x x+3
22.
PROBLEM: 1 1 + 4 x 1 1 − x 2 16 SOLUTION: 1 1 1 x 1 4 x 4 + ⋅ + ⋅ + 4 x = 4 x x 4 = 4x 4x 1 1 x2 1 16 1 x 2 16 − ⋅ − ⋅ − x 2 16 x 2 16 16 x 2 16 x 2 16 x 2 x+4 = 4x 2 16 − x 16 x 2 x+4 4x = ( 4 − x )( 4 + x ) 16 x 2 x+4 16 x 2 = ⋅ 4 x ( 4 − x )( 4 + x ) =
4x 4− x
ANSWER:
4x 4− x
23.
PROBLEM: 1 16 − 2 x 1 −4 x SOLUTION: 1 x 2 1 16 x 2 1 16 x 2 − 1 − 2 16 − 2 16 ⋅ 2 − 2 x = x x = x2 x = x2 1 1 1 4x 1 − 4x x −4 −4 ⋅ − x x x x x x ( 4 x − 1)( 4 x + 1) ( 4 x − 1)( 4 x + 1) x2 x2 = = 1 − 4x 4x −1 − x x ( 4 x − 1)( 4 x + 1) ⋅ ⎛ − x ⎞ = ⎜ ⎟ x2 ⎝ 4x −1 ⎠ 4x +1 =− x ANSWER: −
4x +1 x
24.
PROBLEM: 1 2− y 1 1− 2 4y SOLUTION: 1 2y 1 2 y −1 y 1 − 2− 2⋅ − y y y y y y = = = 2 2 2 1 1 1 − 2 1⋅ 4 y 2 − 1 2 4 y 2 − 1 2 4 y − 2 4y 4y 4y 4y 4y 4y 2 y −1 y = ( 2 y − 1)( 2 y + 1) 4 y2 =
2 y −1 4 y2 ⋅ y ( 2 y − 1)( 2 y + 1)
4y 2 y +1 4y ANSWER: 2 y +1 =
25.
PROBLEM: 1 1 + x y 1 1 − y2 x2 SOLUTION: y 1 1 1 y 1 x + ⋅ + ⋅ + xy x y x y y x = = 2 2 2 1 1 1 1 x y x − ⋅ 2− 2⋅ 2 − 2 y 2 x2 y x x y x2 y 2 x+ y xy = ( x + y )( x − y ) x2 y 2
( x + y) ⋅ = xy
=
xy x− y
ANSWER:
xy x− y
x2 y2 ( x + y )( x − y )
x+ y x xy xy = 2 2 y x − y2 x2 y2 x2 y2
26.
PROBLEM: 1 4 − 2x 3 1 16 − 4x2 9 SOLUTION: 1 4 1 3 4 2x 3 8x − ⋅ − ⋅ − 2x 3 = 2x 3 3 2x = 6x 6x 1 16 1 9 16 4 x 2 9 64 x 2 − ⋅ − ⋅ − 4x2 9 4 x 2 9 9 4 x 2 36 x 2 36 x 2 3 − 8x 6x = 9 − 64 x 2 36 x 2 3 − 8x 6x = ( 3 − 8 x )( 3 + 8 x ) 36 x 2 3 − 8x 36 x 2 = ⋅ 6 x ( 3 − 8 x )( 3 + 8 x ) =
6x 3 + 8x
ANSWER:
6x 3 + 8x
27.
PROBLEM: 2 1 − 2 25 2 x 1 1 − 5 2x SOLUTION: 2 1 2 2x2 1 25 4x2 25 − 2 ⋅ 2− 2⋅ − 2 25 2 x = 25 2 x 2 x 25 = 50 x 50 x 2 1 1 1 2x 1 5 2x 5 − ⋅ − ⋅ − 5 2x 5 2x 2x 5 10 x 2 x 2 4 x − 25 2 = 50 x 2x − 5 10 x ( 2 x − 5)( 2 x + 5) 50 x 2 = 2x − 5 10 x ( 2 x − 5)( 2 x + 5) 10 x = ⋅ 50 x 2 ( 2 x − 5)
=
2x + 5 5x
ANSWER:
2x + 5 5x
28.
PROBLEM: 4 1 − 2 25 4 x 1 1 + 5 4x SOLUTION: 4 1 4 4 x2 1 25 16 x 2 25 − 2 ⋅ 2− 2⋅ − 2 25 4 x = 25 4 x 4 x 25 = 100 x 100 x 2 1 1 1 4x 1 5 4x 5 + ⋅ + ⋅ + 5 4x 5 4x 4x 5 20 x 20 x 2 16 x − 25 2 = 100 x 4x + 5 20 x ( 4 x + 5)( 4 x − 5) 100 x 2 = 4x + 5 20 x ( 4 x + 5)( 4 x − 5) ⋅ 20 x = 100 x 2 4x + 5 4x − 5 = 5x ANSWER:
4x − 5 5x
29.
PROBLEM: 1 1 − y x 2 4− xy SOLUTION: 1 1 1 x 1 y x y x− y − ⋅ − ⋅ − y x y x x y xy xy xy = = = 2 xy 2 4 xy 2 4 xy − 2 4− 4⋅ − − xy xy xy xy xy xy x− y xy = ⋅ xy 4 xy − 2 x− y = 4 xy − 2 ANSWER:
30.
x− y 4 xy − 2
PROBLEM: 1 +2 ab 1 1 + a b SOLUTION: 1 1 ab 1 2ab 1 + 2ab +2 + 2⋅ + ab ab = ab ab = ab = ab 1 1 1 b 1 a b a a+b + ⋅ + ⋅ + a b a b b a ab ab ab 1 + 2ab 1 + 2ab ab ab = = ⋅ a+b ab a+b ab 1 + 2ab = a+b ANSWER:
1 + 2ab a+b
31.
PROBLEM: 1 1 + y x xy SOLUTION: 1 1 1 x 1 y x y x+ y + ⋅ + ⋅ + y x y x x y xy xy xy = = = xy xy xy xy x+ y 1 = ⋅ xy xy x+ y = 2 2 x y ANSWER:
32.
x+ y x2 y2
PROBLEM: 3x 1 1 − 3 x SOLUTION: 3x 3x 3x 3x = = = 1 1 1 x 1 3 x 3 x−3 − ⋅ − ⋅ − 3 x 3 x x 3 3x 3x 3x 3x = 3x ⋅ x−3 2 9x = x −3 ANSWER:
9x2 x−3
33.
PROBLEM: 4 21 1− − 2 x x 2 15 1− − 2 x x SOLUTION: 4 21 ⎛ 4 21 ⎞ 2 1 − − 2 ⎜1 − − 2 ⎟ x 2 x x = ⎝ x x ⎠ = x − 4 x − 21 2 2 15 2 15 1 − − 2 ⎜⎛1 − − 2 ⎟⎞ x 2 x − 2 x − 15 x x ⎝ x x ⎠
34.
=
( x − 7 )( x + 3) ( x − 5)( x + 3)
=
x−7 x −5
ANSWER:
x−7 x−5
PROBLEM: 3 4 1− − 2 x x 16 1− 2 x SOLUTION: 3 4 ⎛ 3 4⎞ 2 1 − − 2 ⎜1 − − 2 ⎟ x 2 x x = ⎝ x x ⎠ = x − 3x − 4 16 x 2 − 16 ⎛ 16 ⎞ 2 1− 2 1 x − ⎜ 2 ⎟ x ⎝ x ⎠ ( x − 4 )( x + 1) = ( x − 4 )( x + 4 ) =
x +1 x+4
ANSWER:
x +1 x+4
35.
PROBLEM: 1 1 3− − 2 2x 2x 2 1 2− + 2 x 2x SOLUTION: 1 1 ⎞ ⎛ 1 1 − 2 ⎟ 2 x2 − 2 ⎜3− 3− 6x2 − x −1 2x 2x ⎠ 2x 2x = ⎝ = 2 2 1 2 1 ⎞ 4x − 4x +1 ⎛ 2− + 2 2 − + 2 ⎟ 2x2 ⎜ x 2x x 2x ⎠ ⎝
ANSWER:
36.
=
( 3x + 1)( 2 x − 1) 2 ( 2 x − 1)
=
3x + 1 2x −1
3x + 1 2x −1
PROBLEM: 1 5 12 − + 2 x x2 1 6 18 − + 2 x x2 SOLUTION: 1 5 12 ⎛ 1 − 5 + 12 ⎞ 2x 2 − + ⎜ 2 ⎟ x 2 − 10 x + 24 2 x x2 = ⎝ 2 x x ⎠ = 1 6 18 ⎛ 1 6 18 ⎞ 2 x 2 − 12 x + 36 − + 2 ⎜ − + 2 ⎟ 2x 2 x x ⎝2 x x ⎠ ( x − 6 )( x − 4 ) = 2 ( x − 6)
=
ANSWER:
x−4 x−6
x−4 x−6
37.
PROBLEM: 1 4 − x 3x 2 8 16 3− + 2 x 3x SOLUTION: ⎛1 4 ⎞ 2 1 4 − 2 ⎜ − 2 ⎟ 3x 3x − 4 1 x 3x = ⎝ x 3x ⎠ = = 2 8 16 8 16 3x − 4 3 − + 2 ⎛⎜ 3 − + 2 ⎞⎟ 3x 2 ( 3x − 4 ) x 3x x 3x ⎠ ⎝ ANSWER:
38.
1 3x − 4
PROBLEM: 3 1 1+ − 10 x 10 x 2 3 1 1 − − 2 5 10 x 5 x SOLUTION: 3 1 ⎞ ⎛ 3 1 − 10 x 2 − 1+ ⎜1 + 2 ⎟ 2 10 x 2 + 3x − 1 10 x 10 x = ⎝ 10 x 10 x ⎠ = 3 1 1 3 1 1 ⎞ 6 x2 − x − 2 − − 2 ⎛⎜ − − 2 ⎟10 x 2 5 10 x 5 x ⎝ 5 10 x 5 x ⎠ ( 2 x + 1)( 5 x − 1) = ( 2 x + 1)( 3x − 2 ) =
ANSWER:
5x − 1 3x − 2
5x − 1 3x − 2
39.
PROBLEM: x −1 4 5 1+ − 2 x x SOLUTION: x − 1) x 2 x 2 ( x − 1) ( x −1 = = 2 4 5 4 5 1 + − 2 ⎛⎜1 + − 2 ⎞⎟ x 2 x + 4 x − 5 x x ⎝ x x ⎠ x 2 ( x − 1) = ( x + 5)( x − 1)
40.
=
x2 x+5
ANSWER:
x2 x+5
PROBLEM: 5 3 2− − 2 2x x 4x + 3 SOLUTION: 5 3 ⎞ 5 3 ⎛ − 2 ⎟ 2x2 − 2 ⎜2− 2− 4 x2 − 5x − 6 2x x ⎠ 2x x = ⎝ = 2 4x + 3 2 x ( 4 x + 3) ( 4 x + 3) 2 x 2
ANSWER:
=
( 4 x + 3)( x − 2 ) 2 x 2 ( 4 x + 3)
=
x−2 2x2
x−2 2 x2
41.
PROBLEM: 1 2 + x −3 x 1 3 − x x −3 SOLUTION:
2x − 6 x 1 2 1 x 2 x −3 + + ⋅ + ⋅ x − 3 x = x − 3 x x x − 3 = x ( x − 3) x ( x − 3) 1 3 1 x −3 3 x 3x x −3 − ⋅ − ⋅ − x x − 3 x x − 3 x − 3 x x ( x − 3) x ( x − 3) 3( x − 2) 3x − 6 x + 2x − 6 x ( x − 3) x ( x − 3) x ( x − 3) = = = 2x + 3 x − 3 − 3x −2 x − 3 − x ( x − 3) x ( x − 3) x ( x − 3) 3 ( x − 2 ) ⎛ x ( x − 3) ⎞ ⋅⎜ − ⎟ x ( x − 3) ⎝ 2 x + 3 ⎠ 3 ( x − 2) =− 2x + 3
=
ANSWER: −
3( x − 2) 2x + 3
42.
PROBLEM: 1 1 + 2 4x − 5 x 1 1 + 2 x 3 x − 10 SOLUTION: 1 1 1 x2 1 4 x − 5 + 2 ⋅ 2+ 2⋅ 4x − 5 x = 4x − 5 x x 4x − 5 1 1 1 3 x − 10 1 x2 + ⋅ + ⋅ x 2 3 x − 10 x 2 3 x − 10 3 x − 10 x 2 x2 4x − 5 + 2 2 x ( 4 x − 5) x ( 4 x − 5) = x2 3x − 10 + x 2 ( 3 x − 10 ) x 2 ( 3 x − 10 )
( x + 5)( x − 1) x2 + 4 x − 5 2 x ( 4 x − 5) x 2 ( 4 x − 5) = 2 = x + 3 x − 10 ( x + 5 ) ( x − 2 ) x 2 ( 3 x − 10 ) x 2 ( 3 x − 10 ) x + 5 )( x − 1) x 2 ( 3 x − 10 ) ( = 2 ⋅ x ( 4 x − 5 ) ( x + 5 )( x − 2 ) ( x − 1)( 3x − 10 ) = ( x − 2 )( 4 x − 5 ) ANSWER:
( x − 1)( 3x − 10 ) ( x − 2 )( 4 x − 5 )
43.
PROBLEM: 1 4 + x+5 x−2 2 1 − x−2 x+5 SOLUTION: 1 4 1 x−2 4 x+5 + ⋅ + ⋅ x+5 x−2 = x+5 x−2 x−2 x+5 2 1 2 x+5 1 x−2 − − x−2 x+5 x−2 x+5 x+5 x−2 4 x + 20 x−2 + ( x + 5)( x − 2 ) ( x − 2 )( x + 5) = 2 x + 10 x−2 − ( x + 5)( x − 2 ) ( x + 5)( x − 2 ) x − 2 + 4 x + 20 ( x + 5)( x − 2 ) = 2 x + 10 − x + 2 ( x + 5)( x − 2 ) 5 x + 18 ( x + 5)( x − 2 ) = x + 12 ( x + 5)( x − 2 ) =
( x + 5)( x − 2 ) 5 x + 18 ⋅ x + 12 ( x + 5)( x − 2 )
=
5 x + 18 x + 12
ANSWER:
5 x + 18 x + 12
44.
PROBLEM: 3 2 − x −1 x + 3 2 1 + x +3 x −3 SOLUTION:
3x + 9 2x − 2 3x + 9 − 2 x + 2 3 2 3 x+3 2 x −1 − − ⋅ − ⋅ x − 1 x + 3 = x − 1 x + 3 x + 3 x − 1 = ( x − 1)( x + 3) ( x − 1)( x + 3) = ( x − 1)( x + 3) 2 1 2 x −3 1 x+3 2x − 6 2x − 6 + x + 3 x+3 + + + x +3 x −3 x +3 x −3 x −3 x +3 ( x + 3)( x − 3) ( x + 3)( x − 3) ( x − 3)( x + 3) x + 11 ( x − 1)( x + 3) = 3x − 3 ( x − 4 )( x + 3) x + 11 ( x − 1)( x + 3) = 3 ( x − 1) ( x − 3)( x + 3) =
( x − 3)( x + 3) x + 11 ⋅ ( x − 1)( x + 3) 3 ( x − 1)
=
( x + 11)( x − 3) 2 3 ( x − 1)
ANSWER:
( x + 11)( x − 3) 2 3 ( x − 1)
45.
PROBLEM: 2 x − x +1 x + 3 x 1 + 3x + 4 x + 1 SOLUTION:
2x + 2 x 2 + 3x 2 2 x +1 x x x+3 − − ⋅ − ⋅ x + 1 x + 3 = x + 1 x + 3 x + 3 x + 1 = ( x + 1)( x + 3) ( x + 1)( x + 3) 1 x x x + 1 1 3x + 4 3x + 4 x2 + x + ⋅ + ⋅ + 3 x + 4 x + 1 3 x + 4 x + 1 x + 1 3 x + 4 ( 3 x + 4 )( x + 1) ( 3x + 4 )( x + 1) x 2 + 3x − 2 x − 2 ( x + 1)( x + 3) = 2 x + x + 3x + 4 ( 3x + 4 )( x + 1) x2 + x − 2 ( x + 1)( x + 3) = x2 + 4x + 4 ( 3x + 4 )( x + 1)
( x + 2 )( x − 1) ( x + 1)( x + 3) = 2 ( x + 2) ( 3x + 4 )( x + 1) ( x + 2 )( x − 1) ⋅ ( 3x + 4 )( x + 1) = ( x + 1)( x + 3) ( x + 2 )2 ( x − 1)( 3x + 4 ) = ( x + 3)( x + 2 ) ANSWER:
( x − 1)( 3x + 4 ) ( x + 3)( x + 2 )
46.
PROBLEM: x 2 + x − 9 x +1 x 1 − 7x − 9 x +1 SOLUTION:
2 x − 18 x2 + x 2 2 x −9 x x x +1 + + ⋅ + ⋅ x − 9 x + 1 = x − 9 x + 1 x + 1 x − 9 = ( x − 9 )( x + 1) ( x − 9 )( x + 1) 1 1 7x − 9 x x x +1 7x − 9 x2 + x − ⋅ − ⋅ − 7 x − 9 x + 1 7 x − 9 x + 1 x + 1 7 x − 9 ( 7 x − 9 )( x + 1) ( 7 x − 9 )( x + 1) x 2 + x + 2 x − 18 ( x − 9 )( x + 1) = 2 x + x − 7x + 9 ( 7 x − 9 )( x + 1) x 2 + 3x − 18 ( x − 9 ) ( x + 1) = x2 − 6x + 9 ( 7 x − 9 )( x + 1)
( x + 6 )( x − 3) ( x − 9 )( x + 1) = 2 ( x − 3) ( 7 x − 9 )( x + 1) ( x + 6 )( x − 3) ⋅ ( 7 x − 9 )( x + 1) = 2 ( x − 9 )( x + 1) ( x − 3) ( x + 6 )( 7 x − 9 ) = ( x − 9 )( x − 3) ANSWER:
( x + 6 )( 7 x − 9 ) ( x − 9 )( x − 3)
47.
PROBLEM: 1 x − 3x + 2 x + 2 2 x − x+2 x+2 SOLUTION:
3x + 2 x2 + 2x 1 1 3x + 2 x x x+2 − − ⋅ − ⋅ 3 x + 2 x + 2 = 3 x + 2 x + 2 x + 2 3 x + 2 = ( 3 x + 2 )( x + 2 ) ( 3x + 2 )( x + 2 ) 2 2 2 x x x − − − x+2 x+2 x+2 x+2 x+2 x+2 2 x + 2 x − 3x − 2 ( 3x + 2 )( x + 2 ) = x−2 x+2 2 x −x−2 ( 3x + 2 )( x + 2 ) = x−2 x+2 ( x − 2 )( x + 1) ( 3x + 2 )( x + 2 ) = x−2 x+2 ( x − 2 )( x + 1) ⋅ x + 2 = ( 3x + 2 )( x + 2 ) x − 2 =
ANSWER:
x +1 3x + 2
x +1 3x + 2
48.
PROBLEM: 1 x + x−4 x+2 1 x + 3x + 4 x + 2 SOLUTION:
x2 + 2x x−4 1 1 x−4 x x x+2 + + ⋅ + ⋅ x − 4 x + 2 = x − 4 x + 2 x + 2 x − 4 = ( x − 4 )( x + 2 ) ( x − 4 )( x + 2 ) 1 1 3x + 4 x x x+2 3x + 4 x2 + 2x + + ⋅ + 3 x + 4 x + 2 3 x + 4 x + 2 x + 2 3 x + 4 ( 3x + 4 )( x + 2 ) ( 3x + 4 )( x + 2 ) x2 + 2x + x − 4 ( x − 4 )( x + 2 ) = 2 x + 2 x + 3x + 4 ( 3x + 4 )( x + 2 ) x 2 + 3x − 4 ( x − 4 )( x + 2 ) = x2 + 5x + 4 ( 3x + 4 )( x + 2 )
( x + 4 )( x − 1) ( x − 4 )( x + 2 ) = ( x + 4 )( x + 1) ( 3x + 4 )( x + 2 ) ( x + 4 )( x − 1) ⋅ ( 3x + 4 )( x + 2 ) = ( x − 4 )( x + 2 ) ( x + 4 )( x + 1) ( 3x + 4 )( x − 1) = ( x − 4 )( x + 1) ANSWER:
( 3x + 4 )( x − 1) ( x − 4 )( x + 1)
49.
PROBLEM: a 3 − 8b3 27 a − 2b SOLUTION: 2 2 a 3 − 8b3 ( a − 2b ) ( a + 2ab + 4b ) 27 = a − 2b
27 a − 2b a 2 + 2ab + 4b 2 = 27
=
( a − 2b ) ( a 2 + 2ab + 4b 2 ) 27
⋅
1 a − 2b
a 2 + 2ab + 4b 2 ANSWER: 27 50.
PROBLEM: 27a 3 + b3 ab 3a + b SOLUTION:
(
)
2 2 27a 3 + b3 ( 3a + b ) 9a − 3ab + b ( 3a + b ) 9a 2 − 3ab + b2 1 ab ab = = ⋅ 3a + b 3a + b ab 3a + b 2 2 9a − 3ab + b = ab
ANSWER:
9a 2 − 3ab + b 2 ab
(
)
51.
PROBLEM: 1 1 + b3 a 3 1 1 + b a SOLUTION: a3 b3 a 3 + b3 1 1 1 a 3 1 b3 + ⋅ + ⋅ + b3 a 3 = b3 a 3 a 3 b3 = a 3b3 a 3b3 = a 3b3 1 1 1 a 1 b a b a+b + ⋅ + ⋅ + b a b a a b ab ab ab
( a + b ) ( a 2 − ab + b2 ) a 3b3 a+b ab
=
=
( a + b ) ( a 2 − ab + b2 ) 3 3
ab a − ab + b 2 = a 2b 2 2
a 2 − ab + b 2 ANSWER: a 2b 2
⋅
ab a+b
52.
PROBLEM: 1 1 − b3 a 3 1 1 − a b SOLUTION: 1 1 1 a 3 1 b3 a3 b3 a 3 − b3 − ⋅ − ⋅ − b3 a 3 = b3 a 3 a 3 b3 = a 3b3 a 3b3 = a3b3 b a b−a 1 1 1 b 1 a − ⋅ − ⋅ − a b a b b a ab ab ab 2 2 ( a − b ) a + ab + b
(
ab b−a ab
=
=
)
3 3
( a − b ) ( a 2 + ab + b2 )
=−
3 3
ab
a + ab + b 2 a 2b 2 2
ANSWER: −
a 2 + ab + b 2 a 2b 2
⎛ ab ⎞ ⋅⎜ − ⎟ ⎝ a −b ⎠
53.
PROBLEM: x2 + y 2 +2 xy x2 − y 2 2 xy SOLUTION: x2 + y 2 x2 + y 2 xy x 2 + y 2 2 xy x 2 + y 2 + 2 xy +2 + 2⋅ + xy xy xy xy xy xy = = = 2 2 2 2 2 2 2 x −y x −y x −y x − y2 2 xy 2 xy 2 xy 2 xy
( x + y) =
2
xy
( x + y )( x − y ) 2 xy
( x + y) =
2
xy
=
ANSWER:
2( x + y) x− y 2( x + y) x− y
⋅
2 xy ( x + y )( x − y )
54.
PROBLEM: x 4y +4+ y x x 2y +3+ y x SOLUTION: 4y x x x xy 4 y y x 2 4 xy 4 y 2 +4+ ⋅ + 4⋅ + ⋅ + + y x y x xy x y xy xy xy = = 2 2y x x x xy 2 y y x 3xy 2 y 2 +3+ ⋅ + 3⋅ + ⋅ + + y x y x xy x y xy xy xy
x 2 + 4 xy + 4 y 2 xy = 2 x + 3xy + 2 y 2 xy
( x + 2y) =
2
xy ( x + 2 y )( x + y ) xy
( x + 2y) = xy
=
x + 2y x+ y
ANSWER:
x + 2y x+ y
2
⋅
xy ( x + 2 y )( x + y )
55.
PROBLEM: 1 1+ 1 1+ 2 SOLUTION: 1 1 1 2 2 1+ = 1+ = 1 + = 1 + 1⋅ = 1 + 1 2 1 3 3 3 1+ 1⋅ + 2 2 2 2 2 3 2 5 = 1 + = 1⋅ + = 3 3 3 3 ANSWER:
56.
5 3
PROBLEM: 1 2− 1 1+ 3 SOLUTION: 1 1 1 3 4 3 = 2− = 2 − = 2 − 1⋅ = 2 ⋅ − 2− 1 3 1 4 4 4 4 1+ 1⋅ + 3 3 3 3 8−3 5 = = 4 4 ANSWER:
5 4
57.
PROBLEM: 1 1 1+ 1+ x SOLUTION: 1 1 1 = = 1 1+ x 1 1+ x 1 + + 1+ 1⋅ 1+ x 1+ x 1+ x 1+ x 1+ x 1 = 2+ x 1+ x x +1 = 1⋅ x+2 x +1 = x+2 ANSWER:
58.
x +1 x+2
PROBLEM: x +1 x 1 1− x +1 SOLUTION: x +1 x +1 x +1 x +1 x x x = = = x x +1 x +1−1 x 1 1 − 1− 1⋅ x +1 x +1 x +1 x +1 x +1 x +1 x +1 ⎛ x +1⎞ = ⋅ =⎜ ⎟ x x ⎝ x ⎠
⎛ x +1 ⎞ ANSWER: ⎜ ⎟ ⎝ x ⎠
2
2
59.
PROBLEM: 1 1− x 1 x− x SOLUTION: 1 x 1 x −1 x −1 1− 1⋅ − x = x x = x = x 1 x 1 x 2 − 1 ( x − 1)( x + 1) x− x⋅ − x x x x x x −1 x = ⋅ x ( x − 1)( x + 1)
= 60.
1 x +1
ANSWER:
1 x +1
PROBLEM: 1 −x x x −1 x2 SOLUTION: 1 1 x 1 x 2 1 − x 2 − ( x − 1)( x + 1) −x − x⋅ − x x = x x = x = x = x x −1 x −1 x −1 x −1 x −1 x2 x2 x2 x2 x2 − ( x − 1)( x + 1) x 2 = ⋅ x x −1 = − x ( x + 1) ANSWER: − x ( x + 1)
Part B: Discussion Board Topics
61.
PROBLEM: Choose a problem from this exercise set and clearly work it out on paper, explaining each step in words. Scan your page and post it on the discussion board. ANSWER: (Students answers may vary.) 1 1 + x y 1 1 − y2 x2 1 1 1 y 1 x y + ⋅ + ⋅ + x y x y y x xy = = 2 2 2 1 1 x y x 1 1 − ⋅ 2− 2⋅ 2 − 2 y 2 x2 y x x y x2 y 2 x+ y xy = ( x + y )( x − y ) x2 y 2
62.
=
( x + y) ⋅
=
xy x− y
xy
x x+ y xy xy = 2 2 y x − y2 x2 y 2 x2 y2
x2 y2 ( x + y )( x − y )
PROBLEM: Explain why we need to simplify the numerator and denominator to a single algebraic fraction before multiplying by the reciprocal of the divisor. ANSWER: The goal of the simplification is to obtain a single algebraic fraction divided by another single algebraic fraction.
63.
PROBLEM: Two methods for simplifying complex rational expressions have been presented in this section. Which of the two methods do you feel is more efficient, and why? ANSWER: (Students answers may vary.) Simplifying complex rational expressions using division is more efficient, as it involves less number of steps compared to the simplifying method using LCD.
7.5 Solving Rational Equations Part A: Rational Equations Solve.
1.
PROBLEM: 1 1 1 + = 2 x 8 SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, 2x. 1 1 1 + = 2 x 8 1 ⎛1 1 ⎞ ⎜ + ⎟ 2x = ⋅ 2x 8 ⎝2 x ⎠ x x+2= 4 4x + 8 = x
4x − x + 8 = x − x 3x + 8 = 0 3x + 8 − 8 = 0 − 8 3x = −8 8 x=− 3 8 We can check our answer by substituting − in for x to see if we obtain a true 3 statement. 1 1 1 + = 2 x 8 1 1 1 + = 2 −8 8 3 1 3 1 − = 2 8 8 4−3 1 = 8 8 1 1 = 8 8 8 ANSWER: The solution is − . 3
2.
PROBLEM: 1 1 2 − = 3 x 9 SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, 3x. 1 1 2 − = 3 x 9 2 ⎛1 1⎞ ⎜ − ⎟ 3x = ⋅ 3x 9 ⎝3 x⎠ 2x x −3 = 3 3x − 9 = 2 x
3x − 2 x − 9 = 2 x − x x −9 = 0 x −9+9 = 0+9 x=9 We can check our answer by substituting 9 in for x to see if we obtain a true statement. 1 1 2 − = 3 x 9 1 1 2 − = 3 9 9 3 −1 2 = 9 9 2 2 = 9 9 ANSWER: The solution is 9 .
3.
PROBLEM: 1 2 1 − = 3x 3 x SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, 3x. ⎛ 1 2⎞ ⎛1⎞ ⎜ − ⎟ 3x = ⎜ ⎟ 3x ⎝ 3x 3 ⎠ ⎝ x⎠ 1 − 2x = 3
1 −1 − 2x = 3 −1 −2 x = 2 x = −1 We can check our answer by substituting −1 in for x to see if we obtain a true statement. 1 2 1 − = 3x 3 x 1 2 1 − = 3 ( −1) 3 ( −1)
−3 = −1 3 −1 = −1 ANSWER: The solution is −1 .
4.
PROBLEM: 2 1 3 − = 5 x x 10 SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, 3x. 3 ⎛ 2 1⎞ ⎜ − ⎟ 5x = ⋅ 5x 10 ⎝ 5x x ⎠ 15 x 2−5 = 10 15 x −3 = 10 3 ⋅10 x=− 15 x = −2 We can check our answer by substituting −2 in for x to see if we obtain a true statement. 2 1 3 − = 5 x x 10 2 1 3 − = 5 ( −2 ) ( −2 ) 10
1 1 3 − + = 5 2 10 5−2 3 = 10 10 3 3 = 10 10 ANSWER: The solution is −2 .
5.
PROBLEM: 1 =5 2x +1
1 SOLUTION: We first note that x ≠ − . 2 Multiply both sides by the LCD, 2 x + 1 . 1 ⋅ 2 x + 1 = 5 ( 2 x + 1) 2x +1 1 = 10 x + 5
1 − 5 = 10 x + 5 − 5 10 x = −4 4 x=− 10 2 x=− 5 We can check our answer by substituting − statement. 1 =5 2x +1 1 =5 ⎛ 2⎞ 2 ⎜ − ⎟ +1 ⎝ 5⎠ 1 =5 4 − +1 5 1 =5 −4 + 5 5 5=5 2 ANSWER: The solution is − . 5
2 in for x to see if we obtain a true 5
6.
PROBLEM: 3 +4=5 3x − 1 SOLUTION: 3 +4=5 3x − 1 3 + 4−4 = 5−4 3x − 1 3 =1 3x − 1 Multiply both sides by the LCD, 3 x − 1 . 3 ⋅ 3x − 1 = 1( 3x − 1) 3x − 1 3 = 3x − 1
3 + 1 = 3x − 1 + 1 4 = 3x x=
4 3
We can check our answer by substituting statement. 3 +4=5 3x − 1 3 +4=5 4 3 ⋅ −1 3 3 +4=5 3 1+ 4 = 5
5=5 4 ANSWER: The solution is . 3
4 in for x to see if we obtain a true 3
7.
PROBLEM: 2x − 3 2 = x+5 x+5 SOLUTION: We first note that x ≠ −5 . Multiply both sides by the LCD, x + 5 . 2x − 3 2 ⋅x+5 = ⋅x+5 x+5 x+5 2x − 3 = 2
2x − 3 + 3 = 2 + 3 2x = 5 5 x= 2 We can check our answer by substituting statement. 2x − 3 2 = x+5 x+5 ⎛5⎞ 2⎜ ⎟ − 3 2 ⎝2⎠ = 5 5 +5 +5 2 2 5−3 2 = 15 15 2 2 2 2 = 15 15 2 2 4 4 = 15 15 5 ANSWER: The solution is . 2
5 in for x to see if we obtain a true 2
8.
PROBLEM: 5x x −1 = 2x −1 2x −1
1 . 2 Multiply both sides by the LCD, 2 x − 1 . x −1 5x = 2x −1 2x −1 x −1 5x ⋅ 2x −1 = ⋅ 2x −1 2x −1 2x −1 5x = x − 1 SOLUTION: We first note that x ≠
5x − x = x − x −1 4 x = −1 1 x=− 4 We can check our answer by substituting − statement. 5x x −1 = 2x −1 2x −1 ⎛ 1⎞ ⎛ 1⎞ 5⎜ − ⎟ ⎜ − ⎟ −1 ⎝ 4⎠ = ⎝ 4⎠ ⎛ 1⎞ ⎛ 1⎞ 2 ⎜ − ⎟ −1 2 ⎜ − ⎟ −1 ⎝ 4⎠ ⎝ 4⎠ 5 5 − − 4= 4 6 6 − − 4 4 5 5 = 6 6 1 ANSWER: The solution is − . 4
1 in for x to see if we obtain a true 4
9.
PROBLEM: 5 6 = x −7 x−9 SOLUTION: We first note that x ≠ 7 and x ≠ 9 . Multiply both sides by the LCD, ( x − 7 )( x − 9 ) .
5 6 ⋅ ( x − 7 )( x − 9 ) = ⋅ ( x − 7 )( x − 9 ) x−7 x −9 5 x − 45 = 6 x − 42 5 x − 6 x − 45 = 6 x − 6 x − 42 − x − 45 = −42 − x − 45 + 45 = −42 + 45 −x = 3 x = −3 We can check our answer by substituting −3 in for x to see if we obtain a true statement. 5 6 = x −7 x −9 5 6 = −3 − 7 −3 − 9 5 6 = −10 −14 5 ⋅ ( −14 ) = 6 ⋅ ( −10 )
−60 = −60 ANSWER: The solution is −3 .
10.
PROBLEM: 5 3 = x + 5 x +1 SOLUTION: We first note that x ≠ −5 and x ≠ −1 . Multiply both sides by the LCD, ( x + 5 )( x + 1) .
5 3 ⋅ ( x + 5 )( x + 1) = ⋅ ( x + 5)( x + 1) x+5 x +1 5 x + 5 = 3x + 15 5 x − 3x + 5 = 3x − 3x + 15 2 x + 5 = 15 2 x + 5 − 5 = 15 − 5 2 x = 10 x=5 We can check our answer by substituting 5 in for x to see if we obtain a true statement. 5 3 = x + 5 x +1 5 3 = 5 + 5 5 +1 5 3 = 10 6 1 1 = 2 2 ANSWER: The solution is 5 .
11.
PROBLEM: x 6 − =0 6 x SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, 6x . x 6 ⋅ 6x − ⋅ 6x = 0 x 6 2 x − 36 = 0
( x − 6 )( x + 6 ) = 0 x = 6 or x = − 6 Check x = 6 6 6 − =0 6 6 1 −1 = 0 0=0 ANSWER: The solution is −6 or 6 .
Check x = −6 −6 6 − =0 6 −6 −1 + 1 = 0 0=0
12.
PROBLEM: 5 x + = −2 x 5 SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, 5x . ⎛5 x⎞ ⎜ + ⎟ 5 x = −2 ⋅ 5 x ⎝ x 5⎠ 25 + x 2 = −10 x
x 2 + 10 x + 25 = −10 x + 10 x x 2 + 10 x + 25 = 0
( x + 5)
2
=0
x+5 = 0 x = −5 We can check our answer by substituting −5 in for x to see if we obtain a true statement. 5 x + = −2 x 5 5 −5 + = −2 −5 5 −1 − 1 = −2
−2 = −2 ANSWER: The solution is −5 .
13.
PROBLEM: x 2 = x + 12 x SOLUTION: We first note that x ≠ 0 and x ≠ 12 . Multiply both sides by the LCD, ( x + 12 ) x .
2 ⎛ x ⎞ ⎜ ⎟ ( x + 12 ) x = ⋅ ( x + 12 ) x x ⎝ x + 12 ⎠ x 2 = 2 x + 24 x 2 − 2 x − 24 = 2 x − 2 x + 24 − 24 x 2 − 2 x − 24 = 0
( x − 6 )( x + 4 ) = 0 x = 6 or x = −4 Check x = 6 6 2 = 6 + 12 6 6 2 = 18 6 1 1 = 3 3 ANSWER: The solution is −4 or 6 .
Check x = −4 −4 2 = −4 + 12 −4 4 2 − =− 8 4 1 1 − =− 2 2
14.
PROBLEM: 2x 1 = x+5 6− x SOLUTION: We first note that x ≠ −5 and x ≠ −6 . Multiply both sides by the LCD, ( x + 5 )( 6 − x ) .
⎛ 2x ⎞ ⎛ 1 ⎞ ⎜ ⎟ ( x + 5 )( 6 − x ) = ⎜ ⎟ ( x + 5)( 6 − x ) ⎝ x+5⎠ ⎝ 6− x ⎠ 12 x − 2 x 2 = x + 5 −2 x 2 + 12 x − x − 5 = x + 5 − x − 5 −2 x 2 + 11x − 5 = 0 − ( 2 x − 1)( x − 5) = 0
( 2 x − 1)( x − 5) = 0 x=
1 and x = 5 2 Check x =
1 2
⎛1⎞ 2⎜ ⎟ ⎝2⎠ = 1 1 1 +5 6− 2 2 1 1 = 11 11 2 2 2 2 = 11 11 1 ANSWER: The solution is or 5 . 2
Check x = 5 2 ( 5) 1 = 5+5 6−5 10 1 = 10 1 1 1 = 1 1
15.
PROBLEM: 1 x + =0 x 2x + 1 SOLUTION: We first note that x ≠ 0 and x ≠ −
Multiply both sides by the LCD, x ( 2 x + 1) .
1 . 2
x 1 ⋅ x ( 2 x + 1) + ⋅ x ( 2 x + 1) = 0 x 2x +1 2 x + 1 + x2 = 0 x2 + 2 x + 1 = 0
( x + 1)
2
=0
x +1 = 0 x = −1 We can check our answer by substituting −1 in for x to see if we obtain a true statement. 1 −1 + =0 −1 2 ( −1) + 1
−1 + 1 = 0 0=0 ANSWER: The solution is −1 .
16.
PROBLEM: 9x 4 − =0 3x − 1 x
1 SOLUTION: We first note that x ≠ and x ≠ 0 . 3 Multiply both sides by the LCD, ( 3 x − 1) x .
9x 4 ⋅ ( 3x − 1) x − ⋅ ( 3 x − 1) x = 0 3x − 1 x 2 9 x − 12 x + 4 = 0
( 3x − 2 )
2
=0
3x − 2 = 0 x=
2 3
We can check our answer by substituting statement. ⎛2⎞ 9⎜ ⎟ ⎝3⎠ − 4 =0 2 ⎛2⎞ 3⎜ ⎟ −1 3 ⎝3⎠ 6−6 = 0
0=0 2 ANSWER: The solution is . 3
2 in for x to see if we obtain a true 3
17.
PROBLEM: 2 48 1− = 2 x x SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, x 2 . ⎛ 2 ⎞ 2 48 2 ⎜1 − ⎟ ⋅ x = 2 ⋅ x x ⎝ x⎠
x 2 − 2 x = 48 x 2 − 2 x − 48 = 0
( x − 8)( x + 6 ) = 0 x = 8 or x = −6 Check x = 8 2 48 1− = 2 8 8 6 48 = 8 64 3 3 = 4 4 ANSWER: The solution is 8 or − 6 .
Check x = −6 −4 2 = −4 + 12 −4 4 2 − =− 8 4 1 1 − =− 2 2
18.
PROBLEM: 9 5 2− = 2 x x SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, x 2 . 9⎞ 2 5 2 ⎛ ⎜2 − ⎟⋅ x = 2 ⋅ x x⎠ x ⎝
2 x2 − 9 x = 5 2 x2 − 9 x − 5 = 0
( 2 x + 1)( x − 5) = 0 x=−
1 or x = 5 2 Check x = −
Check x = 5 9 5 2− = 2 5 5 1 1 = 5 5
1 2 5
9 = 1 ⎛ 1 ⎞2 − 2 ⎜⎝ − 2 ⎟⎠ 2 + 18 = 5 ⋅ 4 20 = 20 2−
ANSWER: The solution is −
1 or 5 . 2
19.
PROBLEM: 12 12 1+ = x x−2 SOLUTION: We first note that x ≠ 0 and x ≠ 2 . Multiply both sides by the LCD, x ( x − 2 ) .
⎛ 12 ⎞ ⎛ 12 ⎞ ⎜1 + ⎟ ⋅ x ( x − 2 ) = ⎜ ⎟ ⋅ x ( x − 2) x⎠ ⎝ ⎝ x−2⎠ x 2 − 2 x + 12 x − 24 = 12 x x 2 − 2 x − 24 = 0
( x − 6 )( x + 4 ) = 0 x = 6 or x = −4 Check x = 6 12 12 1+ = 6 6−2 3=3
ANSWER: The solution is 6 or − 4 .
Check x = −4 12 12 1+ = −4 − 4 − 2 2 − = −2
20.
PROBLEM: 3x − 5 1 1− =− x(3x − 4) x SOLUTION: We first note that x ≠ 0 and x ≠
Multiply both sides by the LCD, x ( 3 x − 4 ) .
1⋅ x(3x − 4) −
4 . 3
3x − 5 1 ⋅ x(3x − 4) = − ⋅ x(3x − 4) x(3x − 4) x
3x 2 − 4 x − 3 x + 5 = −3x + 4 3x 2 − 4 x + 1 = 0
( 3x − 1)( x − 1) = 0 x=
1 or x = 1 3 Check x =
1 3
⎛1⎞ 3⎜ ⎟ − 5 1 ⎝3⎠ 1− =− ⎛1⎞ ⎛ 1 ⎞⎛ ⎛ 1 ⎞ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 3⎜ ⎟ − 4 ⎟ ⎝3⎠ ⎝ 3 ⎠⎝ ⎝ 3 ⎠ ⎠ −4 1− = −3 −1 −3 = −3 ANSWER: The solution is
1 or 1 . 3
Check x = 1 3 (1) − 5 1 1− =− (1) (3 (1) − 4) 1 −2 = −1 −1 − 1 = −1
1−
21.
PROBLEM: x 14 = 2 x+3 SOLUTION: We first note that x ≠ −3 . Multiply both sides by the LCD, 2 ( x + 3) .
14 x ⋅ 2 ( x + 3) 2 ( x + 3) ) = ( x+3 2 x 2 + 3x = 28 x 2 + 3x − 28 = 0
( x + 7 )( x − 4 ) = 0 x = −7 or x = 4 Check x = −7 7 14 − = 2 −7 + 3 7 14 − =− 2 4 7 7 − =− 2 2 ANSWER: The solution is −7 or 4 .
Check x = 4 4 14 = 2 4+3 14 2= 7 2=2
22.
PROBLEM: 3x x + 1 = 2 3− x SOLUTION: We first note that x ≠ 3 . Multiply both sides by the LCD, 2 ( 3 − x ) .
x +1 3x ⋅ 2 (3 − x ) = ⋅ 2 (3 − x ) 2 3− x 9 x − 3x 2 = 2 x + 2 3x 2 − 7 x + 2 = 0
( 3x − 1)( x − 2 ) = 0 x=
1 or x = 2 3 Check x =
Check x = 2 3⋅ 2 2 +1 = 2 3− 2 3=3
1 3
⎛1⎞ 3⎜ ⎟ 1 +1 ⎝3⎠ = 3 1 2 3− 3 4 1 3 = 2 8 3 1 1 = 2 2
ANSWER: The solution is
1 or 2 . 3
23.
PROBLEM: −3x + 3 6= x −1 SOLUTION: We first note that x ≠ 1 . Multiply both sides by the LCD, x − 1 . −3x + 3 6 ( x − 1) = ⋅ x −1 x −1 6 x − 6 = −3x + 3
9x = 9 x =1 x ≠ 1. ANSWER: Hence, the solution is ∅.
24.
PROBLEM: 12 6(4 − x) = 2+ x−2 x−2 SOLUTION: We first note that x ≠ 2 . Multiply both sides by the LCD, x − 2 . 12 6(4 − x) ⎞ ⎛ ⋅x −2 = ⎜2+ ⎟ ( x − 2) x−2 x−2 ⎠ ⎝ 12 = 2 x − 4 + 24 − 6 x
4x = 8 x=2 x ≠ 2. ANSWER: Hence, the solution is ∅.
25.
PROBLEM: 2x 3( x − 1) = 2+ x −3 x −3 SOLUTION: We first note that x ≠ 3 . Multiply both sides by the LCD, x − 3 . 2x ⎞ 3( x − 1) ⎛ ⋅x −3 ⎜2+ ⎟ ( x − 3) = x −3⎠ x −3 ⎝ 2 x − 6 + 2 x = 3x − 3
x=3 x ≠ 3. ANSWER: Hence, the solution is ∅.
26.
PROBLEM: x 1 x + = x − 1 6 x − 1 ( x − 1)( 6 x − 1)
1 . 6 Multiply both sides by the LCD, ( x − 1)( 6 x − 1) . SOLUTION: We first note that x ≠ 1 and x ≠
1 ⎞ x ⎛ x + ⋅ ( x − 1)( 6 x − 1) ⎜ ⎟ ( x − 1)( 6 x − 1) = ( x − 1)( 6 x − 1) ⎝ x −1 6x −1 ⎠ 6 x2 − x + x − 1 = x 6 x2 − x − 1 = 0
( 3x + 1)( 2 x − 2 ) = 0 x=−
1 1 or x = 3 2 Check x = −
1 3
Check x =
1 2
1 1 1 ⎛ 1⎞ ⎛ 1⎞ + = − ⎟ ⎜− ⎟ ⎜ 1 1 ⎛1⎞ ⎞ 1 ⎛ 1 ⎝ 3⎠ + ⎝ 3⎠ 1− 6 ⎜ ⎟ − 1 ⎛⎜ 1 − ⎞⎟ ⎜ 6 ⎛⎜ ⎞⎟ − 1⎟ = 2 ⎝2⎠ ⎛ 1⎞ ⎛ 1⎞ ⎛⎛ 1 ⎞ ⎞⎛ ⎛ 1 ⎞ ⎞ ⎝ 2 ⎠⎝ ⎝ 2 ⎠ ⎠ ⎜ − ⎟ − 1 6 ⋅ ⎜ − ⎟ − 1 ⎜ ⎜ − ⎟ − 1⎟ ⎜ 6 ⋅ ⎜ − ⎟ − 1⎟ ⎝ 3⎠ ⎝ 3⎠ 1 1 ⎝⎝ 3 ⎠ ⎠⎝ ⎝ 3 ⎠ ⎠ 2+ = 2 1 1 1 − − 2 3 −1 = 3 4 3 4 2=2 − − ⋅ ( −3) 3 3 1 1 1 − =− 4 3 3⋅ 4 1 1 − =− 12 12
ANSWER: The solution is −
1 1 or . 3 2
27.
PROBLEM: 12 1 2 = − 2 x − 81 x + 9 x − 9 SOLUTION: 12 1 2 = − 2 x − 81 x + 9 x − 9 12 1 2 = − ( x + 9 )( x − 9 ) x + 9 x − 9
We first note that x ≠ ±9 . Multiply both sides by the LCD, ( x + 9 )( x − 9 ) .
12 2 ⎞ ⎛ 1 ⋅ ( x + 9 )( x − 9 ) = ⎜ − ⎟ ( x + 9 )( x − 9 ) ( x + 9 )( x − 9 ) ⎝ x +9 x −9 ⎠ 12 = x − 9 − 2 x − 18 x = −39 We can check our answer by substituting −39 in for x to see if we obtain a true statement. 12 1 2 = − 2 ( −39 ) − 81 −39 + 9 −39 − 9 12 1 2 =− + 1440 30 48 12 −8 + 10 = 1440 240 12 2 = 1440 240 1 1 = 120 120 ANSWER: The solution is −39 .
28.
PROBLEM: 14 2 3 = − 2 x − 49 x − 7 x + 7 SOLUTION: 14 2 3 = − 2 x − 49 x − 7 x + 7 14 2 3 = − ( x − 7 )( x + 7 ) x − 7 x + 7
1 . 6 Multiply both sides by the LCD, ( x − 7 )( x + 7 ) .
We first note that x ≠ ±7 and x ≠
14 3 ⎞ ⎛ 2 ⋅ ( x − 7 )( x + 7 ) = ⎜ − ⎟ ( x − 7 )( x + 7 ) ( x − 7 )( x + 7 ) ⎝ x−7 x+7⎠ 14 = 2 x + 14 − 3x + 21 x = 21 We can check our answer by substituting 21 in for x to see if we obtain a true statement. 14 2 3 = − 2 21 − 49 21 − 7 21 + 7 14 2 3 = − 392 14 28 1 1 = 28 28 ANSWER: The solution is 21.
29.
PROBLEM: 6x 4 3x + = 2 x +3 x −3 x −9 SOLUTION: 6x 4 3x + = 2 x +3 x −3 x −9 6x 4 3x + = x + 3 x − 3 ( x − 3)( x + 3)
We first note that x ≠ ±3 . Multiply both sides by the LCD, ( x + 3)( x − 3) .
4 ⎞ 3x ⎛ 6x + ⋅ ( x − 3)( x + 3) ⎜ ⎟ ( x − 3)( x + 3) = ( x − 3)( x + 3) ⎝ x +3 x −3⎠ 6 x 2 − 18 x + 4 x + 12 = 3 x 6 x 2 − 17 x + 12 = 0
( 3x − 4 )( 2 x − 3) = 0 x=
4 3 or x = 3 2 Check x =
4 3
Check x =
3 2
6⋅
4 4 3⋅ 3 + 4 = 3 2 4 4 +3 −3 ⎛ 4 ⎞ −9 ⎜ ⎟ 3 3 ⎝3⎠
3 3 3⋅ 2 + 4 = 2 2 3 3 +3 −3 ⎛ 3 ⎞ −9 ⎜ ⎟ 2 2 ⎝2⎠
24 12 4⋅9 − =− 13 5 65 120 − 12 ⋅13 4⋅9 =− 65 65 36 36 − =− 65 65
9 8 2− = 2 3 − 27 4 2 2 − =− 3 3
ANSWER: The solution is
4 3 or . 3 2
6⋅
30.
PROBLEM: 3x 17 48 − =− 2 x+2 x−2 x −4 SOLUTION: 3x 17 48 − =− 2 x+2 x−2 x −4 3x 17 48 − =− x+2 x−2 ( x + 2 )( x − 2 )
We first note that x ≠ ±2 . Multiply both sides by the LCD, ( x + 2 )( x − 2 ) .
17 ⎞ 48 ⎛ 3x − ⋅ ( x + 2 )( x − 2 ) ⎜ ⎟ ( x + 2 )( x − 2 ) = − ( x + 2 )( x − 2 ) ⎝ x+2 x−2⎠ 3x 2 − 6 x − 17 x − 34 = −48 3x 2 − 23x + 14 = 0
Check x = 7 21 17 −48 − = 9 5 45 48 48 =− =− 45 45
Check x =
3 2 − 17 = −48 2 3 3 +2 −3 ⎛ 3 ⎞ −4 ⎜ ⎟ 2 2 ⎝2⎠ 3
=
ANSWER: The solution is 7 or
3 2
3 . 2
192 192 = 7 7
31.
PROBLEM: x −1 + 3 = 0 SOLUTION: x −1 + 3 = 0
1 +3= 0 x We first note that x ≠ 0 . Multiply both sides by the LCD, x . ⎛1 ⎞ ⎜ + 3⎟ x = 0 ⋅ x ⎝x ⎠ 1 + 3x = 0
x=−
1 3
We can check our answer by substituting − statement. 1 +3= 0 1 − 3 −3 + 3 = 0
0=0 1 ANSWER: The solution is − . 3
1 in for x to see if we obtain a true 3
32.
PROBLEM: 4 − y −1 = 0 SOLUTION: 4 − y −1 = 0
1 =0 y We first note that y ≠ 0 . Multiply both sides by the LCD, y . 4−
⎛ 1⎞ ⎜4 − ⎟ y = 0⋅ y y⎠ ⎝ 4 y −1 = 0 1 y= 4 We can check our answer by substituting statement. 1 4− = 0 1 4 4−4 = 0
0=0 1 ANSWER: The solution is . 4
1 in for x to see if we obtain a true 4
33.
PROBLEM: y −2 − 4 = 0 SOLUTION: y −2 − 4 = 0
1 −4 =0 y2 We first note that y ≠ 0 . Multiply both sides by the LCD, y 2 .
⎛ 1 ⎞ 2 2 ⎜ 2 − 4⎟ y = 0⋅ y y ⎝ ⎠ 2 1− 4 y = 0
(1 − 2 y )(1 + 2 y ) = 0 y=
1 1 or y = − 2 2 Check y = 1 2
1 2
−4 = 0
⎛1⎞ ⎜ ⎟ ⎝2⎠ 4−4 = 0 0=0
1 ANSWER: The solution is ± . 2
Check y = − 1
1 2
−4=0 2 ⎛ 1⎞ ⎜− ⎟ ⎝ 2⎠ 4−4 = 0 0=0
34.
PROBLEM: 9 x −2 − 1 = 0 SOLUTION: 9 x −2 − 1 = 0
9 −1 = 0 x2 We first note that x ≠ 0 . Multiply both sides by the LCD, x 2 . ⎛ 9 ⎞ 2 2 ⎜ 2 − 1⎟ x = 0 ⋅ x ⎝x ⎠ 2 9− x = 0
( 3 − x )( 3 + x ) = 0 x = 3 or x = −3 Check x = 3 9 −1 = 0 32 1 −1 = 0 0=0 ANSWER: The solution is ±3 .
Check x = −3 9 −1 = 0 2 ( −3 ) 1−1 = 0 0=0
35.
PROBLEM: −1 3 ( x − 1) + 5 = 0 SOLUTION: −1 3 ( x − 1) + 5 = 0 3 +5 = 0 x −1 We first note that x ≠ 1 . Multiply both sides by the LCD, x − 1 . ⎛ 3 ⎞ + 5 ⎟ ( x − 1) = 0 ⋅ ( x − 1) ⎜ ⎝ x −1 ⎠ 3 + 5x − 5 = 0
5x − 2 = 0 5x = 2 2 x= 5 We can check our answer by substituting statement. 3 +5 = 0 2 −1 5 3 +5 = 0 3 − 5 −5 + 5 = 0
0=0 2 ANSWER: The solution is . 5
2 in for x to see if we obtain a true 5
36.
PROBLEM: −1 5 − 2 ( 3x + 1) = 0 SOLUTION: −1 5 − 2 ( 3x + 1) = 0 5−
2 =0 3x + 1
1 We first note that x ≠ − . 3 Multiply both sides by the LCD, 3 x + 1 . 2 ⎞ ⎛ ⎜5 − ⎟ ( 3x + 1) = 0 ⋅ ( 3x + 1) ⎝ 3x + 1 ⎠ 15 x + 5 − 2 = 0
x=−
1 5
We can check our answer by substituting − statement. 2 5− =0 ⎛ 1⎞ 3⎜ − ⎟ +1 ⎝ 5⎠ 2 5− = 0 2 5 5−5 = 0
0=0 1 ANSWER: The solution is − . 5
1 in for x to see if we obtain a true 5
37.
PROBLEM: 2 2 3+ = x −3 x −3 SOLUTION: We first note that x ≠ 3 . Multiply both sides by the LCD, x − 3 . 2 ⎞ 2 ⎛ ⋅ ( x − 3) ⎜3+ ⎟ ( x − 3) = x −3⎠ x −3 ⎝ 3x − 9 + 2 = 2
3x = 9 x=3 x ≠ 3. ANSWER: Hence, the solution is ∅.
38.
PROBLEM: 1 1 = x x +1 SOLUTION: We first note that x ≠ 0 and x ≠ 1 . Multiply both sides by the LCD, x ( x + 1) .
1 1 ⋅ x ( x + 1) = ⋅ x ( x + 1) x x +1 x +1 = x 1≠ 0 ANSWER: The solution is ∅.
39.
PROBLEM: x x +1 = x +1 x SOLUTION: We first note that x ≠ 0 and x ≠ −1 . Multiply both sides by the LCD, x ( x + 1) .
⎛ x ⎞ ⎛ x +1 ⎞ ⎜ ⎟ ⋅ x ( x + 1) = ⎜ ⎟ ⋅ x ( x + 1) ⎝ x +1 ⎠ ⎝ x ⎠ x2 = x2 + 2 x + 1 2 x = −1 x=−
1 2
We can check our answer by substituting − statement. 1 1 − − +1 2 = 2 1 1 − +1 − 2 2 1 1 − 2= 2 1 1 − 2 2 −1 = −1 1 ANSWER: The solution is- . 2
1 in for x to see if we obtain a true 2
40.
PROBLEM: 3x − 1 x = 3x x+3 SOLUTION: We first note that x ≠ 0 and x ≠ −3 . Multiply both sides by the LCD, x ( x + 1) .
x ⎛ 3x − 1 ⎞ ⋅ 3 x ( x + 3) ⎜ ⎟ ⋅ 3 x ( x + 3) = x+3 ⎝ 3x ⎠ 3x 2 + 8 x − 3 = 3x 2 8x − 3 = 0 x=
3 8
We can check our answer by substituting statement. ⎛3⎞ 3 3⎜ ⎟ − 1 ⎝8⎠ = 8 3 ⎛ 3⎞ +3 3⎜ ⎟ 8 ⎝8⎠ 1 3 8= 8 9 27 8 8 1 1 = 9 9 3 ANSWER: The solution is . 8
3 in for x to see if we obtain a true 8
41.
PROBLEM: 4 x − 7 3x − 2 = x−5 x −5 SOLUTION: We first note that x ≠ 5 . Multiply both sides by the LCD, x − 5 . 4x − 7 3x − 2 ⋅x−5 = ⋅x−5 x−5 x −5 4 x − 7 = 3x − 2
x=5 x ≠ 5. ANSWER: Hence, the solution is ∅.
42.
PROBLEM: x 1 = 2 x −9 x −3 SOLUTION: x 1 = 2 x −9 x−3 x 1 = ( x − 3)( x + 3) x − 3
We first note that x ≠ ±3 . Multiply both sides by the LCD, ( x − 3)( x + 3) .
x
( x − 3)( x + 3)
⋅ ( x − 3)( x + 3) =
x = x+3 0≠3 ANSWER: The solution is ∅.
1 ⋅ ( x − 3)( x + 3) x−3
43.
PROBLEM: 3x + 4 2 − =1 x −8 8− x SOLUTION: 3x + 4 2 − =1 x −8 8− x 3x + 4 2 + =1 x −8 x −8 We first note that x ≠ 8 . Multiply both sides by the LCD, x − 8 . ⎛ 3x + 4 ⎞ ⎛ 2 ⎞ ⎜ ⎟ ( x − 8) + ⎜ ⎟ ( x − 8) = 1 ⋅ ( x − 8) ⎝ x −8 ⎠ ⎝ x −8⎠ 3x + 4 + 2 = x − 8
2 x = −14 x = −7 We can check our answer by substituting −7 in for x to see if we obtain a true statement. 3 ( −7 ) + 4 2 + =1 −7 − 8 −7 − 8 17 2 − =1 15 15 1=1 ANSWER: The solution is −7 .
44.
PROBLEM: 1 6 = x x ( x + 3) SOLUTION: We first note that x ≠ 0 and x ≠ 3 . Multiply both sides by the LCD, x ( x − 3) .
1 6 ⋅ x ( x + 3) = ⋅ x ( x + 3) x x ( x + 3)
x+3= 6 x=3 We can check our answer by substituting 3 in for x to see if we obtain a true statement. 1 6 = 3 3 ( 3 + 3) 1 6 = 3 18 1 1 = 3 3 ANSWER: The solution is 3 .
45.
PROBLEM: 3 1 13 = + x x + 1 x ( x + 1) SOLUTION: We first note that x ≠ 0 and x ≠ −1 . Multiply both sides by the LCD, x ( x + 1) .
⎛ 1 3 13 ⎞ x x + 1) ⋅ x ( x + 1) = ⎜ + ⎜ x + 1 x ( x + 1) ⎟⎟ ( x ⎝ ⎠ 3x + 3 = x + 13 2 x = 10 x=5 We can check our answer by substituting 5 in for x to see if we obtain a true statement. 3 1 13 = + 5 5 + 1 5 ( 5 + 1) 3 1 13 = + 5 6 30 3 3 = 5 5 ANSWER: The solution is 5 .
46.
PROBLEM: x 3 9x − = x − 1 4 x − 1 ( 4 x − 1)( x − 1)
1 . 4 Multiply both sides by the LCD, ( 4 x − 1)( x − 1) .
SOLUTION: We first note that x ≠ 1 and x ≠
⎛ ⎞ 3 ⎞ 9x ⎛ x x x 4 1 1 4 x − 1)( x − 1) − − − = ( )( ) ⎜ ⎜ ⎟ ⎜ ( 4 x − 1)( x − 1) ⎟⎟ ( ⎝ x − 1 4x − 1 ⎠ ⎝ ⎠ 4 x 2 − x − 3x + 3 = 9 x 4 x 2 − 13x + 3 = 0
( 4 x − 1)( x − 3) = 0 1 or x = 3 4 1 x≠ . 4
x=
ANSWER: Hence, the only solution is 3. We can check our answer by substituting 3 in for x to see if we obtain a true statement. 9 ( 3) 3 3 − = 3 − 1 4 ( 3) − 1 ( 4 ( 3) − 1) ( 3 − 1)
3 3 27 − = 2 11 11 ⋅ 2 27 27 = 22 22
47.
PROBLEM: 1 x 2 + = 2 x − 4 x − 2 x − 6x + 8 SOLUTION: 1 x 2 + = 2 x − 4 x − 2 x − 6x + 8 1 x 2 + = x − 4 x − 2 ( x − 4 )( x − 2 )
We first note that x ≠ 4 and x ≠ 2 . Multiply both sides by the LCD, ( x − 4 )( x − 2 ) .
⎛ ⎞ x ⎞ 2 ⎛ 1 + ⎟⎟ ( x − 4 )( x − 2 ) ⎜ ⎟ ( x − 4 )( x − 2 ) = ⎜⎜ ⎝ x−4 x−2⎠ ⎝ ( x − 4 )( x − 2 ) ⎠ x − 2 + x2 − 4 x = 2 x 2 − 3x − 4 = 0
( x − 4 )( x + 1) = 0 x = 4 or x = −1 x ≠ 4. ANSWER: Hence, the only solution is −1 . We can check our answer by substituting −1 in for x to see if we obtain a true statement. 1 1 2 + = −1 − 4 2 + 1 ( −1 − 4 )( −1 − 2 )
1 1 2 − = 3 5 15 2 2 = 15 15
48.
PROBLEM: x x −1 5 + 2 = x − 5 x − 11x + 30 x − 6 SOLUTION: x x −1 5 + 2 = x − 5 x − 11x + 30 x − 6 x x −1 5 + = x − 5 ( x − 6 )( x − 5) x − 6 We first note that x ≠ 5 and x ≠ 6 . Multiply both sides by the LCD, ( x − 6 )( x − 5 ) .
⎛ x ⎞ x −1 ⎛ 5 ⎞ + ⎜⎜ ⎟⎟ ( x − 6 )( x − 5) = ⎜ ⎟ ( x − 6 )( x − 5) x x x x 5 6 5 6 − − − − ( )( ) ⎝ ⎠ ⎝ ⎠ x 2 − 6 x + x − 1 = 5 x − 25 x 2 − 10 x + 24 = 0
( x − 6 )( x − 4 ) = 0 x = 6 or x = 4 x ≠ 6. ANSWER: Hence, the only solution is 4. We can check our answer by substituting 3 in for x to see if we obtain a true statement. 4 4 −1 5 + = 4 − 5 ( 4 − 6 )( 4 − 5) 4 − 6
3 5 =− 2 2 5 5 − =− 2 2 −4 +
49.
PROBLEM: x 6 5 − 2 =− 5x − 1 x + 1 5x + 4 x − 1 SOLUTION: x 6 5 − 2 =− x + 1 5x + 4x − 1 5x − 1 x 6 5 − =− 5x − 1 x + 1 ( x + 1)( 5 x − 1)
1 and x ≠ −1 . 5 Multiply both sides by the LCD, ( x + 1)( 5 x − 1) .
We first note that x ≠
⎛ x ⎞ 6 5 ⎞ ⎛ − ⎜⎜ ⎟⎟ ( x + 1)( 5 x − 1) = ⎜ − ⎟ ( x + 1)( 5 x − 1) 1 5 1 x x x x + + 1 5 − 1 − ( )( ) ⎝ ⎠ ⎝ ⎠ 2 5 x − x − 6 = −5 x − 5 5x2 + 4 x − 1 = 0
( 5x − 1)( x + 1) = 0 1 or x = −1 5 1 x ≠ and x ≠ −1 . 5
x=
ANSWER: Hence, the solution is ∅.
50.
PROBLEM: −8 2( x + 2) 1 + 2 = 2 x − 4 x − 12 x + 4 x − 60 x + 2 SOLUTION: −8 2( x + 2) 1 + 2 = 2 x − 4 x − 12 x + 4 x − 60 x + 2 −8 2( x + 2) 1 + = ( x − 6 )( x + 2 ) ( x + 10 )( x − 6 ) x + 2
We first note that x ≠ 6, x ≠ −2, and x ≠ −10 . Multiply both sides by the LCD, ( x + 10 )( x − 6 )( x + 2 ) .
⎛ ⎞ −8 2( x + 2) 1 + ⋅ ( x + 10 )( x − 6 )( x + 2 ) ⎜⎜ ⎟⎟ ( x + 10 )( x − 6 )( x + 2 ) = x+2 ⎝ ( x − 6 )( x + 2 ) ( x + 10 )( x − 6 ) ⎠ −8 x − 80 + 2 x 2 + 8 x + 8 = x 2 + 4 x − 60 x 2 − 4 x − 12 = 0
( x − 6 )( x + 2 ) = 0 x = 6 or x = −2 x ≠ 6 and x ≠ −2 .
ANSWER: Hence, the solution is∅.
51.
PROBLEM: x 20 4 − 2 =− x+2 x − x−6 x−3 SOLUTION: x 20 4 − 2 =− x+2 x − x−6 x−3 x 20 4 − =− x + 2 ( x + 2 )( x − 3) x −3 We first note that x ≠ −2 and x ≠ 3 . Multiply both sides by the LCD, ( x + 2 )( x − 3) .
⎛ x ⎞ 20 4 ⎞ ⎛ − ⎜⎜ ⎟⎟ ( x + 2 )( x − 3) = ⎜ − ⎟ ( x + 2 )( x − 3) ⎝ x −3⎠ ⎝ x + 2 ( x + 2 )( x − 3) ⎠ x 2 − 3x − 20 = −4 x − 8 x 2 + x − 12 = 0
( x + 4 )( x − 3) = 0 x = −4 or x = 3 x ≠ 3. ANSWER: Hence, the only solution is –4. We can check our answer by substituting −4 in for x to see if we obtain a true statement. −4 20 4 − =− −4 + 2 ( −4 + 2 )( −4 − 3) −4 − 3
10 4 = 7 7 4 4 = 7 7
2−
52.
PROBLEM: x + 7 x −1 4 + = 2 x −1 x + 1 x −1 SOLUTION: x + 7 x −1 4 + = 2 x −1 x +1 x −1 x + 7 x −1 4 + = x − 1 x + 1 ( x − 1)( x + 1) We first note that x ≠ ±1 . Multiply both sides by the LCD, ( x + 1)( x − 1) .
⎛ ⎞ 4 ⎛ x + 7 x −1⎞ + ⎟⎟ ( x + 1)( x − 1) ⎜ ⎟ ( x + 1)( x − 1) = ⎜⎜ − + 1 1 x x ( )( ) ⎝ x −1 x +1⎠ ⎝ ⎠ x2 + 8x + 7 + x2 − 2 x + 1 = 4 2 x2 + 6 x + 4 = 0 2 ( x + 2 )( x + 1) = 0 x = −2 or x = −1 x ≠ ±1 . ANSWER: Hence, the only solution is –2.
We can check our answer by substituting −2 in for x to see if we obtain a true statement. −2 + 7 −2 − 1 4 + = −2 − 1 −2 + 1 ( −2 − 1)( −2 + 1)
5 4 − +3= 3 3 4 4 = 3 3
53.
PROBLEM: x − 1 x − 3 −x + 5 + = x − 3 x −1 x−3 SOLUTION: We first note that x ≠ 3 and x ≠ 1 . Multiply both sides by the LCD, ( x − 3)( x − 1) .
⎛ x −1 x − 3 ⎞ ⎛ −x + 5 ⎞ + ⎜ ⎟ ( x − 3)( x − 1) = ⎜ ⎟ ( x − 3)( x − 1) ⎝ x − 3 x −1 ⎠ ⎝ x−3 ⎠ x2 − 2 x + 1 + x2 − 6 x + 9 = − x2 + 6 x − 5 3x 2 − 14 x + 15 = 0
( x − 3)( 3x − 5) = 0 x = 3 or x = x ≠ 3.
5 3
5 ANSWER: Hence, the only solution is . 3 We can check our answer by substituting statement. 5 5 5 −1 −3 − +5 3 +3 = 3 5 5 5 −3 −1 −3 3 3 3 2 4 10 − 3 + 3= 3 4 2 4 − − 3 3 3 1 5 − −2= − 2 2 5 5 − =− 2 2
5 in for x to see if we obtain a true 3
54.
PROBLEM: x−2 x−5 8− x − = x−5 x−2 x−5 SOLUTION: We first note that x ≠ 5 and x ≠ 2 . Multiply both sides by the LCD, ( x − 5 )( x − 2 ) .
⎛ x−2 x −5⎞ ⎛8− x⎞ − ⎜ ⎟ ( x − 5)( x − 2 ) = ⎜ ⎟ ( x − 5)( x − 2 ) ⎝ x−5 x−2⎠ ⎝ x−5⎠ x 2 − 4 x + 4 − x 2 + 10 x − 25 = − x 2 + 10 x − 16
x2 − 4 x − 5 = 0
( x − 5)( x + 1) = 0 x = 5 or x = −1 x ≠ 5. ANSWER: Hence, the only solution is −1 .
We can check our answer by substituting −1 in for x to see if we obtain a true statement. −1 − 2 −1 − 5 8 − ( −1) − = −1 − 5 −1 − 2 ( −1) − 5
1 3 −2= − 2 2 3 3 − =− 2 2
55.
PROBLEM: x+7 81 9 − 2 = x − 2 x + 5 x − 14 x + 7 SOLUTION: x+7 81 9 − 2 = x − 2 x + 5 x − 14 x + 7 x+7 81 9 − = x − 2 ( x − 2 )( x + 7 ) x + 7 We first note that x ≠ 2 and x ≠ −7 . Multiply both sides by the LCD, ( x − 2 )( x + 7 ) .
⎛ x+7 ⎞ 81 ⎛ 9 ⎞ − ⎜⎜ ⎟⎟ ( x − 2 )( x + 7 ) = ⎜ ⎟ ( x − 2 )( x + 7 ) x x x x 2 2 7 7 − − + + ( )( ) ⎝ ⎠ ⎝ ⎠ x 2 + 14 x + 49 − 81 = 9 x − 18 x 2 + 5 x − 14 = 0
( x + 7 )( x − 2 ) = 0 x = −7 or x = 2 x ≠ 2 and x ≠ −7 . ANSWER: Hence, the solution is ∅.
56.
PROBLEM: x 5 x + 30 +1 = x−6 36 − x 2 SOLUTION: x 5 x + 30 +1 = x−6 36 − x 2 x 5 x + 30 +1 = − x−6 ( x + 6 )( x − 6 ) We first note that x ≠ ±6 . Multiply both sides by the LCD, ( x − 6 )( x + 6 ) .
⎛ 5 x + 30 ⎞ ⎛ x ⎞ + 1⎟ ( x + 6 )( x − 6 ) = ⎜ − x + 6 )( x − 6 ) ⎜ ⎜ ( x + 6 )( x − 6 ) ⎟⎟ ( ⎝ x−6 ⎠ ⎝ ⎠
x 2 + 6 x + x 2 − 36 = −5 x − 30 2 x 2 + 11x − 6 = 0
( 2 x − 1)( x + 6 ) = 0 1 or x = −6 2 x ≠ ±6 .
x=
1 ANSWER: Hence, the only solution is . 2 We can check our answer by substituting −1 in for x to see if we obtain a true statement. ⎛1⎞ 1 5 ⎜ ⎟ + 30 ⎝2⎠ 2 +1 = − 1 ⎛1 ⎞⎛ 1 ⎞ −6 + 6 ⎟⎜ − 6 ⎟ ⎜ 2 ⎝2 ⎠⎝ 2 ⎠ 1 130 − +1 = 11 143 10 10 = 11 11
57.
PROBLEM: 2x 4 −7 − = 2 x +1 4x − 3 4x + x − 3 SOLUTION: 2x 4 −7 − = 2 x + 1 4x − 3 4x + x − 3 2x 4 −7 − = x + 1 4 x − 3 ( 4 x − 3)( x + 1)
3 . 4 Multiply both sides by the LCD, ( 4 x − 3)( x + 1) .
We first note that x ≠ −1 and x ≠
⎛ ⎞ −7 4 ⎞ ⎛ 2x x x − − + = 4 3 1 ( )( ) ⎜ ⎟⎟ ( 4 x − 3)( x + 1) ⎜ ⎟ ⎜ ⎝ x +1 4x − 3 ⎠ ⎝ ( 4 x − 3)( x + 1) ⎠ 8 x 2 − 6 x − 4 x − 4 = −7 8 x 2 − 10 x + 3 = 0
( 4 x − 3)( 2 x − 1) = 0 3 1 or x = 4 2 3 x≠ . 4
x=
1 . 2 1 We can check our answer by substituting in for x to see if we obtain a true 2 statement. ⎛1⎞ 2⎜ ⎟ 4 −7 ⎝2⎠ − = 1 1 + 1 4 ⎜⎛ ⎟⎞ − 3 ⎜⎛ 4 ⎜⎛ 1 ⎟⎞ − 3 ⎟⎞ ⎜⎛ 1 + 1⎟⎞ 2 ⎝2⎠ ⎝ ⎝ 2 ⎠ ⎠⎝ 2 ⎠
ANSWER: Hence, the only solution is
2 14 +4= 3 3 14 14 = 3 3
58.
PROBLEM: x −5 5 5x + =− 2 x − 10 x − 5 x − 15 x + 50 SOLUTION: x−5 5 5x + =− 2 x − 10 x − 5 x − 15 x + 50 x−5 5 5x + =− x − 10 x − 5 ( x − 10 )( x − 5) We first note that x ≠ 10 and x ≠ 5 . Multiply both sides by the LCD, ( x − 10 )( x − 5 ) .
⎛ ⎞ 5 ⎞ 5x ⎛ x−5 + ⎟⎟ ( x − 10 )( x − 5) ⎜ ⎟ ( x − 10 )( x − 5 ) = ⎜⎜ − ⎝ x − 10 x − 5 ⎠ ⎝ ( x − 10 )( x − 5) ⎠ x 2 − 10 x + 25 + 5 x − 50 = −5 x x 2 − 25 = 0
( x − 5)( x + 5) = 0 x = ±5 x ≠ 5. ANSWER: Hence, the only solution is −5 . We can check our answer by substituting −5 in for x to see if we obtain a true statement. 5 ( −5) −5 − 5 5 + =− −5 − 10 −5 − 5 ( −5 − 10)( −5 − 5)
2 1 1 − = 3 2 6 1 1 = 6 6
59.
PROBLEM: 5 x +1 5 + 2 = 2 2 x + 5 x + 4 x + 3x − 4 x − 1 SOLUTION: 5 x +1 5 + 2 = 2 2 x + 5 x + 4 x + 3x − 4 x − 1 5 x +1 5 + = ( x + 1)( x + 4 ) ( x + 4 )( x − 1) ( x − 1)( x + 1) We first note that x ≠ ±1 and x ≠ −4 . Multiply both sides by the LCD, ( x − 1)( x + 1)( x + 4 ) .
⎛ ⎞ ⎛ ⎞ x +1 5 5 + ⎜⎜ ⎟⎟ ( x − 1)( x + 1)( x + 4 ) = ⎜⎜ ⎟⎟ ( x − 1)( x + 1)( x + 4 ) x x x x x x 1 4 4 1 1 1 + + + − − + ( )( ) ( )( ) ( )( ) ⎝ ⎠ ⎝ ⎠ 2 5 x − 5 + x + 2 x + 1 = 5 x + 20 x 2 + 2 x − 24 = 0
( x + 6 )( x − 4 ) = 0 x = −6 or x = 4 Check x = −6 5
( −6 + 1)( −6 + 4 )
+
Check x = 4
5 4 +1 5 5 −6 + 1 + = = ( −6 + 4 )( −6 − 1) ( −6 − 1)( − ( 4 + 1)( 4 + 4 ) ( 4 + 4 )( 4 − 1) ( 4 − 1)( 4
1 5 1 − = 2 14 7 2 1 = 14 7 1 1 = 7 7
ANSWER: The solution is −6 or 4 .
1 5 1 + = 8 24 3 8 1 = 24 3 1 1 = 3 3
60.
PROBLEM: 1 x −9 1 + 2 = 2 2 x − 2 x − 63 x + 10 x + 21 x − 6 x − 27 SOLUTION: x −9 1 1 + 2 = 2 2 x − 2 x − 63 x + 10 x + 21 x − 6 x − 27 x −9 1 1 + = ( x + 7 )( x − 9 ) ( x + 3)( x + 7 ) ( x − 9 )( x + 3)
We first note that x ≠ −7, x ≠ 9, and x ≠ −3 . Multiply both sides by the LCD, ( x − 9 )( x + 3)( x + 7 ) .
⎛ ⎞ ⎛ ⎞ x −9 1 1 + ⎜⎜ ⎟⎟ ⋅ ( x − 9 )( x + 3)( x + 7 ) = ⎜⎜ ⎟⎟ ( x − 9 )( x + 3)( x + 7 ) ⎝ ( x + 7 )( x − 9 ) ( x + 3)( x + 7 ) ⎠ ⎝ ( x − 9 )( x + 3) ⎠ x + 3 + x 2 − 18 x + 81 = x + 7 x 2 − 18 x + 77 = 0
( x − 11)( x − 7 ) = 0 x = 11 or x = 7 Check x = 11 1
(11 + 7 )(11 − 9 )
+
11 − 9 1 = (11 + 3)(11 + 7 ) (11 − 9 )(11
1 1 1 + = 36 126 28 1 1 = 28 28
ANSWER: The solution is 11 or 7 .
Check x = 7 1
( 7 + 7 )( 7 − 9 )
+
7−9 1 = ( 7 + 3)( 7 + 7 ) ( 7 − 9 )( 7 +
1 1 1 − =− 28 70 20 1 1 − =− 20 20 −
61.
PROBLEM: 2 ( x − 2) x+2 4 + 2 = 2 2 x − 4 x − 4 x − 12 x − 8 x + 12 SOLUTION: 2 ( x − 2) x+2 4 + 2 = 2 2 x − 4 x − 4 x − 12 x − 8 x + 12 2 ( x − 2) x+2 4 + = ( x + 2 )( x − 2 ) ( x − 6 )( x + 2 ) ( x − 6 )( x − 2 ) We first note that x ≠ ±2 and x ≠ 6 . Multiply both sides by the LCD, ( x − 6 )( x − 2 )( x + 2 ) .
⎛ ⎛ ⎞ 2 ( x − 2) ⎞ 4 x+2 + ⎜⎜ ⎟⎟ ( x − 6 )( x − 2 )( x + 2 ) = ⎜⎜ ⎟⎟ ( x − 6 )( x − 2 )( x + 2 ) ⎝ ( x + 2 )( x − 2 ) ( x − 6 )( x + 2 ) ⎠ ⎝ ( x − 6 )( x − 2 ) ⎠ 4 x − 24 + 2 x 2 − 8 x + 8 = x 2 + 4 x + 4 x 2 − 8 x − 20 = 0
( x − 10 )( x + 2 ) = 0 x = 10 or x = −2 x ≠ ±2 .
ANSWER: Hence, the only solution is 10. We can check our answer by substituting 10 in for x to see if we obtain a true statement. 2 (10 − 2 ) 4 10 + 2 + = (10 + 2 )(10 − 2 ) (10 − 6 )(10 + 2 ) (10 − 6 )(10 − 2 )
1 1 3 + = 24 3 8 3 3 = 8 8
62.
PROBLEM: x+2 x+2 x −1 + 2 = 2 2 x − 5x + 4 x + x − 2 x − 2x − 8 SOLUTION: x+2 x+2 x −1 + 2 = 2 2 x − 5x + 4 x + x − 2 x − 2 x − 8 x+2 x+2 x −1 + = ( x − 4 )( x − 1) ( x + 2 )( x − 1) ( x + 2 )( x − 4 )
We first note that x ≠ 4, x ≠ 1, and x ≠ −2 . Multiply both sides by the LCD, ( x + 2 )( x − 4 )( x − 1) .
⎛ ⎞ ⎛ ⎞ x+2 x+2 x −1 + ⎜⎜ ⎟⎟ ( x + 2 )( x − 4 )( x − 1) = ⎜⎜ ⎟⎟ ( x + 2 )( x − 4 )( x − 1) x x x x x x 4 1 2 1 2 4 − − + − + − ( )( ) ( )( ) ( )( ) ⎝ ⎠ ⎝ ⎠ 2 2 2 x + 4x + 4 + x − 2x − 8 = x − 2x +1 x2 + 4 x − 5 = 0
( x + 5)( x − 1) = 0 x = −5 or x = 1 x ≠ 1. ANSWER: Hence, the only solution is –5. We can check our answer by substituting −5 in for x to see if we obtain a true statement. −5 + 2 −5 + 2 −5 − 1 + = ( −5 − 4 )( −5 − 1) ( −5 + 2 )( −5 − 1) ( −5 + 2 )( −5 − 4 )
1 1 2 − =− 18 6 9 2 2 − =− 9 9 −
63.
PROBLEM: 6x 11x + 1 6x − 2 = x −1 2x − x −1 2x +1 SOLUTION: 6x 11x + 1 6x − 2 = x −1 2x − x −1 2x +1 6x 11x + 1 6x − = x − 1 ( 2 x + 1)( x − 1) 2 x + 1
1 . 2 Multiply both sides by the LCD, ( 2 x + 1)( x − 1) .
We first note that x ≠ 1 and x ≠ −
⎛ 6x ⎞ 11x + 1 ⎛ 6x ⎞ − ⎜⎜ ⎟⎟ ( 2 x + 1)( x − 1) = ⎜ ⎟ ( 2 x + 1)( x − 1) − + − + 1 2 1 1 2 1 x x x x ( )( ) ⎝ ⎠ ⎝ ⎠ 2 12 x + 6 x − 11x − 1 = 6 x 2 − 6 x 6 x2 + x − 1 = 0
( 3x − 1)( 2 x + 1) = 0 1 1 or x = − 3 2 1 x≠− . 2
x=
1 ANSWER: Hence, the only solution is . 3 We can check our answer by substituting statement. 1 1 1 6⋅ 11⋅ + 1 6⋅ 3− 3 3 = 1 1 1 1 − 1 ⎛⎜ 2 ⋅ + 1⎞⎟ ⎛⎜ − 1⎞⎟ 2 ⋅ + 1 3 3 ⎝ 3 ⎠⎝ 3 ⎠ 14 6 3 −3 − = 5 ⎛ 2⎞ 5 ⋅⎜ − ⎟ 3 ⎝ 3⎠ 21 6 −3 + = 5 5 6 6 = 5 5
1 in for x to see if we obtain a true 3
64.
PROBLEM: 8x 4x 1 + 2 = 2x − 3 2x − 7 x + 6 x − 2 SOLUTION: 8x 4x 1 + 2 = 2x − 3 2x − 7x + 6 x − 2 8x 4x 1 + = 2 x − 3 ( 2 x − 3)( x − 2 ) x − 2
3 and x ≠ 2 . 2 Multiply both sides by the LCD, ( 2 x − 3)( x − 2 ) .
We first note that x ≠
⎛ 8x ⎞ 4x ⎛ 1 ⎞ + ⎜⎜ ⎟⎟ ( 2 x − 3)( x − 2 ) = ⎜ ⎟ ( 2 x − 3)( x − 2 ) ⎝ x−2⎠ ⎝ 2 x − 3 ( 2 x − 3)( x − 2 ) ⎠ 8 x 2 − 16 x + 4 x = 2 x − 3 8 x 2 − 14 x + 3 = 0
( 4 x − 1)( 2 x − 3) = 0 1 3 or x = 4 2 3 x≠ . 2
x=
1 ANSWER: Hence, the only solution is . 4 1 We can check our answer by substituting in for x to see if we obtain a true 4 statement. 1 1 8⋅ 4⋅ 1 4 + 4 = 1 1 1 1 2 ⋅ − 3 ⎛⎜ 2 ⋅ − 3 ⎞⎟ ⎛⎜ − 2 ⎞⎟ −2 4 ⎝ 4 ⎠⎝ 4 ⎠ 4 4 28 4 − + =− 5 35 7 4 4 − =− 7 7
Part B: Literal Equations Solve for the indicated variable.
65.
PROBLEM:
Solve for r: t =
D r
SOLUTION: Multiply both sides by the LCD, r . D t ⋅r = ⋅r r tr = D
tr D = t t D r= t ANSWER: r =
66.
D t
PROBLEM:
Solve for b: h =
2A b
SOLUTION: Multiply both sides by the LCD, b . 2A h ⋅b = ⋅b b hb = 2 A
hb 2 A = h h 2A b= h ANSWER: b =
2A h
67.
PROBLEM:
Solve for P: t =
I Pr
SOLUTION: Multiply both sides by the LCD, Pr . I t ⋅ Pr = ⋅ Pr Pr Ptr = I
Ptr I = tr tr I P= tr ANSWER: P =
68.
I tr
PROBLEM:
Solve for π : r =
C 2π
SOLUTION: Multiply both sides by the LCD, 2π . C r= 2π C r ⋅ 2π = ⋅ 2c 2π 2π r = C
2π r C = 2r 2r C π= 2r ANSWER: π =
C 2r
69.
PROBLEM: 1 1 1 Solve for c: = + a b c SOLUTION: Multiply both sides by the LCD, abc. 1 ⎛1 1⎞ ⋅ abc = ⎜ + ⎟ ⋅ abc a ⎝b c⎠ bc = ac + ab
bc − ac = ac − ac + ab bc − ac = ab c ( b − a ) = ab c (b − a )
(b − a )
c=
=
ab b−a
ab b−a
ANSWER: c =
70.
PROBLEM:
Solve for y: m =
ab b−a y − y1 x − x1
SOLUTION: Multiply both sides by the LCD, x − x1 . ⎛ y − y1 ⎞ m ( x − x1 ) = ⎜ ⎟ ( x − x1 ) ⎝ x − x1 ⎠ m ( x − x1 ) = y − y1 y − y1 + y1 = m ( x − x1 ) + y1 y = m ( x − x1 ) + y1
ANSWER: y = m ( x − x1 ) + y1
71.
PROBLEM: Solve for w: P = 2(l + w) SOLUTION: P = 2(l + w)
P = 2l + 2w P − 2l = 2l − 2l + 2w 2w = P − 2l 2w P − 2l = 2 2 P − 2l w= 2 ANSWER: w =
72.
P − 2l 2
PROBLEM: Solve for t: A = P(1 + rt ) SOLUTION: A = P(1 + rt )
A P(1 + rt ) = P P A = 1 + rt P A − 1 = 1 − 1 + rt P A− P rt = p rt A − P = r r⋅ p A− P t= pr ANSWER: t =
A− P pr
73.
PROBLEM:
1 n+m
Solve for m: s =
SOLUTION: Multiply both sides by the LCD, n + m . 1 s ( n + m) = ⋅ ( n + m) n+m sn + sm = 1
ms + ns − ns = 1 − ns ms = 1 − ns ms 1 − ns = s s 1 − ns m= s ANSWER: m =
74.
1 − ns s
PROBLEM:
Solve for S: h =
S −r 2π r
SOLUTION: Multiply both sides by the LCD, 2π r . ⎛ S ⎞ h ⋅ 2π r = ⎜ − r ⎟ 2π r ⎝ 2π r ⎠
2π rh = S − 2π r 2
S − 2π r 2 + 2π r 2 = 2π rh + 2π r 2 S = 2π rh + 2π r 2 ANSWER: S = 2π rh + 2π r 2
75.
PROBLEM:
Solve for x: y =
x x+2
SOLUTION: Multiply both sides by the LCD, x + 2 . x ⋅x+2 y ( x + 2) = x+2 xy + 2 y = x
xy − xy + 2 y = x − xy 2 y = x (1 − y ) x (1 − y ) 2y = 1 − y (1 − y ) x=
2y 1− y
ANSWER: x =
76.
PROBLEM:
Solve for x: y =
2y 1− y
2x +1 5x
SOLUTION: Multiply both sides by, 5x . 2x +1 y ⋅ 5x = ⋅ 5x 5x 5 xy = 2 x + 1 5 xy − 2 x = 2 x − 2 x + 1
x (5 y − 2) = 1 x (5 y − 2)
(5 y − 2)
x=
=
1 5y − 2
1 5y − 2
ANSWER: x =
1 5y − 2
77.
PROBLEM:
Solve for R :
1 1 1 = + R R1 R2
SOLUTION: Multiply both sides by LCD, RR1 R2 . ⎛ 1 1 1 ⎞ ⋅ RR1 R2 = ⎜ + ⎟ RR1 R2 R ⎝ R1 R2 ⎠ R1 R2 = RR2 + RR1 R1 R2 = R ( R2 + R1 )
R ( R2 + R1 ) R1 R2 = R2 + R1 ( R2 + R1 ) R=
R1 R2 R2 + R1
ANSWER: R =
78.
R1 R2 R2 + R1
PROBLEM: 1 1 1 = + f S1 S 2 SOLUTION: Multiply both sides by LCD, fS1S2 . ⎛1 1 ⎞ 1 ⋅ fS1S 2 = ⎜ + ⎟ fS1S 2 f ⎝ S1 S 2 ⎠ S1S 2 = fS 2 + fS1 S1S 2 − fS1 = fS 2 + fS1 − fS1 S1 ( S 2 − f ) = fS 2 S1 ( S 2 − f )
( S2 − f )
S1 =
=
fS 2 S2 − f
fS 2 S2 − f
ANSWER: S1 =
fS2 S2 − f
Part C: Discussion Board Solve for the indicated variable.
79.
PROBLEM: Explain why multiplying both sides of an equation by the LCD sometimes produces extraneous solutions. ANSWER: When both the sides are multiplied by the LCD, the solutions that go missing while arriving at the rational expression, emerges again.
80.
PROBLEM: Explain the connection between the technique of cross multiplication and multiplying both sides by the LCD. ANSWER: Cross multiplication is a simplified version of the LCD method. 5 3 = For e.g.: x + 5 x +1 Cross multiplication method. 5 3 = x + 5 x +1 5 ( x + 1) = 3 ( x + 5)
5 x + 5 = 3x + 15 2 x = 10 x=5 The LCD method. We first note that x ≠ −5 and x ≠ −1 . Multiply both sides by the LCD, ( x + 5 )( x + 1) .
5 3 ⋅ ( x + 5 )( x + 1) = ⋅ ( x + 5 )( x + 1) x+5 x +1 5 x + 5 = 3x + 15 5 x − 3x + 5 = 3x − 3x + 15 2 x + 5 = 15 2 x + 5 − 5 = 15 − 5 2 x = 10 x=5
81.
PROBLEM: Explain how we can tell the difference between a rational expression and a rational equation. How do we treat them differently? ANSWER: A rational expression is in the form X/Y, where Y is not equal to zero A rational equation equates two rational expressions. A/B=C/D, where B and D are not equal to zero.
7.6 Applications of Rational Equations
Part A: Number Problems Use algebra to solve the following applications.
1.
PROBLEM: A positive integer is twice that of another. The sum of the reciprocals of the two positive integers is 3/10. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: 2n 1 1 3 + = n 2n 10 We first note that n ≠ 0. Multiply both sides by the LCD, 20n. 3 ⎛1 1 ⎞ ⎜ + ⎟ ⋅ 20n = ⋅ 20n 10 ⎝ n 2n ⎠ 20 + 10 = 6n 6n = 30 n=5
Larger integer: 2n = 2 ⋅ 5 = 10 ANSWER: The two integers are 5 and 10.
2.
PROBLEM: A positive integer is twice that of another. The sum of the reciprocals of the two positive integers is 3/12. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: 2n 1 1 3 + = n 2n 12 We first note that n ≠ 0. Multiply both sides by the LCD, 12n. 3 ⎛1 1 ⎞ ⎜ + ⎟ ⋅12n = ⋅12n 12 ⎝ n 2n ⎠ 12 + 6 = 3n 3n = 18 n=6 Larger integer: 2n = 2 ⋅ 6 = 12 ANSWER: The two integers are 6 and 12.
3.
PROBLEM: A positive integer is twice that of another. The difference of the reciprocals of the two positive integers is 1/8. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: 2n 1 1 1 − = n 2n 8 We first note that n ≠ 0. Multiply both sides by the LCD, 16n. 1 ⎛1 1 ⎞ ⎜ − ⎟ ⋅16n = ⋅16n 8 ⎝ n 2n ⎠ 16 − 8 = 2n 2n = 8 n=4 Larger integer: 2n = 8 ANSWER: The two integers are 4 and 8.
4.
PROBLEM: A positive integer is twice that of another. The difference of the reciprocals of the two positive integers is 1/18. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: 2n 1 1 1 − = n 2n 18 We first note that n ≠ 0. Multiply both sides by the LCD, 36n. 1 ⎛1 1 ⎞ ⎜ − ⎟ ⋅ 36n = ⋅ 36n 18 ⎝ n 2n ⎠ 36 − 18 = 2n 2n = 18 n=9 Larger integer: 2n = 2 ⋅ 9 = 18 ANSWER: The two integers are 9 and 18.
5.
PROBLEM: A positive integer is 2 less than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is 5/12, then find the two integers. SOLUTION: Larger positive integer is represented by n. Smaller integer: n − 2 1 2 5 + = n − 2 n 12 We first note that n ≠ 0 and n ≠ 2 . Multiply both sides by the LCD, 12n ( n − 2 ) .
2⎞ 5 ⎛ 1 + ⎟12n ( n − 2 ) = ⋅12n ( n − 2 ) ⎜ 12 ⎝ n−2 n⎠ 12n + 24n − 48 = 5n 2 − 10n 5n 2 − 46n + 48 = 0
( 5n − 6 )( n − 8) = 0 6 or n = 8 5 The problem speaks about positive integer. Hence, the larger integer is 8. Smaller integer: n − 2 = 8 − 2 = 6 n=
ANSWER: The integers are 6 and 8.
6.
PROBLEM: A positive integer is 2 more than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is 17/35, then find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer is n + 2. 1 2 17 + = n n + 2 35 We first note that n ≠ 0 and n ≠ −2. Multiply both sides by the LCD, 35n ( n + 2 ) .
2 ⎞ 17 ⎛1 ⎜ + ⎟ 35n ( n + 2 ) = ⋅ 35n ( n + 2 ) 35 ⎝n n+2⎠ 2 35n + 70 + 70n = 17 n + 34n 17 n 2 − 71n + 70 = 0
(17n + 14 )( n − 5) 14 or n = 5 17 The problem speaks about positive integers. Hence, the smaller integer is 5. Larger integer: n + 2 = 5 + 2 = 7 n=−
ANSWER: The two integers are 5 and 7.
7.
PROBLEM: The sum of the reciprocals of two consecutive positive even integers is 11/60. Find the two even integers. SOLUTION: The first even integer is represented by n. The second even integer is n + 2. 1 1 11 + = n n + 2 60 We first note that n ≠ 0 or n ≠ −2. Multiply both the sides by the LCD, 60n ( n + 2 ) .
1 ⎞ 11 ⎛1 ⎜ + ⎟ 60n ( n + 2 ) = ⋅ 60n ( n + 2 ) 60 ⎝n n+2⎠ 2 60n + 120 + 60n = 11n + 22n 11n 2 − 98n − 120 = 0
( n − 10 )(11n + 12 ) = 0 12 11 The problem speaks about positive integers. Hence, the first even integer is 10. The next even integer: n + 2 = 10 +2 = 12 n = 10 or n = −
ANSWER: The two consecutive even integers are 10 and 12.
8.
PROBLEM: The sum of the reciprocals of two consecutive positive odd integers is 16/63. Find the integers. SOLUTION: The first odd integer is represented by n. The consecutive odd integer: n + 2 1 1 16 + = n n + 2 63 We first note that n ≠ 0 and n ≠ −2 . Multiply both the sides by the LCD, 63n ( n + 2 ) . 1 ⎞ 16 ⎛1 ⎜ + ⎟ 63n ( n + 2 ) = ⋅ 63n ( n + 2 ) 63 ⎝n n+2⎠ 2 63n + 126 + 63n = 16n + 32n 16n 2 − 94n − 126 = 0
(
)
2 8n 2 − 47 n − 63 = 0 8n 2 − 47 n − 63 = 0
( n − 7 )(8n + 9 ) = 0 9 8 The problem speaks about positive integers. Hence, the first odd integer is 7. Consecutive odd integer: n + 2 = 7 + 2 = 9 n = 7 or n = −
ANSWER: The two consecutive odd integers are 7 and 9.
9.
PROBLEM: The difference of the reciprocals of two consecutive positive even integers is 1/24. Find the two even integers. SOLUTION: The first even integer is represented by n. Consecutive even integer: n + 2 1 1 1 − = n n + 2 24 We first note that n ≠ 0 or n ≠ −2. Multiply both the sides by the LCD, 24n ( n + 2 ) .
1 ⎞ 1 ⎛1 ⋅ 24n ( n + 2 ) ⎜ − ⎟ 24n ( n + 2 ) = 24 ⎝n n+2⎠ 24n + 48 − 24n = n 2 + 2n n 2 + 2n − 48 = 0
( n + 8)( n − 6 ) = 0 n = −8 or n = 6 The problem speaks about positive integers. Hence, the first even integer is 8. The next even integer: n + 2 = 6 +2 = 8 ANSWER: The two consecutive even integers are 6 and 8.
10.
PROBLEM: The difference between the reciprocals of two consecutive positive odd integers is 2/99. Find the integers. SOLUTION: The first odd integer is represented by n. The consecutive odd integer: n + 2 1 1 2 − = n n + 2 99 We first note that n ≠ 0 and n ≠ −2 . Multiply both the sides by the LCD, 99n ( n + 2 ) .
1 ⎞ 2 ⎛1 ⎜ − ⎟ 99n ( n + 2 ) = ⋅ 99n ( n + 2 ) 99 ⎝n n+2⎠ 2 99n + 198 − 99n = 2n + 4n 2n 2 + 4n − 198 = 0
(
)
2 n 2 + 2n − 99 = 0 n 2 + 2n − 99 = 0
( n + 11)( n − 9 ) = 0 n = −11 or n = 9 The problem speaks about positive integers. Hence, the first odd integer is 9. Consecutive odd integer: 9 + 2 = 9 + 2 = 11 ANSWER: The two consecutive odd integers are 9 and 11.
11.
PROBLEM: If 3 times the reciprocal of the larger of two consecutive integers is subtracted from 2 times the reciprocal of the smaller, then the result is 1/2. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: n + 1 2 3 1 − = n n +1 2 We first note that n ≠ 0 and n = −1 . Multiply both sides by the LCD, 2n ( n + 1) . 3 ⎞ 1 ⎛2 ⎜ − ⎟ 2n ( n + 1) = ⋅ 2n ( n + 1) 2 ⎝ n n +1 ⎠ 4n + 4 − 6 n = n 2 + n n 2 + 3n − 4 = 0
( n + 4 )( n − 1) = 0 n = −4 or n = 1 n + 1 = −4 + 1 or n + 1 = 1 + 1 n + 1 = −3 or n + 1 = 2
ANSWER: The two consecutive integers are –4 and –3 or 1 and 2.
12.
PROBLEM: If 3 times the reciprocal of the smaller of two consecutive integers is subtracted from 7 times the reciprocal of the larger, then the result is 1/2. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: n + 1 7 3 1 − = n +1 n 2 We first note that n ≠ 0 and n = −1 . Multiply both sides by the LCD, 2n ( n + 1) . 3⎞ 1 ⎛ 7 − ⎟ 2n ( n + 1) = ⋅ 2n ( n + 1) ⎜ 2 ⎝ n +1 n ⎠ 14n − 6n − 6 = n 2 + n n 2 − 7n + 6 = 0
( n − 1)( n − 6 ) = 0 n = 1 or n = 6 n + 1 = 1 + 1 or n + 1 = 6 + 1 n +1 = 2
or n + 1 = 7
ANSWER: The two consecutive integers are 1 and 2 or 6 and 7.
13.
PROBLEM: A positive integer is 5 less than another. If the reciprocal of the smaller integer is subtracted from 3 times the reciprocal of the larger, then the result is 1/12. Find the two integers. SOLUTION: Larger integer is represented by n. Smaller integer: n – 5 3 1 1 − = n n − 5 12 We first note that n ≠ 0 and n = 5 . Multiply both sides by the LCD, 12n ( n − 5 ) . 1 ⎞ 1 ⎛3 ⎜ − ⎟12n ( n − 5 ) = ⋅12n ( n − 5 ) 12 ⎝ n n−5⎠ 36n − 180 − 12n = n 2 − 5n n 2 − 29n + 180
( n − 20 )( n − 9 ) = 0 n = 20 or n = 9 n − 5 = 20 − 5 or n − 5 = 9 − 5 n − 5 = 15
or n − 5 = 4
ANSWER: The two consecutive integers are 20 and 15 or 9 and 4.
14.
PROBLEM: A positive integer is 6 less than another. If the reciprocal of the smaller integer is subtracted from 10 times the reciprocal of the larger, then the result is 3/7. Find the two integers. SOLUTION: Larger integer is represented by n. Smaller integer: n – 6 10 1 3 − = n n−6 7 We first note that n ≠ 0 and n ≠ 6. Multiply both sides by the LCD, 7 n ( n − 6 ) . 1 ⎞ 3 ⎛ 10 ⎜ − ⎟ 7n ( n − 6 ) = ⋅ 7n ( n − 6 ) 7 ⎝ n n−6⎠ 70n − 420 − 7 n = 3n 2 − 18n 3n 2 − 81n + 420 = 0
(
)
3 n 2 − 27 n + 140 = 0 n 2 − 27 n + 140 = 0
( n − 20 )( n − 7 ) = 0 n = 20 or n = 7 n − 6 = 20 − 6 = 14 or n − 6 = 7 − 6 = 1
ANSWER: The two integers are 20 and 14 or 7 and 1.
Part B: Uniform Motion Problems Use algebra to solve the following applications.
15.
PROBLEM: James can jog twice as fast as he can walk. He was able to jog the first 9 miles to his grandmother’s house, but then he tired and walked the remaining 1.5 miles. If the total trip took 2 hours, then what was his average jogging speed? SOLUTION: Let x represent James’ average walking speed. James average jogging speed: 2x D 9 = Time taken by James to jog 9 miles: r 2x D 1.5 Time taken by James to walk 1.5 miles: = r x Total time taken = 2 hours Time taken by jogging + Time taken by walking = Total time taken 9 1.5 + =2 2x x 9 15 + =2 2 x 10 x We first note that x ≠ 0 . Multiply both sides with the LCD, 10x. ⎛ 9 15 ⎞ ⎜ + ⎟10 x = 2 ⋅10 x ⎝ 2 x 10 x ⎠ 45 + 15 = 20 x 60 = 20 x x = 3 miles/hour
Jogging speed = 2x = 2 ⋅ 3 = 6 miles/hour ANSWER: Jogging speed = 2x = 2 ⋅ 3 = 6 miles/hour
16.
PROBLEM: On a business trip, an executive traveled 720 miles by jet aircraft and then another 80 miles by helicopter. If the jet averaged 3 times the speed of the helicopter and the total trip took 4 hours, then what was the average speed of the jet? SOLUTION: Speed of the helicopter is represented by x. Speed of the jet: 3x D 80 Time taken by the helicopter to cover 80 miles: t = = r x D 720 240 Time taken by the jet to cover 720 miles: t = = = r 3x x Total time taken = 4 hours We first note that x ≠ 0. Time taken by the helicopter + Time taken by the jet = Total time taken 80 240 + =4 x x 320 =4 x 320 x= 4 x = 80 miles/hour Jet speed: 3x = 3 ⋅ 80 = 240 miles/hour ANSWER: Jet speed: 3x = 3 ⋅ 80 = 240 miles/hour
17.
PROBLEM: Sally was able to drive an average of 20 mph faster in her car when the traffic cleared. She drove 23 miles in traffic before it cleared and then drove another 99 miles. If the total trip took 2 hours, then what was her average speed in traffic? SOLUTION: Speed of Sally’s car in traffic is represented by x. Speed of Sally’s car when the traffic clears: x + 20
Time taken by Sally to drive 99 miles when the traffic clears: t = Time taken by Sally to drive 23 miles in traffic: t =
D 99 = r x + 20
D 23 = r x
Total time taken = 2 hours Time taken in traffic + Time taken when the traffic clears = Total time taken 23 99 + =2 x x + 20 We first note that x ≠ 0 and x ≠ −20 . Multiplying both sides with the LCD, x ( x + 20 ) . 99 ⎞ ⎛ 23 ⎜ + ⎟ x ( x + 20 ) = 2 ⋅ x ( x + 20 ) ⎝ x x + 20 ⎠ 23 x + 460 + 99 x = 2 x 2 + 40 x 2 x 2 − 82 x − 460 = 0
(
2 x 2 − 41x − 230
)
x 2 − 41x − 230 = 0
( x + 5 )( x − 46 ) = 0 x = −5 or x = 46 The problem speaks about time taken. Hence, only postive values are considered. ANSWER: Average speed of Sally’s car in traffic is 46 miles/hour.
18.
PROBLEM: Harry traveled 15 miles on the bus and then another 72 miles on a train. If the train was 18 mph faster than the bus and the total trip took 2 hours, then what was the average speed of the train? SOLUTION: Speed of the bus is represented by x. Speed of the train is represented by x + 18. D 72 Time taken by the train to cover 72 miles: t = = r x + 18 D 15 Time taken by the bus to cover 15 miles: t = = r x Total time taken = 2 hours
Time taken in the rain + Time taken in the bus = Total time taken 72 15 + =2 x + 18 x We first note that x ≠ 0 and x = −18 . Multiply both sides with the LCD, x ( x + 18 ) . 15 ⎞ ⎛ 72 + ⎟ x ( x + 18 ) = 2 ⋅ x ( x + 18 ) ⎜ ⎝ x + 18 x ⎠ 72 x + 15 x + 270 = 2 x 2 + 36 x 2 x 2 − 51x − 270 = 0
( x − 30 )( 2 x + 9 ) = 0 9 2 The problem speaks about time. Hence, the positive integer is the solution. x = 30 or x = −
ANSWER: Speed of the train: x + 18 = 30 + 18 = 48 miles/hour.
19.
PROBLEM: A bus is on average 6 mph faster than a trolley. If the bus can travel 90 miles in the same time it takes the trolley to travel 75 miles, then what was the speed of each? SOLUTION: Speed of the trolley is represented by x. Speed of the bus: x + 6 D 75 Time taken by the trolley to cover 75 miles: t = = r x D 90 Time taken by the bus to cover 90 miles: t = = r x+6 Time taken by the trolley to cover 75 miles = Time taken by the bus to cover 90 miles 75 90 = x x+6 We first note that x ≠ 0 and x ≠ −6 . 75 90 = x x+6 75 ( x + 6 ) = 90 ⋅ x
75 x + 450 = 90 x 15 x = 450 x = 30 Speed of the trolley: x = 30 miles/hour Speed of the bus: x + 6 = 30 + 6 = 36 miles/hour ANSWER: Speed of the trolley: x = 30 miles/hour
Speed of the bus: x + 6 = 30 + 6 = 36 miles/hour
20.
PROBLEM: A passenger car is on average 16 mph faster than the bus. If the bus can travel 56 miles in the same time it takes the passenger car to travel 84 miles, then what was the speed of each? SOLUTION: Speed of the bus is represented by x. Speed of the passenger car: x + 16 D 56 Time taken by the bus to cover 56 miles: t = = r x
D 84 = r x + 16 Time taken by the passenger car to = Time taken by the bus to cover 56 miles cover 84 miles 84 56 = x + 16 x 84 x = 56 ( x + 16 )
Time taken by the passenger car to cover 84 miles: t =
84 x = 56 x + 896 28 x = 896 x = 32 Speed of the bus: x = 32 miles/hour Speed of the passenger car: x + 16 = 32 + 16 = 48 miles/hour ANSWER: Speed of the bus: x = 32 miles/hour Speed of the passenger car: x + 16 = 32 + 16 = 48 miles/hour
21.
PROBLEM: A light aircraft travels 2 mph less than twice as fast as a passenger car. If the passenger car can travel 231 miles in the same time it takes the aircraft to travel 455 miles, then what was the average speed of each? SOLUTION: Speed of the passenger car is represented by x. Speed of the light aircraft: 2x – 2 D 231 Time taken by passenger car to cover 231 miles: t = = r x D 455 Time taken by light aircraft to cover 455 miles: t = = r 2x − 2 Time taken by passenger car to cover 231 miles = Time taken by light aircraft to cover 455 miles 231 455 = x 2x − 2 231( 2 x − 2 ) = 455 x
462 x − 462 = 455 x 7 x = 462 x = 66 Speed of the passenger car: x = 66 miles/hour Speed of the light aircraft: 2x − 2 = 2 ⋅ 66 − 2 = 130 miles/hour ANSWER: Speed of the passenger car: x = 66 miles/hour Speed of the light aircraft: 2x − 2 = 2 ⋅ 66 − 2 = 130 miles/hour
22.
PROBLEM: Mary can run 1 mph more than twice as fast as Bill can walk. If Bill can walk 3 miles in the same time it takes Mary to run 7.2 miles, then what is Bill’s average walking speed? SOLUTION: Bill’s average walking speed is represented by x. Mary’s running speed: 2x + 1 D 3 Time taken by Bill to walk 3 miles: t = = r x D 7.2 Time taken by Mary to run 7.2 miles: t = = r 2x +1 Time taken by Bill to walk 3 miles = Time taken by Mary to run 7.2 miles 3 7.2 = x 2x +1 3 ( 2 x + 1) = 7.2 x
6 x + 3 = 7.2 x 60 x + 30 = 72 x 72 x − 60 x = 30 12 x = 30 30 = 2.5 x= 12 Bill's walking speed: x = 2.5 miles/hour ANSWER: Bill's walking speed: x = 2.5 miles/hour
23.
PROBLEM: An airplane, traveling with a 20 mph tailwind, was able to cover 270 miles. On the return trip, against the wind, it was only able to cover 190 miles in the same amount of time. What was the speed of the airplane in still air? SOLUTION: Speed of the airplane in still air is represented by x. Speed of the airplane along the direction of the tailwind: x + 20 Speed of the airplane against the direction of the tailwind: x – 20
D 270 = r x + 20 D 190 Time taken to cover 190 miles against the direction of the wind: t = = r x − 20 Time taken to cover 270 miles = Time taken to cover 190 miles against the along the direction of the wind direction of the wind 270 190 = x + 20 x − 20 270 ( x − 20 ) = 190 ( x + 20 ) Time taken to cover 270 miles along the direction of the wind: t =
270 x − 5400 = 190 x + 3800 80 x = 9200 x = 115 Speed of the airplane in still air: x = 115 miles/hour ANSWER: Speed of the airplane in still air: x = 115 miles/hour
24.
PROBLEM: A jet airliner, traveling with a 30 mph tailwind, was able to cover 525 miles in the same amount of time it was able to travel 495 miles after the tailwind eased to 10 mph. What was the speed of the airliner in still air? SOLUTION: Speed of the jet airliner in still air is represented by x. Speed of the jet airliner along the direction of the tailwind 30 mph: x + 30 Speed of the jet airliner after the tail wind eases to 10 mph: x + 10 D 525 Time taken to cover 525 miles along the direction of the: t = = r x + 30 tailwind 30 mph D 495 Time taken to cover 495 miles along the direction of the: t = = r x + 10 tailwind 10 mph Time taken to cover 525 miles along the = Time taken to cover 495 miles along along the direction of the tailwind 30 mph the direction of the tailwind 10 mph 525 495 = x + 30 x + 10 525 ( x + 10 ) = ( x + 30 ) 495
525 x + 5250 = 495 x + 14850 525 x − 495 x = 14850 − 5250 30 x = 9600 x = 320 Speed of the jet airliner in still air : x = 320 miles/hour ANSWER: Speed of the jet airliner in still air : x = 320 miles/hour
25.
PROBLEM: A boat can average 16 mph in still water. With the current, the boat can travel 95 miles in the same time it can travel 65 miles against it. What is the speed of the current? SOLUTION: Speed of the current is represented by x. Speed of the boat in the direction of the current: 16 + x Speed of the boat against the direction of the current: 16 – x D 95 along the direction of Time taken by the boat to cover 95 miles: t = = r 16 + x the current
Time taken by the boat to cover 65 miles: t =
D 65 against the direction of = r 16 − x
the current Time taken by the boat to cover 95 miles = Time taken by the boat to cover 65 miles along the direction of the current against the direction of the current 95 65 = 16 + x 16 − x 95 (16 − x ) = 65 (16 + x ) 1520 − 95 x = 65 x + 1040 160 x = 480 x=3 ANSWER: Speed of the current is 3 miles/hour.
26.
PROBLEM: A river tour boat can average 7 mph in still water. If the total 24 mile tour downriver and 24 miles back took 7 hours, then how fast was the river current? SOLUTION: Speed of the river current is represented by x. Speed of the tour boat along the direction of the river current: 7 + x Speed of the tour boat against the direction of the river current: 7 – x 24 Time taken by the tour boat to cover 24 miles downstream: 7+ x
Time taken by the tour boat to cover 24 miles upstream:
24 7−x
Total time taken = 7 hours Time taken to travel upstream + Time taken to travel downstream = Total time taken 24 24 + =7 7+ x 7−x We first note that x = ±7 . Multiply both sides with the LCD, ( 7 + x )( 7 − x ) . 24 ⎞ ⎛ 24 + ⎜ ⎟ ( 7 + x )( 7 − x ) = 7 ⋅ ( 7 + x )( 7 − x ) ⎝7+ x 7−x⎠ 168 − 24 x + 168 + 24 x = 343 − 7 x 2 7 x2 − 7 = 0 7 ( x + 1)( x − 1) = 0 x = ±1 The solution is the positive integer. ANSWER: Speed of the river current: x = 1 mile/hour
27.
PROBLEM: If the river current flows at an average 3 mph, then a tour boat can make the 9 mile tour downstream with the current and back the 9 miles against the current in 4 hours. What is the average speed of the boat in still water? SOLUTION: Speed of the boat in still water is represented by x. Speed of the current = 3 mph Speed of the boat along the direction of the current: x + 3 Speed of the boat against the direction of the current: x – 3 D 9 Time taken by the boat to cover 9 miles downstream: t = = r x+3 D 9 Time taken by the boat to cover 9 miles upstream: t = = r x −3 Total time taken = 4 hours Time taken by the boat to + Time taken by the boat to = Total time taken cover 9 miles downstream cover 9 miles upstream 9 9 + =4 x+3 x−3 We first note that x ≠ ±3 . Multiply both the sides with the LCD, ( x + 3)( x − 3) . 9 ⎞ ⎛ 9 + ⎜ ⎟ ( x + 3)( x − 3) = 4 ⋅ ( x + 3)( x − 3) ⎝ x+3 x−3⎠ 9 x − 27 + 9 x + 27 = 4 x 2 − 36 4 x 2 − 18 x − 36 = 0
(
)
2 2 x 2 − 9 x − 18 = 0 2 x 2 − 9 x − 18 = 0
( 2 x + 3)( x − 6 ) = 0 3 or x = 6 2 The positive integer is the solution. x=−
ANSWER: Speed of the boat in still water is 6 miles/hour.
28.
PROBLEM: Jane rowed her canoe, against a 1 mph current, upstream 12 miles and then returned the 12 miles back downstream. If the total trip took 5 hours, then at what speed can Jane row in still water? SOLUTION: Speed at which Jane can row in still water is represented by x. Speed of the current = 1 mph Speed at which Jane can row upstream: x – 1 Speed at which Jane can row downstream: x + 1 D 12 Time taken by Jane to cover 12 miles against the current: t = = r x −1 D 12 Time taken by Jane to cover 12 miles along the current: t = = r x +1 Total time taken = 5 hours Time taken by Jane to cover + Time taken by Jane to cover = Total time taken 12 miles against the current 12 miles along the current 12 12 + =5 x −1 x +1 We first note that x = ±1 . Multiply both sides with the LCD, ( x − 1)( x + 1) .
12 ⎞ ⎛ 12 + ⎜ ⎟ ( x − 1)( x + 1) = 5 ⋅ ( x − 1)( x + 1) ⎝ x −1 x +1 ⎠ 12 x + 12 + 12 x − 12 = 5 x 2 − 5 5 x 2 − 24 x − 5 = 0
( 5 x + 1)( x − 5) = 0 1 or x = 5 5 The solution is the positive integer. x=−
ANSWER: Speed at which Jane can row in still water is 5 miles/hour.
29.
PROBLEM: Jose drove 15 miles to pick up his sister and then returned home. On the return trip, he was able to average 15 mph faster than he did on the trip to pick her up. If the total trip took 1 hour, then at what average speed was Jose able to average on the return trip? SOLUTION: Speed at which Jose drives to pick his sister is represented by x
The speed Jose was able to average on the return trip: x + 15. D 15 Time taken by Jose to reach the destination: t = = r x D 15 Time taken by Jose on his return trip: t = = r x + 15 Total time taken = 1 hour Time taken by Jose + Time taken by Jose on his return trip = Total time taken to reach the destination 15 15 + =1 x x + 15 We first note that x ≠ 0 and x ≠ −15 . Multiply both the sides with the LCD, x ( x + 15 ) . 15 ⎞ ⎛ 15 ⎜ + ⎟ x ( x + 15 ) = 1 ⋅ x ( x + 15 ) ⎝ x x + 15 ⎠ 15 x + 225 + 15 x = x 2 + 15 x x 2 − 15 x − 225 = 0 a = 1; b = −15; c = −225 x= x=
−b ± b 2 − 4ac 2a − ( −15 ) ±
( −15)
2
− 4 (1)( −225 )
2 ⋅1 x = 24.25 or x = −9.125 The positive number is the solution.
=
15 ± 33.5 2
ANSWER: The speed Jose was able to average on the return trip: x + 15 = 24.25 + 15 = 39.25 ~ 40 miles/hour.
30.
PROBLEM: Barry drove the 24 miles to town and then back in 1 hour. On the return trip, he was able to average 14 mph faster than he was on the trip to town. What was his average speed on the trip to town? SOLUTION: Speed at which Barry drove to town is represented by x. Speed of the return trip: x + 14 D 24 Time taken by Barry to reach the town: t = = r x D 24 Time taken by Barry on his return trip: t = = r x + 14 Total time taken = 1 hour Time taken by Barry + Time taken by Barry = Total time taken to reach the town on his return trip 24 24 + =1 x x + 14 We first note that x ≠ 0 and x ≠ −14 . Multiply both the sides with the LCD, x ( x + 14 ) .
24 ⎞ ⎛ 24 ⎜ + ⎟ x ( x + 14 ) = 1 ⋅ x ( x + 14 ) ⎝ x x + 14 ⎠ 24 x + 336 + 24 x = x 2 + 14 x x 2 − 34 x − 336 = 0
( x − 42 )( x + 8 ) = 0 x = 42 or x = −8 The positive integer is the solution. Speed of the return trip: x + 14 = 42 + 14 = 56 mile/hour ANSWER: Speed of the trip to the town : x = 42
31.
PROBLEM: Jerry paddled his kayak, upstream against a 1 mph current, for 12 miles. The return trip, downstream with the 1 mph current, took one hour less time. How fast can Jerry paddle the kayak in still water? SOLUTION: The speed at which Jerry can paddle the kayak in still water is represented by x. Speed of the current = 1 mph Speed of the kayak along the direction of the current: x +1 Speed of the kayak against the direction of the current: x – 1 12 Time taken by Jerry to cover 12 miles upstream: x −1 12 Time taken by Jerry to cover 12 miles downstream: x +1 Time taken by Time taken by Jerry to cover = Jerry to cover – 1 12 miles downstream 12 miles upstream 12 12 = −1 x −1 x +1 We first note that x ≠ ±1 . Multiply both the side with the LCD, ( x + 1)( x − 1) .
⎛ 12 ⎞ ⎛ 12 ⎞ − 1⎟ ( x + 1)( x − 1) ⎜ ⎟ ( x + 1)( x − 1) = ⎜ ⎝ x −1 ⎠ ⎝ x +1 ⎠ 12 x + 12 = 12 x − 12 − x 2 + 1 x 2 − 25 = 0
( x + 5)( x − 5) = 0 x = ±5 The positive integer is the solution. ANSWER: The speed at which Jerry can paddle the kayak in still water is 5 miles/hour.
32.
PROBLEM: It will take a light aircraft 1 hour more time to fly 360 miles against a 30 mph headwind than it will to fly the same distance with it. What is the speed of the aircraft in calm air? SOLUTION: The speed of the aircraft in calm air is represented by x. Speed of the headwind = 30 mph Speed of the aircraft along the direction of the headwind: x + 30 Speed of the aircraft against the direction of the headwind: x – 30 D 360 Time taken by the aircraft to cover 360 miles: t = = r x + 30 along the direction of the wind D 360 Time taken by the aircraft to cover 360 miles: t = = r x − 30 against the direction of the wind
Time taken by the aircraft to = Time taken by the aircraft cover 360 miles against the to cover 360 miles along direction of the wind the direction of the wind 360 360 = +1 x − 30 x + 30 We first note that x ≠ ±30 . Multiply both sides with LCD, ( x − 30 )( x + 30 ) .
+
⎛ 360 ⎞ ⎛ 360 ⎞ + 1⎟ ( x − 30 )( x + 30 ) ⎜ ⎟ ( x − 30 )( x + 30 ) = ⎜ ⎝ x − 30 ⎠ ⎝ x + 30 ⎠ 360 x + 10800 = 360 x − 10800 + x 2 − 900 x 2 − 22500 = 0
( x + 150 )( x − 150 ) = 0 x = ±150 The positive integer is the solution. ANSWER: Speed of the aircraft in calm air is 150 miles/hour.
1
Part C: Work Rate Problems Use algebra to solve the following applications.
33.
PROBLEM: James can paint the office, by himself, in 7 hours and Manny can do it in 10 hours. How long will it take them to paint the office working together? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 7 hours; t2 = 10 hours 1 1 ⋅t + ⋅t = 1 7 10 17t =1 70 17t = 70 70 t= 17 2 t=4 17
ANSWER: James and Manny can paint the office in 4
2 hours. 17
34.
PROBLEM: Barry can lay a brick driveway, by himself, in 12 hours and Robert can do it in 10 hours. How long will it take them to lay the brick driveway working together? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 12 hours; t2 = 10 hours 1 1 ⋅t + ⋅t =1 12 10 22t =1 120 22t = 120 120 t= 22 60 t= 11 5 t =5 11
ANSWER: Barry and Robert can lay the brick driveway in 5
together.
5 hours working 11
35.
PROBLEM: Jerry can detail a car, by himself, in 50 minutes and Sally can do it in one hour. How long will it take them to detail a car working together? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 50 minutes; t2 = 1 hour = 60 minutes 1 1 ⋅t + ⋅t = 1 50 60 110t =1 3000 110t = 3000 300 t= 11 3 t = 27 11
ANSWER: Jerry and Sally can detail the car together in 27
36.
3 minutes. 11
PROBLEM: Jose can build a small shed, by himself, in 26 hours. Alex can build the same small shed, but it will take him 2 days to do so. How long would it take them to build the shed working together? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 26 hours; t2 = 48 hours 1 1 ⋅t + ⋅t = 1 26 48 48t + 26t =1 26 ⋅ 48 74t = 1248 624 t= 37 32 t = 16 37
ANSWER: Jose and Alex can build the small shed in 16
together.
32 hours working 37
37.
PROBLEM: Allison can complete a sales route, by herself, in 6 hours. Working with an associate, they can complete the route in 4 hours. How long would it take her associate to complete the route by herself? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 6 hours; t = 4 hours 1 1 ⋅4+ ⋅4 =1 t2 6 4 4 = 1− t2 6 4 2 = t2 6 t2 =
4⋅6 = 12 2
ANSWER: The associate will take 12 hours to complete the sales route on her own.
38.
PROBLEM: James can prepare and paint a house by himself in 5 days. Working with his brother, Bryan, they can do it in 3 days. How long would it take Bryan to prepare and paint the house by himself? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 5 days; t = 3 days 1 1 ⋅3 + ⋅3 =1 5 t2 3 3 + =1 5 t2 3 3 = 1− 5 t2 3 2 = t2 5 t2 =
5 ⋅ 3 15 1 = =7 2 2 2
1 ANSWER: Bryan will take 7 days to prepare and paint the house himself. 2
39.
PROBLEM: Joe can assemble a computer, by himself, in 1 hour. Working with an assistant, he can assemble a computer in 40 minutes. How long would it take his assistant to assemble a computer working alone? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 1 hour = 60 minutes; t = 40 minutes 1 1 ⋅ 40 + ⋅ 40 = 1 60 t2 2 40 + =1 3 t2 40 2 = 1− 3 t2 40 1 = t2 3 t2 = 40 ⋅ 3 = 120
ANSWER: The assistant will take 120 minutes = 2 hours to assemble a computer working alone.
40.
PROBLEM: The teacher’s assistant can grade class homework assignment by herself in one hour. If the teacher helps, then the grading can be completed in 20 minutes. How long would it take the teacher to grade the papers working alone? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t2 = 1 hour = 60 minutes; t = 20 minutes 1 1 ⋅ 20 + ⋅ 20 = 1 60 t1 20 1 + =1 t1 3 20 1 = 1− 3 t1 20 2 = t1 3 t1 =
3 ⋅ 20 = 30 2
ANSWER: The teacher will take 30 minutes grading the paper alone.
41.
PROBLEM: A larger pipe can fill a water tank twice as fast as a smaller pipe. When both pipes are on, they can fill the tank in 5 hours. If the larger pipe is left off, then how long would it take the smaller pipe to fill the tank? SOLUTION: If we let x represent the time it takes for the small pipe to fill the 1 tank, then the small pipe has a work rate of . x 2 The large pipe has a work rate of . x When both the pipes are on, the tank is filled in 5 hours. Multiply the individual work rates by 5. The sum of these two products is equal to 1. 1 2 ⋅5 + ⋅5 = 1 x x We first note that x ≠ 0 . 1 2 ⋅5 + ⋅5 = 1 x x 5 + 10 =1 x x = 15 ANSWER: The smaller pipe takes 15 hours to fill the tank all alone.
42.
PROBLEM: A newer printer can print twice as fast as an older printer. If both printers, working together, can print a batch of flyers in 45 minutes, then how long would it take the newer printer to print the batch working alone? SOLUTION: If we let x represent the time it takes for the older printer to print a 1 batch of flyers, then the old printer has a work rate of . x x The new printer can print the same batch of flyers in . Hence the new printer 2 2 has a work rate of . x When both the printers are used the batch of flyers can be printed in 45 minutes. Multiply the individual work rates by 45. The sum of these two products is equal to 1. 1 2 ⋅ 45 + ⋅ 45 = 1 x x We first note that x ≠ 0 . 45 + 90 =1 x x = 135 ANSWER: The new printer can print the same batch of flyers in x 135 = = 67.5 minutes . 2 2
43.
PROBLEM: Working alone, Henry would take 9 hours longer than Mary to clean the carpets in the entire office. Working together they can clean the carpets in 6 hours. How long would it take Mary to clean the office carpets if Henry was not there to help? SOLUTION: If we let x represent the time it takes Mary to clean the carpets in 1 the entire office, then Mary has a work rate of . x Henry takes 9 hours longer than Mary, x + 9. Hence, Henry has a work rate of 1 . x+9 Mary and Henry can clean the carpets in 6 hours together. Multiply the individual work rates by 6. The sum of these two products is equal to 1. 1 1 ⋅6 + ⋅6 =1 x x+9 6 6 + =1 x x+9 We first note that x ≠ 0 and x ≠ −9 . Multiply both sides with the LCD, x ( x + 9 ) .
6 ⎞ ⎛6 ⎜ + ⎟ x ( x + 9 ) = 1⋅ x ( x + 9 ) ⎝ x x+9⎠ 6 x + 54 + 6 x = x 2 + 9 x
x 2 − 3 x − 54 = 0
( x − 9 )( x + 6 ) = 0 x = 9 or x = −6 The positive integer is the solution. ANSWER: Mary will take 9 hours to clean the carpets on her own.
44.
PROBLEM: Working alone, Monique would take 4 hours longer than Audrey to record the inventory of the entire shop. Working together they can take inventory in 1.5 hours. How long would it take Audrey to record the inventory working alone? SOLUTION: If we let x represent the time it takes for Audrey to record the 1 inventory of the entire shop, then Audrey has a work rate of . x Monique takes 4 hours longer than Audrey, x + 4 hours. Hence, Monique has a 1 . work rate of x+4 Working together they can take inventory in 1.5 hours. Multiply the individual work rates by 1.5. The sum of these two products is equal to 1. 1 1 ⋅1.5 + ⋅1.5 = 1 x x+4 We first note that x ≠ 0 and x ≠ −4 . 15 15 + = 10 x x+4 Multiply both the sides with the LCD, x ( x + 4 ) .
⎛ 15 15 ⎞ ⎜ + ⎟ x ( x + 4 ) = 10 ⋅ x ( x + 4 ) ⎝ x x+4⎠ 15 x + 60 + 15 x = 10 x 2 + 40 x 10 x 2 + 10 x − 60 = 0 2 ( x − 2 )( x + 3) = 0
x = 2 or x = −3 The positive integer is the solution. ANSWER: Audrey takes 2 hours to record the inventory of the entire shop.
45.
PROBLEM: Jerry can lay a tile floor in 3 hours less time than Jake. If they work together, the floor will take 2 hours. How long would it take Jerry to lay the floor by himself? SOLUTION: If we let x represent the time it takes for Jake to lay a tile floor, then 1 Jake has a work rate of . x Jerry takes 3 hours less than Jake, x – 3 hours. Hence, Jerry has a work rate of 1 . x−3 Working together they can tile the floor in 2 hours. Multiply the individual work rates by 2. The sum of these two products is equal to 1. 1 1 ⋅2+ ⋅2 =1 x x −3 2 2 + =1 x x −3 We first note that x ≠ 0 and x ≠ 3 . Multiply both the sides with the LCD, x ( x − 3) .
2 ⎞ ⎛2 ⎜ + ⎟ x ( x − 3) = 1 ⋅ x ( x − 3) ⎝ x x −3⎠ 2 x − 6 + 2 x = x 2 − 3x
x2 − 7 x + 6 = 0
( x − 6 )( x − 1) = 0 x = 6 or x = 1 x ≠ 1 (Jerry takes less than 3 hours.) ANSWER: Jerry takes x – 3 hours = 6 – 3 = 3 hours
46.
PROBLEM: Jeremy can build a model airplane in 5 hours less time than his brother could. Working together it took 6 hours. How long would it have taken Jeremy to build the model airplane working alone? SOLUTION: If we let x represent the time it takes for Jeremy to build the model 1 airplane, then Jeremy has a work rate of . x Jeremy’s brother takes 5 hours longer than Jeremy, x + 5 hours. Hence, Jeremy’s 1 . brother has a work rate of x+5 Working together they can build the model in 6 hours. Multiply the individual work rates by 6. The sum of these two products is equal to 1. 1 1 ⋅6 + ⋅6 =1 x x+5 We first note that x ≠ 0 and x ≠ −5 . 6 6 + =1 x x+5 Multiply both the sides with the LCD, x ( x + 5 ) .
6 ⎞ ⎛6 ⎜ + ⎟ x ( x + 5) = 1⋅ x ( x + 5) ⎝ x x+5⎠ 6 x + 30 + 6 x = x 2 + 5 x
x 2 − 7 x − 30 = 0
( x − 10 )( x + 3) = 0 x = 10 or x = −3 The positive integer is the solution. ANSWER: Jeremy takes 10 hours to build the model all alone.
47.
PROBLEM: Harry can paint a shed, by himself, in 6 hours and Jeremy can do it in 8 hours. How long will it take them to paint two sheds working together? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 6 hours; t2 = 8 hours 1 1 ⋅t + ⋅t = 1 6 8 6t + 8t =1 48 14t = 48 48 t= 14 24 t= 7
24 hours. 7 24 48 6 ANSWER: They can paint two sheds in , ⋅2 = = 6 hours. 7 7 7
Working together, Harry and Jeremy can paint a shed in
48.
PROBLEM: Joe can assemble a computer, by himself, in 1 hour. Working with an assistant, he can assemble 10 computers in 6 hours. How long would it take his assistant to assemble one computer working alone? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2
Joe and his assistant can assemble 10 computers in 6 hours. 6 hour. Joe and his assistant can assemble a computer in 10 6 t1 = 1 hour; t = hour 10 1 6 1 6 ⋅ + ⋅ =1 1 10 t2 10 6 ⎛ 1⎞ ⎜1 + ⎟ = 1 10 ⎝ t2 ⎠ 6 ⎛ t2 + 1 ⎞ ⎜ ⎟ =1 10 ⎝ t2 ⎠ 6t2 + 6 = 10t2 4t2 = 6 6 4 3 1 t2 = = 1 2 2 t2 =
1 ANSWER: Joe’s assistant will take 1 hours to assemble one computer. 2
49.
PROBLEM: Jerry can lay a tile floor in 3 hours and his assistant can do the same job in 4 hours. If Jerry starts the job and his assistant joins him 1 hour later, then how long will it take to lay the floor?
1 SOLUTION: Jerry can lay a tile floor in 3 hours. Hence, Jerry’s work rate is . 3 Jerry’s assistant can lay the same tile floor in 4 hours. Hence, assistant work rate 1 is . 4 1 1 Work done by Jerry in a hour is 1 ⋅ = 3 3 1 2 Remaining floor to be tiled: 1 − = 3 3 1 1 2 Using the formula, ⋅ t + ⋅ t = (Remaining floor to be tiled) t1 t2 3 t1 = 3 hours; t2 = 4 hours 1 1 2 ⋅t + ⋅t = 3 3 4 7t 2 = 12 3 2 12 t= ⋅ 3 7 8 t= 7 Jerry starts one hour earlier. ANSWER:
1+ Hence, the total time taken to lay the floor = 1 + t =
8 15 1 = = 2 hours. 7 7 7
50.
PROBLEM: Working alone, Monique would take 6 hours to record the inventory of the entire shop while it would take Audrey only 4 hours to do the same job. How long will it take them working together if Monique leaves 2 hours early? SOLUTION: Monique takes 6 hours to record the inventory of the entire shop. 1 Hence, Monique’s work rate is . 6 Audrey takes 4 hours to record the inventory of the entire shop. Hence, Audrey’s 1 work rate is . 4 Audrey leaves 2 hours early. Hence, Monique has to do Audrey’s 2 hour work. 1 1 Audrey’s work to be completed by Monique 2 ⋅ = . Hence, Monique takes 2 4 2 hours to complete Audrey’s work.
Work done together by Monique and Audrey: 1 −
1 1 = 2 2
1 1 1 Using the formula, ⋅ t + ⋅ t = (Work done by Audrey and Monique) t1 t2 2 t1 = 6 hours; t2 = 4 hours 1 1 1 ⋅t + ⋅t = 6 4 2 4t + 6t 1 = 24 2 1 24 t= ⋅ 2 10 6 t= 5 Total time taken = Time taken by Audrey and Monique + Time taken by Audrey 1 to complete of the work to complete Monique’s work 2 6 16 1 = +2= = 3 hours 5 5 5
ANSWER: 3
1 hours 5
7.7 Variation
Part A: Variation Problems Translate the following sentences into a mathematical formula.
1.
PROBLEM: The distance D an automobile can travel is directly proportional to the time t that it travels at a constant speed. ANSWER: D = kt
2.
PROBLEM: The extension of a hanging spring d is directly proportional to the weight w attached to it. ANSWER: d = kw
3.
PROBLEM: The breaking distance of an automobile d is directly proportional to the square of its speed v. ANSWER: d = kv2
4.
PROBLEM: The volume of a sphere V varies directly as the cube of its radius r. ANSWER: V = kr3
5.
6.
PROBLEM: The volume V of a given mass of gas is inversely proportional to the pressure p exerted on it. k ANSWER: V = p PROBLEM: The intensity of light from a light source I is inversely proportional to the square of the distance from the source. ANSWER: The distance from the source is represented by d. k I= 2 d
7.
PROBLEM: Every particle of matter in the universe attracts every other particle with a force F that is directly proportional to the product of the masses m1 and m2 of the particles and inversely proportional to the square of the distance d between them. ANSWER: F = k
8.
m1 ⋅ m2 d2
PROBLEM: Simple interest I is jointly proportional to the annual interest rate r and the time t, in years, a fixed amount of money is invested. ANSWER: I = krt
9.
PROBLEM: The period of a pendulum T is directly proportional to the square root of its length L. ANSWER: T = k L
10.
PROBLEM: The time t it takes an object to fall is directly proportional to the square root of the distance d it falls. ANSWER: t = k d
Construct a mathematical model given the following:
11.
PROBLEM: y varies directly as x and y = 30 when x = 6. SOLUTION: y = kx Substitute the values of x and y in the expression above. 30 = k6 k=5 ANSWER: The expression becomes y = 5x.
12.
PROBLEM: y varies directly as x and y = 52 when x = 4. SOLUTION: y = kx Substitute the values of x and y in the expression above. 52 = k4 k = 13 ANSWER: The expression becomes y = 13x.
13.
PROBLEM: y is directly proportional to x and y = 12 when x = 3. SOLUTION: y = kx Substitute the values of x and y in the expression above. 12 = k3 k=4 ANSWER: The expression becomes y = 4x.
14.
PROBLEM: y is directly proportional to x and y = 120 when x = 20. SOLUTION: y = kx Substitute the values of x and y in the expression above. 120 = k20 k=6 ANSWER: The expression becomes y = 6x.
15.
PROBLEM: y varies directly as x and y = 14 when x = 10. SOLUTION: y = kx Substitute the values of x and y in the expression above. 14 = k10 14 7 k= = 10 5 7 ANSWER: The expression becomes y = x. 5
16.
PROBLEM: y varies directly as x and y = 2 when x = 8. SOLUTION: y = kx Substitute the values of x and y in the expression above. 2 = k8 2 1 k= = 8 4 x ANSWER: The expression becomes y = . 4
17.
PROBLEM: y varies inversely as x and y = 5 when x = 7.
k x Substitute the values of x and y in the expression above. k 5= 7 k = 35 35 . ANSWER: The expression becomes y = x SOLUTION: y =
18.
PROBLEM: y varies inversely as x and y = 12 when x = 2.
k x Substitute the values of x and y in the expression above. k 12 = 2 k = 24 24 . ANSWER: The expression becomes y = x SOLUTION: y =
19.
PROBLEM: y is inversely proportional to x and y = 3 when x = 9.
k x Substitute the values of x and y in the expression above. k 3= 9 k = 27 27 ANSWER: The expression becomes y = . x SOLUTION: y =
20.
PROBLEM: y is inversely proportional to x and y = 21 when x = 3.
k x Substitute the values of x and y in the expression above. k 21 = 3 k = 63 63 ANSWER: The expression becomes y = . x SOLUTION: y =
21.
PROBLEM: y varies inversely as x and y = 2 when x = 1/8.
k x Substitute the values of x and y in the expression above. k 2= 1 8 1 k = 2⋅ 8 1 k= 4 1 ANSWER: The expression becomes y = . 4x SOLUTION: y =
22.
PROBLEM: y varies inversely as x and y = 3/2 when x = 1/9.
k x Substitute the values of x and y in the expression above. 3 k = 2 1 9 3 1 k= ⋅ 2 9 1 k= 6 1 ANSWER: The expression becomes y = . 6x SOLUTION: y =
23.
PROBLEM: y varies jointly as x and z where y = 8 when x = 4 and z = 1/2. SOLUTION: y = kxz Substitute the values of x, y, and z in the expression above. 1 8 = k ⋅4⋅ 2 k =4 ANSWER: The expression becomes y = 4xz.
24.
PROBLEM: y varies jointly as x and z where y = 24 when x = 1/3 and z = 9. SOLUTION: y = kxz Substitute the values of x, y, and z in the expression above. 1 24 = k ⋅ ⋅ 9 3 k =8 ANSWER: The expression becomes y = 8xz.
25.
PROBLEM: y is jointly proportional to x and z where y = 2 when x = 1 and z = 3. SOLUTION: y = kxz Substitute the values of x, y, and z in the expression above. 2 = k ⋅1 ⋅ 3
k=
2 3
ANSWER: The expression becomes y =
26.
2 xz . 3
PROBLEM: y is jointly proportional to x and z where y = 15 when x = 3 and z = 7. SOLUTION: y = kxz Substitute the values of x, y, and z in the expression above. 15 = k ⋅ 3 ⋅ 7
k=
5 7
ANSWER: The expression becomes y =
27.
5 xz. 7
PROBLEM: y varies jointly as x and z where y = 2/3 when x = 1/2 and z = 12. SOLUTION: y = kxz Substitute the values of x, y, and z in the expression above. 2 1 = k ⋅ ⋅12 3 2 2 1 k= = 3⋅ 6 9 xz . ANSWER: The expression becomes y = 9
28.
PROBLEM: y varies jointly as x and z where y = 5 when x = 3/2 and z = 2/9. SOLUTION: y = kxz Substitute the values of x, y, and z in the expression above. 3 2 5=k⋅ ⋅ 2 9 k = 5 ⋅ 3 = 15 ANSWER: The expression becomes y = 15xz.
29.
PROBLEM: y varies directly as the square of x where y = 45 when x = 3. SOLUTION: y = kx2 Substitute the values of x and y in the expression above. 45 = k 32 45 =5 9 ANSWER: The expression becomes y = 5 x 2 . k=
30.
PROBLEM: y varies directly as the square of x where y = 3 when x = 1/2. SOLUTION: y = kx2 Substitute the values of x and y in the expression above. 2 ⎛1⎞ 3= k⎜ ⎟ ⎝2⎠ k = 3 ⋅ 4 = 12 ANSWER: The expression becomes y = 12 x 2 .
31.
PROBLEM: y is inversely proportional to the square of x where y = 27 when x = 1/3.
k x2 Substitute the values of x and y in the expression above. k 27 = 2 ⎛1⎞ ⎜ ⎟ ⎝3⎠ 1 k = ⋅ 27 = 3 9 3 ANSWER: The expression becomes y = 2 . x SOLUTION: y =
32.
PROBLEM: y is inversely proportional to the square of x where y = 9 when x = 2/3.
k x2 Substitute the values of x and y in the expression above. k 9= 2 ⎛2⎞ ⎜ ⎟ ⎝3⎠ 4 k = ⋅9 = 4 9 4 ANSWER: The expression becomes y = 2 . x SOLUTION: y =
33.
PROBLEM: y varies jointly as x and the square of z where y = 54 when x = 2 and z = 3. SOLUTION: y = kxz 2 Substitute the values of x, y, and z in the expression above. 54 = k ⋅ 2 ⋅ 32 54 =3 18 ANSWER: The expression becomes y = 3xz 2 k=
34.
PROBLEM: y varies jointly as x and the square of z where y = 6 when x = 1/4 and z = 2/3. SOLUTION: y = kxz 2 Substitute the values of x, y, and z in the expression above. 2 1 ⎛2⎞ 6 = k ⋅ ⋅⎜ ⎟ 4 ⎝3⎠ k = 6 ⋅ 9 = 54 ANSWER: The expression becomes y = 54 xz 2
35.
PROBLEM: y varies jointly as x and z and inversely as the square of w, where y = 30 when x = 8, z = 3, and w = 2.
kxz w2 Substitute the values of x, y, and z in the expression above. k ⋅8⋅3 30 = 22 30 k= =5 6 5xz ANSWER: The expression becomes y = 2 w SOLUTION: y =
36.
PROBLEM: y varies jointly as x and z and inversely as the square of w, where y = 5 when x = 1, z = 3, and w = 1/2.
kxz w2 Substitute the values of x, y, and z in the expression above. k ⋅1 ⋅ 3 5= 2 ⎛1⎞ ⎜ ⎟ ⎝2⎠ 5 k= 12 5 xz ANSWER: The expression becomes y = 12w2 SOLUTION: y =
37.
PROBLEM: y varies directly as the square root of x and inversely as z where y = 12 when x = 9 and z = 5.
x z Substitute the values of x, y, and z in the expression above. 9 12 = k 5 5 ⋅12 k= = 20 3 20 x ANSWER: The expression becomes y = z SOLUTION: y = k
38.
PROBLEM: y varies directly as the square root of x and inversely as the square of z where y = 15 when x = 25 and z = 2. SOLUTION: y = k
x 2
z Substitute the values of x, y, and z in the expression above. 25 15 = k 2 2 15 ⋅ 4 k= = 12 5 12 x ANSWER: The expression becomes y = 2 z
39. PROBLEM: y varies directly as the square of x and inversely as z and the square of w where y = 14 when x = 4, w = 2, and z = 2. x2 SOLUTION: y = k 2 zw Substitute the values of x, y, z, and w in the expression above. 42 14 = k 2 ⋅ 22 14 =k 2 k =7 7x 2 ANSWER: The expression becomes y = 2 zw 40.
PROBLEM: y varies directly as the square root of x and inversely as z and the square of w where y = 27 when x = 9, w = 1/2, and z = 4.
x zw2 Substitute the values of x, y, z, and w in the expression above. 9 27 = k 2 ⎛1⎞ 4⋅⎜ ⎟ ⎝2⎠ 27 k= =9 3 9 x ANSWER: The expression becomes y = zw2
SOLUTION: y = k
Part B: Variation Problems Applications involving variation.
41.
PROBLEM: The revenue, in dollars, is directly proportional to the number of branded sweatshirts sold. If the revenue earned from selling 25 sweatshirts was $318.75, then determine the revenue if 30 sweatshirts are sold. SOLUTION: Revenue in dollars is represented by R. Number of sweatshirts sold is represented by x. R = kx Revenue earned from selling 25 sweatshirts: $318.75 R = kx 318.75 = k ⋅ 25 318.75 k= = 12.75 25 Revenue for 30 sweatshirts: R = kx R = 12.75 ⋅ 30 = 382.5 ANSWER: Revenue for 30 sweatshirts sold is $382.50.
42.
PROBLEM: The sales tax on the purchase of a new car varies directly as the price of the car. If a $18,000 new car is purchased, then the sales tax would be $1,350. How much sales tax would be required if the new car was priced at $22,000? SOLUTION: Sales tax in dollars is represented by S. Price of the car is represented by x. S = kx If a $18,000 new car is purchased, then the sales tax would be $1,350. S = kx 1350 = k ⋅18000 k = 0.075 Sales tax required if the new car was priced at $22,000: S = kx S = 0.075 ⋅ 22, 000 = 1650
ANSWER: Sales tax required if the new car was priced at $22,000 is $1,650.
43.
PROBLEM: The price of a share of common stock in a company is directly proportional to the earnings per share (EPS) of the previous 12 months. If the price of a share of common stock in a company is $22.55 and the EPS is published to be $1.10, then determine the value of the stock if the earnings per share increased by $0.20. SOLUTION: Price of a share of common stock in the company is represented by P. Earnings per share is represented by x. P = kx Price of a share of common stock in a company is $22.55 and the EPS is published to be $1.10. P = kx 22.55 = k ⋅1.10 k = 20.5 If the earnings per share increased by $0.20, x = 1.10 + 0.20 = 1.3. Value of the stock: P = kx P = 20.5 ⋅1.3 = 26.65
ANSWER: Value of the stock if the earnings per share increased by $0.20 is $26.65.
44.
PROBLEM: The distance traveled on a road trip varies directly as the time spent on the road. If a 126 mile trip can be made in 3 hours, then what distance could be traveled in 4 hours? SOLUTION: Distance traveled on a road trip D. Time spent on the road is represented by t. D = kt A 126 mile trip can be made in 3 hours. D = kt 126 = k ⋅ 3 k = 42 Distance travelled in 4 hours: D = kt D = 42 ⋅ 4 = 168
ANSWER: Distance travelled in 4 hours is 168 miles.
45.
PROBLEM: The circumference of a circle is directly proportional to its radius. If the circumference of a circle with radius 7 cm is measured to be 14π cm , then find the constant of proportionality. SOLUTION: Circumference of the circle is represented by C. Radius of the circle is represented by r. C = kr Circumference of a circle with radius 7 cm is measured to be 14π cm . C = kr 14π = k ⋅ 7 k = 2π
ANSWER: k = 2π
46.
PROBLEM: The area of circle varies directly as the square of its radius. If the area of a circle with radius 7 cm is determined to be 49π cm 2 , then find the constant of proportionality. SOLUTION: Area of the circle is represented by A. Radius of the circle is represented by r. A = kr 2 Area of a circle with radius 7 cm is measured to be 49π cm 2 .
A = kr 2 49π = k ⋅ 7 2 49π = k ⋅ 49 k =π ANSWER: k = π
47.
PROBLEM: The surface area of a sphere varies directly as the square of its radius. When the radius of a sphere measures 2 m, the surface area measures 16π m 2 . Find the surface area of a sphere with radius 3 m. SOLUTION: Surface area of the sphere is represented by S. Radius of the sphere is represented by r. S = kr 2 When the radius of a sphere measures 2 m, the surface area measures 16π m 2 .
S = kr 2 16π = k ⋅ 22 4k = 16π k = 4π Surface area of a sphere with radius 3 m: S = kr 2 S = 4π ⋅ 32 = 36π ANSWER: Surface area of a sphere with radius 3 m is 36π m2.
48.
PROBLEM: The volume of a sphere varies directly as the cube of its radius. When the radius of a sphere measures 3 m, the volume measures 36π m3 . Find the volume of a sphere with radius 1 m. SOLUTION: Volume of a sphere is represented by V. Radius of the sphere is represented by r. V = kr 3 When the radius of a sphere measures 3 m, the volume measures 36π m3 . V = kr 3
36π = k ⋅ 33 27k = 36π 4 k= π 3 Volume of a sphere with radius 1 m: V = kr 3 4 V = π ⋅13 3 4π V= 3 ANSWER: Volume of a sphere with radius 1 m is
4π 3 m. 3
49.
PROBLEM: With a fixed height, the volume of a cone is directly proportional to the square of the radius at the base. When the radius at the base measures 10 cm, the volume is determined to be 200 cubic centimeters. Determine the volume of the cone if the radius of the base is halved. SOLUTION: Volume of the cone is represented by V. Radius of the base of the cone is represented by r. V = kr 2 When the radius at the base measures 10 cm, the volume is determined to be 200 cubic centimeters. V = kr 2
200 = k ⋅102 k =2 Volume of the cone if the radius of the base is halved, r =
10 = 5 cm . 2
V = kr 2 V = 2 ⋅ 52 = 50 cm3 ANSWER: Volume of the cone if the radius of the base is halved is 50 cm3. Hooke’s law suggests that the extension of a hanging spring is directly proportional to the weight attached to it. The constant of variation is called the spring constant.
50.
PROBLEM: The distance d an object in freefall falls varies directly with the square of the time t that it has been falling. If an object in freefall falls 36 feet in 1.5 seconds, then how far will it have fallen in 3 seconds? SOLUTION: d = kt 2 An object in freefall falls 36 feet in 1.5 seconds. 36 = k ⋅1.52 k = 16 The object in freefall for 3 seconds falls: d = kt 2 d = 16 ⋅ 9 = 144 ANSWER: The object in freefall for 3 seconds falls 144 feet.
Hooke’s law suggests that the extension of a hanging spring is directly proportional to the weight attached to it. The constant of variation is called the spring constant.
51.
PROBLEM: If a hanging spring is stretched 5 inches when a 20 lb weight is attached to it, then determine its spring constant. SOLUTION: The extension of the hanging spring is represented by L. The weight attached to the hanging spring is represented by w. L = kw A hanging spring is stretched 5 inches when a 20 lb weight is attached to it. L = kw 5 = k ⋅ 20 1 k= 4 ANSWER: k =
52.
1 4
PROBLEM: If a hanging spring is stretched 3cm when a 2 kg weight is attached to it, then determine the spring constant. SOLUTION: The extension of the hanging spring is represented by L. The weight attached to the hanging spring is represented by w. L = kw A hanging spring is stretched 3cm when a 2 kg weight is attached to it. L = kw 3 = k ⋅2 3 k= 2 3 ANSWER: k = 2
53.
PROBLEM: If a hanging spring is stretched 3 inches when a 2 pound weight is attached, then how far will it stretch with a 5 lb weight attached? SOLUTION: The extension of the hanging spring is represented by L. The weight attached to the hanging spring is represented by w. L = kw A hanging spring is stretched 3 inches when a 2 pound weight is attached. L = kw 3 = k ⋅2 3 k= 2 If a 5 lb weight is attached: L = kw 3 L = ⋅5 2 15 1 L= =7 2 2 ANSWER: If a 5 lb weight is attached, the hanging spring will stretch by 7
1 2
inches. The breaking distance of an automobile is directly proportional to the square of its speed.
54.
PROBLEM: If a hanging spring is stretched 6 cm when a 4 kg weight is attached to it, then how far will it stretch with a 2 kg weight attached? SOLUTION: The extension of the hanging spring is represented by L. The weight attached to the hanging spring is represented by w. L = kw A hanging spring is stretched 6 cm when a 4 kg weight is attached. L = kw 6 = k ⋅4 6 3 k= = 4 2 If a 2 kg weight is attached: L = kw 3 L = ⋅2 2 L=3 ANSWER: If a 2 kg weight is attached, the hanging spring will stretch by 3 cm.
The breaking distance of an automobile is directly proportional to the square of its speed.
55.
PROBLEM: If it took 36 feet to stop a particular automobile moving at a speed of 30 miles per hour, then how much breaking distance is required if the speed is 35 miles per hour? SOLUTION: Breaking distance of the automobile is represented by d. Speed of the vehicle is represented by s. d = ks 2 It takes 36 feet to stop the automobile moving at a speed of 30 miles per hour. d = ks 2 36 = k ⋅ 302 36 1 = 900 25 If the speed is 35 miles per hour, breaking distance required: d = ks 2 1 d = ⋅ 352 = 49 25 k=
ANSWER: If the speed is 35 miles per hour, breaking distance required is 49 feet. Boyle's Law states that if the temperature remains constant, the volume V of a given mass of gas is inversely proportional to the pressure p exerted on it.
56.
PROBLEM: After an accident, it was determined that it took a driver 80 feet to stop his car. In an experiment, under similar conditions, it took 45 feet to stop the car moving at a speed of 30 miles per hour. Estimate how fast the driver was moving before the accident. SOLUTION: Breaking distance of the automobile is represented by d. Speed of the vehicle is represented by s. d = ks 2 In an experiment, under similar conditions, it took 45 feet to stop the car moving at a speed of 30 miles per hour. d = ks 2 45 = k ⋅ 302 1 20 After an accident, it was determined that it took a driver 80 feet to stop his car. d = ks 2 k=
1 2 s 20 s 2 = 1600
80 =
s = 40
ANSWER: The driver was moving at 40 miles/hour before the accident. Boyle's Law states that if the temperature remains constant, the volume V of a given mass of gas is inversely proportional to the pressure p exerted on it.
57.
PROBLEM: A balloon is filled to a volume of 216 cubic inches on a diving boat under 1 atmosphere (atm) of pressure. If the balloon is taken under water approximately 33 ft where the pressure measures 2 atm, then what is the volume of the balloon?
k p The balloon is filled to a volume of 216 cubic inches on a diving boat under 1 atmosphere (atm) of pressure. k V= p k 216 = 1 k = 216 The balloon is taken under water approximately 33 ft where the pressure measures 2 atm. k V= p 216 V= = 108 2 SOLUTION: Boyle’s law, V =
ANSWER: If the balloon is taken under water approximately 33 ft where the pressure measures 2 atm, the volume of the balloon is 108 in.3
58.
PROBLEM: If a balloon is filled to 216 cubic inches under a pressure of 3 atmospheres at a depth of 66 ft, then what would the volume be at the surface where the pressure is 1 atm?
k p The balloon is filled to a volume of 216 cubic inches under a pressure of 3 atmospheres at a depth of 66 ft. k V= p k 216 = 3 k = 648 The volume at the surface where the pressure is 1 atm. k V= p 648 V= 1 V = 648 SOLUTION: Boyle’s law, V =
ANSWER: The volume at the surface where the pressure is 1 atm is 648 in.3
59.
PROBLEM: To balance a seesaw, the distance from the fulcrum that a person must sit is inversely proportional to his weight. If a 72 lb boy is sitting 3 feet from the fulcrum, then how far from the fulcrum must a 54 lb boy sit to balance the seesaw? SOLUTION: The distance from the fulcrum that a person must sit is represented by d. Weight of the person is represented by w. k d= w A 72 lb boy is sitting 3 feet from the fulcrum. k d= w k 3= 72 k = 216
If a 54 lb boy sits on these saw, the distance from the fulcrum that a person must k sit: d = w 216 d= =4 54 ANSWER: If a 54 lb boy sits on these saw, the person must sit 4 feet away from the fulcrum.
60.
PROBLEM: The current ( I ) in an electrical conductor is inversely proportional to its resistance ( R ). If the current is 1/4 amp when the resistance is 100 ohms, then what is the current when the resistance is 150 ohms?
k R The current is 1/4 amp when the resistance is 100 ohms. 1 k = 4 100 k = 25 k When the resistance is 150 ohms: I = R 25 1 I= = 150 6 SOLUTION: I =
ANSWER: When the resistance is 150 ohms, the current is
1 amp. 6
61.
PROBLEM: The number of men y needed to lay a cobblestone driveway is directly proportional to the area A of the driveway and inversely proportional to the amount of time t allowed to complete the job. Typically, 3 men can lay 1200 square feet of cobblestone in 4 hours. How many men will be required to lay 2400 square feet of cobblestone given 6 hours?
kA t 3 men can lay 1200 square feet of cobblestone in 4 hours. k ⋅1200 3= 4 1 k= 100 SOLUTION: y =
Men required to lay 2400 square feet of cobblestone given 6 hours: y =
kA t
1 ⋅ 2400 100 y= =4 6 ANSWER: To lay 2400 square feet of cobblestone given 6 hours, 4 men are required.
62.
PROBLEM: The volume of a right circular cylinder varies jointly as the square of its radius and its height. A right circular with a 3 cm radius and a 4 cm height has a volume of 36π cm3 . Find a formula for the volume of a right circular cylinder in terms of its radius and height. SOLUTION: Volume of the right circular cylinder is represented by V. Radius of the right circular cylinder is represented by r. V = kr 2 h Right circular with a 3 cm radius and a 4 cm height has a volume of 36π cm3 .
36π = k 32 ⋅ 4 k =π ANSWER: The formula for a right circular cylinder in terms of its radius and height is V = π r 2 h
63.
PROBLEM: The period of a pendulum T is directly proportional to the square root of its length L. If the length of a pendulum is 1 meter, then the period is approximately 2 seconds. Approximate the period of a pendulum that is 0.5 meter in length. SOLUTION: T = k L When the length of a pendulum is 1 meter, the period is approximately 2 seconds. T =k L 2=k 1 k=2 The period of a pendulum that is 0.5 meter in length: T = k L T = 2 0.5 = 1.4
ANSWER: The period of a pendulum that is 0.5 meter in length is 1.4 seconds. Newton's universal law of gravitation states that every particle of matter in the universe attracts every other particle with a force F that is directly proportional to the product of the masses m1 and m2 of the particles and inversely proportional to the square of the distance d between them. The constant of proportionality is called the gravitational constant.
64.
PROBLEM: The time t it takes an object to fall is directly proportional to the square root of the distance d it falls. An object dropped from 4 feet will take 1/2 second to hit the ground. How long will it take an object dropped from 16 feet to hit the ground? SOLUTION: T = k d An object dropped from 4 feet will take 1/2 second to hit the ground. 1 =k 4 2 1 k= 4 If an object is dropped from 16 feet to hit the ground: T = k d 1 T= 16 = 1 4 ANSWER: It takes 1 second for an object dropped from 16 feet to hit the ground.
Newton's universal law of gravitation states that every particle of matter in the universe attracts every other particle with a force F that is directly proportional to the product of the masses m1 and m2 of the particles and inversely proportional to the square of the distance d between them. The constant of proportionality is called the gravitational constant.
65.
PROBLEM: If two objects with masses 50 kg and 100 kg are 1/2 meter apart, then they produce approximately 1.34 ×10−6 Newtons (N) of force. Calculate the gravitational constant.
km1m2 d2 Two objects with masses 50 kg and 100 kg 1/2 meter apart, produce approximately 1.34 ×10−6 Newtons (N) of force. km m F = 12 2 d k 50 ⋅100 1.34 ×10−6 = 2 ⎛1⎞ ⎜ ⎟ ⎝2⎠ k = 6.7 × 10−11 SOLUTION: F =
ANSWER: The gravitational constant is 6.7 × 10−11 N m 2 / kg 2 .
66.
PROBLEM: Use the gravitational constant from the previous exercise to write a formula that approximates the force F, in Newtons, between two masses m1 and m2 , in kilograms, given the distance d between them, in meters. ANSWER: F =
(
F = 6.7 × 10−11
)
km1m2 d2 m1m2 d2
67.
PROBLEM: Calculate the force, in Newtons, between the Earth and Moon given that the mass of the Moon is approximately 7.3 ×1022 kg , the mass of the Earth is approximately 6.0 ×1024 kg , and the distance between them is on average 1.5 × 1011 m .
(
SOLUTION: F = 6.7 × 10−11
) mdm 1
2
2
The mass of the Moon is approximately 7.3 ×1022 kg , the mass of the Earth is approximately 6.0 ×1024 kg , and the distance between them is on average 1.5 × 1011 m . 7.3 × 1022 × 6.0 ×1024 F = 6.7 × 10−11 2 1.5 × 1011
(
)
(
)
F = 1.3 ×1015 ANSWER: The force between the Earth and Moon is 1.3 × 1015 N.
68.
PROBLEM: Calculate the force, in Newtons, between the Earth and Sun given that the mass of the Sun is approximately 2.0 ×1030 kg , the mass of the Earth is approximately 6.0 ×1024 kg , and the distance between them is on average 3.85 ×108 m . SOLUTION: The mass of the Sun is approximately 2.0 ×1030 kg , the mass of the Earth is approximately 6.0 ×1024 kg , and the distance between them is on average 3.85 ×108 m . mm F = 6.7 × 10−11 1 2 2 d 2.0 × 1030 × 6.0 ×1024 F = 6.7 × 10−11 2 3.85 × 108
(
)
(
)
(
)
F = 5.4 ×1027 ANSWER: Force between the Earth and Sun is 5.4 × 1027 N.
69.
PROBLEM: If y varies directly as the square of x, then how does y change if x is doubled? SOLUTION: y = kx 2
If x is doubled, y = k ( 2 x ) = 4kx 2
ANSWER: y changes by a factor of 4
70.
PROBLEM: If y varies inversely as square of t, then how does y change if t is doubled? SOLUTION: y =
If t is doubled, y =
k t2 k
( 2t )
2
=
k 4t 2 1 4
ANSWER: y changes by a factor of
71.
PROBLEM: If y varies directly as the square of x and inversely as the square of t, then how does y change if both x and t are doubled? SOLUTION: y =
x2 t2
( 2x) y= 2 ( 2t )
2
If both x and t are doubled,
ANSWER: y remains unchanged
=
4 x2 x2 = 4t 2 t 2
7.8 Review Exercises and Sample Exam Chapter 7 Review Exercises 7.1 Simplifying Rational Expressions Evaluate for the given set of x‐values.
1.
PROBLEM: 25 ; {−5, 0, 5} 2x 2 ANSWER: Substitute the values in for x. x = −5
x=0
25 25 25 1 = = = 2 2 2x 2 ⋅ 25 2 2 ( −5 ) 2.
25 25 25 = = = undefined 2 2 2x 0 2 ( 0)
x=5
25 25 25 1 = = = 2 2 2x 2 ⋅ 25 2 2 ( 5)
PROBLEM: x−4 ; {1/2, 2, 4} 2x −1 ANSWER: Substitute the values in for x. x=
x=2
1 2
1 7 −4 − 2 = 2 = undefined 1 2 ⋅ −1 0 2 3.
2−4 2 =− 2 ( 2) −1 3
x=4
4−4 0 = =0 2 ( 4) −1 7
PROBLEM: 1 ; {−3, 0, 3} 2 x +9 ANSWER: Substitute the values in for x. x = −3
1
( −3)
2
+9
=
1 18
x=0
1 1 = 0 +9 9 2
x=3
1 1 = 3 + 9 18 2
4.
PROBLEM: x+3 ; {−3, 0, 3} x2 − 9 ANSWER: Substitute the values in for x. x = −3
−3 + 3
( −3)
2
−9
=−
x=0
0 =0 18
0+3 3 1 =− =− 0−9 9 3
x=3
3+3 = undefined 32 − 9
State the restrictions to the domain. 5. PROBLEM: 5 x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x=0 ANSWER: The domain consists of any real number x where x ≠ 0 .
6.
PROBLEM: 1 x ( 3x + 1) SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x ( 3x + 1) = 0
x = 0 or 3x + 1 = 0 1 x = 0 or x = − 3 1 ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ − . 3
7.
PROBLEM: x+2 x 2 − 25 SOLUTION: x+2 x+2 = 2 x − 25 ( x − 5)( x + 5 ) To find the restrictions to the domain, set the denominator equal to zero and solve: ( x − 5)( x + 5) = 0 x − 5 = 0 or x + 5 = 0 x=5
or x = −5
ANSWER: The domain consists of any real number x where x ≠ ±5 .
8.
PROBLEM: x −1 ( x − 1)( 2 x − 3) SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: ( x − 1)( 2 x − 3) = 0
x − 1 = 0 or 2 x − 3 = 0 3 x = 1 or x = 2 ANSWER: The domain consists of any real number x where x ≠ 1 and x ≠
3 . 2
State the restrictions and simplify.
9.
PROBLEM: x −8 x 2 − 64 SOLUTION: x −8 x −8 1 = = 2 x − 64 ( x − 8)( x + 8) x + 8 To find the restrictions to the domain, set the denominator equal to zero and solve: ( x − 8)( x + 8) = 0 x − 8 = 0 or x + 8 = 0 x=8
or x = −8
ANSWER: The domain consists of any real number x where x ≠ ±8 .
10.
PROBLEM: 3x 2 + 9 x 2 x 3 − 18 x SOLUTION: 3 x ( x + 3) 3x 2 + 9 x 3 = = 3 2 x − 18 x 2 x ( x + 3)( x − 3) 2 ( x − 3) To find the restrictions to the domain, set the denominator equal to zero and solve: 2 x ( x − 3)( x + 3) = 0 x = 0 or x − 3 = 0 or x + 3 = 0 x = 0 or x = 3
or x = −3
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ ±3 .
11.
PROBLEM: x 2 − 5 x − 24 x 2 − 3x − 40 SOLUTION: x 2 − 5 x − 24 ( x − 8 )( x + 3) x + 3 = = x 2 − 3x − 40 ( x − 8 )( x + 5 ) x + 5 To find the restrictions to the domain, set the denominator equal to zero and solve: ( x − 8)( x + 5) = 0 x − 8 = 0 or x + 5 = 0 x=8 or x = −5
ANSWER: The domain consists of any real number x where x ≠ 8 and x ≠ −5 .
12.
PROBLEM: 2x2 + 9x − 5 4x2 −1 SOLUTION: 2 x 2 + 9 x − 5 ( 2 x − 1)( x + 5 ) x+5 = = 2 4x −1 ( 2 x − 1)( 2 x + 1) 2 x + 1 To find the restrictions to the domain, set the denominator equal to zero and solve: ( 2 x − 1)( 2 x + 1) = 0
2 x − 1 = 0 or 2x + 1 = 0 1 1 x= or x = − 2 2 1 ANSWER: The domain consists of any real number x where x ≠ ± . 2
13.
PROBLEM: x 2 − 144 12 − x SOLUTION: x 2 − 144 ( x − 12 )( x + 12 ) = = − x − 12 − ( x − 12 ) 12 − x To find the restrictions to the domain, set the denominator equal to zero and solve: 12 − x = 0 x = 12
ANSWER: The domain consists of any real number x where x ≠ 12 .
14.
PROBLEM: 8 x 2 − 10 x − 3 9 − 4 x2 SOLUTION: 8 x 2 − 10 x − 3 ( 4 x + 1)( 2 x − 3) 4x +1 = =− 2 − ( 2 x − 3)( 2 x + 3) 9 − 4x 2x + 3 To find the restrictions to the domain, set the denominator equal to zero and solve: − ( 2 x − 3)( 2 x + 3)
2 x − 3 = 0 or 2x + 3 = 0 3 3 x= or x = − 2 2 3 ANSWER: The domain consists of any real number x where x ≠ ± . 2
15.
PROBLEM:
Given f ( x) =
x −3 find f ( −3) , f ( 0 ) , and f ( 3) . x2 + 9
ANSWER: Substitute the values in for x. f ( x ) = f ( −3 )
f ( x) = 16.
−3 − 3
( −3)
2
+9
=−
f ( x ) = f (0)
6 1 =− 18 3
f ( x) =
0−3
3 1 =− =− (0) + 9 9 3 2
f ( x ) = f ( 3)
f ( x) =
3−3
( 3)
2
+9
=
0 =0 9
PROBLEM:
Simplify g ( x ) =
x 2 − 2 x − 24 and state the restrictions. 2 x 2 − 9 x − 18
SOLUTION: ( x − 6 )( x + 4 ) = x + 4 x 2 − 2 x − 24 g ( x) = 2 = 2 x − 9 x − 18 ( 2 x + 3)( x − 6 ) 2 x + 3 To find the restrictions to the domain, set the denominator equal to zero and solve: ( 2 x + 3)( x − 6 ) = 0
2 x + 3 = 0 or x − 6 = 0 3 x=− or x = 6 2 ANSWER: The domain consists of any real number x where x ≠ −
3 and x ≠ 6 . 2
7.2 Multiplying and Dividing Rational Expressions Multiply. (Assume all denominators are nonzero.)
17.
PROBLEM: 3x5 x − 3 ⋅ x − 3 9 x2 SOLUTION: 3x5 x − 3 3x5 x3 ⋅ = = x − 3 9 x2 9 x2 3 ANSWER:
18.
x3 3
PROBLEM: ( 2 y − 1) 12 y 2 ⋅ 3 3y y ( 2 y − 1) SOLUTION: ( 2 y − 1) = 12 y 2 = 4 12 y 2 ⋅ y 3 ( 2 y − 1) 3y 3y4 y2 ANSWER:
19.
4 y2
PROBLEM: 3x 2 x 2 − 4 x + 4 ⋅ x−2 5 x3 SOLUTION:
3( x − 2) 3x 2 x 2 − 4 x + 4 3x 2 ( x − 2 ) ⋅ = ⋅ = 3 3 x−2 x − 2 5x 5x 5x 2
ANSWER:
3( x − 2) 5x
20.
PROBLEM: x 2 − 8 x + 15 12 x 2 ⋅ 9 x5 x−3 SOLUTION: x 2 − 8 x + 15 12 x 2 ( x − 3)( x − 5 ) 12 x 2 4 ( x − 5 ) ⋅ = ⋅ = 9 x5 x−3 9 x5 x−3 3x3 ANSWER:
21.
4 ( x − 5) 3x3
PROBLEM: x 2 − 36 2 x 2 + 10 x ⋅ x 2 − x − 30 x 2 + 5 x − 6 SOLUTION: x 2 − 36 2 x 2 + 10 x ( x + 6 )( x − 6 ) 2 x ( x + 5 ) ⋅ = ⋅ x 2 − x − 30 x 2 + 5 x − 6 ( x − 6 )( x + 5 ) ( x + 6 )( x − 1)
= ANSWER:
22.
2x x −1
2x x −1
PROBLEM: 9 x 2 + 11x + 2 9 x − 2 ⋅ 2 4 − 81x 2 ( x + 1) SOLUTION: ( 9 x + 2 )( x + 1) ⋅ 9 x − 2 9 x 2 + 11x + 2 9 x − 2 ⋅ = 2 2 2 4 − 81x ( x + 1) − ( 9 x − 2 )( 9 x + 2 ) ( x + 1) =−
ANSWER: −
1 x +1
1 x +1
Divide. (Assume all denominators are nonzero.)
23.
PROBLEM: 9 x 2 − 25 3 x + 5 ÷ 5 x3 15 x 4 SOLUTION: 9 x 2 − 25 3x + 5 9 x 2 − 25 15 x 4 ( 3x + 5 )( 3x − 5 ) 15 x 4 ÷ = ⋅ = ⋅ 5 x3 15 x 4 5 x3 3x + 5 5 x3 3x + 5 = 3x ( 3x − 5) ANSWER: 3 x ( 3 x − 5 )
24.
PROBLEM: 4x2 2x ÷ 2 4x −1 2x −1 SOLUTION: 4x2 2x 4 x2 2 x −1 4 x2 2x −1 ÷ = ⋅ = ⋅ 2 2 4x −1 2x −1 4x −1 2x ( 2 x − 1)( 2 x + 1) 2 x
=
ANSWER:
25.
2x 2x +1
2x 2x +1
PROBLEM: 3x 2 − 13 x − 10 9 x 2 + 12 x + 4 ÷ 2 x 2 − x − 20 x + 8 x + 16 SOLUTION: 3x 2 − 13x − 10 9 x 2 + 12 x + 4 3 x 2 − 13 x − 10 x 2 + 8 x + 16 ÷ 2 = 2 ⋅ x 2 − x − 20 x + 8 x + 16 x − x − 20 9 x 2 + 12 x + 4
( 3x + 2 )( x − 5) ⋅ ( x + 4 ) = ( x − 5)( x + 4 ) ( 3x + 2 )2 2
= ANSWER:
x+4 3x + 2
x+4 3x + 2
26.
PROBLEM: 2 x 2 + xy − y 2 4x2 − y 2 ÷ x 2 + 2 xy + y 2 3x 2 + 2 xy − y 2 SOLUTION: 2 x 2 + xy − y 2 4 x2 − y 2 2 x 2 + xy − y 2 3x 2 + 2 xy − y 2 ÷ = ⋅ x 2 + 2 xy + y 2 3 x 2 + 2 xy − y 2 x 2 + 2 xy + y 2 4 x2 − y 2
ANSWER:
27.
=
( 2 x − y )( x + y ) ⋅ ( 3x − y )( x + y ) 2 ( 2 x − y )( 2 x + y ) ( x + y)
=
3x − y 2x + y
3x − y 2x + y
PROBLEM: 2 x 2 − 6 x − 20 ÷ 8 x 2 + 39 x − 5 2 8 x + 17 x + 2
(
)
SOLUTION: 2 x 2 − 6 x − 20 2 x 2 − 6 x − 20 1 2 ÷ 8 x + 39 x − 5 = ⋅ 2 2 2 8 x + 17 x + 2 8 x + 17 x + 2 8 x + 39 x − 5
(
)
(
=
ANSWER:
2 ( x − 5)
2 ( x − 5 )( x + 2 )
⋅
1 (8 x − 1)( x + 5)
(8 x + 1)( x + 2 ) 2 ( x − 5) = (8 x + 1)(8 x − 1)( x + 5)
(8 x + 1)(8 x − 1)( x + 5)
)
28.
PROBLEM: 12 x 2 − 27 x 4 ÷ 3x 2 + x − 2 4 3 15 x + 10 x
(
)
SOLUTION: 12 x 2 − 27 x 4 12 x 2 − 27 x 4 1 2 ÷ 3x + x − 2 = ⋅ 4 3 4 3 2 15 x + 10 x 15 x + 10 x 3x + x − 2
(
)
(
3x ( 2 + 3 x )( 2 − 3 x ) 2
=
5 x ( 3x + 2 ) 3
=− ANSWER: −
29.
⋅
)
1 ( 3x − 2 )( x + 1)
3 5 x ( x + 1)
3 5 x ( x + 1)
PROBLEM: 25 y 2 − 1 1 10 y 2 ⋅ ÷ 5 y 4 ( y − 2 ) 5 y − 1 ( y − 2 )2 SOLUTION:
( y − 2 ) = ( 5 y − 1)( 5 y + 1) ⋅ 1 ⋅ ( y − 2 ) 25 y 2 − 1 1 10 y 2 25 y 2 − 1 1 ⋅ ÷ = ⋅ ⋅ 2 4 4 5 y ( y − 2) 5 y −1 ( y − 2) 5 y ( y − 2 ) 5 y − 1 10 y 2 5 y4 ( y − 2) 5 y − 1 10 y 2 2
=
ANSWER:
( 5 y + 1)( y − 2 ) 50 y 6
( 5 y + 1)( y − 2 ) 50 y 6
2
30.
PROBLEM: 10 x 4 5x2 x −1 ÷ ⋅ 2 2 1 − 36 x 6 x − 7 x + 1 2 x SOLUTION: x −1 10 x 4 5x2 10 x 4 6 x 2 − 7 x + 1 x − 1 ÷ ⋅ = ⋅ ⋅ 1 − 36 x 2 6 x 2 − 7 x + 1 2 x 1 − 36 x 2 5x2 2x 4 ( 6 x − 1)( x − 1) ⋅ x − 1 10 x = ⋅ − (1 + 6 x )( 6 x − 1) 5x2 2x
=− ANSWER: −
31.
x ( x − 1)
x ( x − 1)
2
1+ 6x
2
1+ 6x
PROBLEM:
16 x 2 − 9 x 2 + 3x − 10 and g ( x ) = 2 , calculate ( f ⋅ g )( x ) and state x+5 4x + 5x − 6 the restrictions. Given f ( x ) =
SOLUTION: 16 x 2 − 9 x 2 + 3x − 10 f ( x) = and g ( x ) = 2 x+5 4x + 5x − 6 2 2 16 x − 9 x + 3x − 10 ( 4 x + 3)( 4 x − 3) ( x + 5 )( x − 2 ) ⋅ = ⋅ ( f ⋅ g ) ( x) = x + 5 4x2 + 5x − 6 ( x + 5) ( 4 x − 3)( x + 2 )
=
( 4 x + 3)( x − 2 )
x+2 In this case, the domain of f ( x) consists of all real numbers except –5, and the 3 domain of g ( x) consists all real numbers except and –2. 4 ANSWER: ( f ⋅ g ) ( x) =
( 4 x + 3)( x − 2 ) x+2
3 The domain consists of any real number x where x ≠ −5, x = , and x ≠ −2. 4
32.
PROBLEM:
Given f ( x ) =
x+7 x 2 − 49 and g ( x ) = , calculate ( f / g )( x ) and state the 5x − 1 25 x 2 − 5 x
restrictions. SOLUTION: x+7 x 2 − 49 f ( x) = and g ( x ) = 5x − 1 25 x 2 − 5 x x+7 x + 7 25 x 2 − 5 x x + 7 5 x ( 5 x − 1) ⋅ = ⋅ ( f / g ) ( x) = 52x − 1 = x − 49 5 x − 1 x 2 − 49 5 x − 1 ( x + 7 )( x − 7 ) 25 x 2 − 5 x 5x = x−7
In this case, the domain of f ( x) consists of all real numbers except domain of g ( x) consists all real numbers except 0 and reciprocal of g ( x) has a restriction of ±7 . ANSWER: ( f / g ) ( x) =
1 . In addition, the 5
5x x−7
The domain consists of any real number x where x ≠ 0 , x ≠ 7.3 Adding and Subtracting Rational Expressions Simplify. (Assume all denominators are nonzero.)
33.
PROBLEM: 5x 3 − y y SOLUTION: 5x 3 5x − 3 − = y y y ANSWER:
5x − 3 y
1 , and the 5
1 , and x ≠ ±7 . 5
34.
PROBLEM: x 3 − 2 2 x − x−6 x − x−6 SOLUTION: x 3 x −3 x−3 − 2 = 2 = 2 x − x − 6 x − x − 6 x − x − 6 ( x − 3)( x + 2 )
= ANSWER:
35.
1 x+2
1 x+2
PROBLEM: 2x 1 + 2x + 1 x − 5 SOLUTION: LCD = ( 2 x + 1)( x − 5 )
To obtain equivalent terms with this common denominator, we will multiply the x−5 2x +1 first term by . and the second term by x−5 2x +1 2x 1 2x x − 5 1 2 x + 1 2 x 2 − 10 x + 2 x + 1 + = ⋅ + ⋅ = 2x +1 x − 5 2x +1 x − 5 x − 5 2x +1 ( 2 x + 1)( x − 5)
2 x2 − 8x + 1 = ( 2 x + 1)( x − 5) 2 x2 − 8x + 1 ANSWER: ( 2 x + 1)( x − 5)
36.
PROBLEM: 3 1 − 2x + 2 x−7 x SOLUTION: LCD = x 2 ( x − 7 )
To obtain equivalent terms with this common denominator, we will multiply the x2 x−7 . first term by 2 and the second term by x−7 x 3 1− 2x 3 x 2 1 − 2 x x − 7 3x 2 − 2 x 2 + 15 x − 7 + 2 = ⋅ + 2 ⋅ = x−7 x x − 7 x2 x x−7 x2 ( x − 7 )
= ANSWER:
37.
x 2 + 15 x − 7 x2 ( x − 7 )
x 2 + 15 x − 7 x2 ( x − 7 )
PROBLEM: 7x 2 − 2 4x − 9x + 2 x − 2 SOLUTION: 7x 2 7x 2 − = − 2 4 x − 9 x + 2 x − 2 ( 4 x − 1)( x − 2 ) x − 2
LCD = ( 4 x − 1)( x − 2 ) To obtain equivalent terms with this common denominator, we will multiply the 1 4x −1 first term by and the second term by . 1 4x −1 7x 2 7x 1 2 4x −1 − = ⋅ − ⋅ ( 4 x − 1)( x − 2 ) x − 2 ( 4 x − 1)( x − 2 ) 1 x − 2 4 x − 1 =
ANSWER: −
1 4x −1
7 x − 8x + 2 ( 4 x − 1)( x − 2 )
=−
x−2 ( 4 x − 1)( x − 2 )
=−
1 4x −1
38.
PROBLEM: 5 20 − 9 x + 2 x − 5 2 x − 15 x + 25 SOLUTION: 5 20 − 9 x 5 20 − 9 x + 2 = + x − 5 2 x − 15 x + 25 x − 5 ( 2 x − 5 )( x − 5 )
LCD = ( 2 x − 1)( x − 5 ) To obtain equivalent terms with this common denominator, we will multiply the 2x − 5 1 first term by and the second term by . 2x − 5 1 5 20 − 9 x 5 2x − 5 20 − 9 x 1 + = ⋅ + ⋅ x − 5 ( 2 x − 5 )( x − 5 ) x − 5 2 x − 5 ( 2 x − 5 )( x − 5 ) 1
ANSWER:
1 2x − 5
=
10 x − 25 + 20 − 9 x ( 2 x − 5)( x − 5)
=
x −5 ( 2 x − 5)( x − 5)
=
1 2x − 5
39.
PROBLEM: 5 ( x − 3) x 2 − − 2 x − 5 x − 3 x − 8 x + 15 SOLUTION: 5 ( x − 3) 5 ( x − 3) x 2 x 2 − − 2 = − − x − 5 x − 3 x − 8 x + 15 x − 5 x − 3 ( x − 3)( x − 5 )
LCD = ( x + 3)( x − 5 )( x − 3) To obtain equivalent terms with this common denominator, we will multiply the x−3 x −5 1 first term by , and the third term by . , the second term by x−3 x −5 1 5 ( x − 3) 5 ( x − 3) x 2 x x−3 2 x−5 1 − − = ⋅ − ⋅ − ⋅ x − 5 x − 3 ( x + 3)( x − 5 ) x − 5 x − 3 x − 3 x − 5 ( x − 3)( x − 5 ) 1 = =
x 2 − 3x − 2 x + 10 − 5 x + 15 ( x − 3)( x − 5) x 2 − 10 x + 25 ( x − 3)( x − 5)
( x − 5) = ( x − 3)( x − 5) 2
=
ANSWER:
x−5 x−3
x −5 x−3
40.
PROBLEM: x − 4 12 ( 2 − x ) 3x − + 2 x − 1 x + 4 2 x2 + 7 x − 4 SOLUTION: 12 ( 2 − x ) 3x x − 4 12 ( 2 − x ) 3x x−4 − + 2 = − + 2 x − 1 x + 4 2 x + 7 x − 4 2 x − 1 x + 4 ( 2 x − 1)( x + 4 )
LCD = ( 2 x − 1)( x + 4 ) To obtain equivalent terms with this common denominator, we will multiply the x+4 2x −1 1 first term by , and the third term by . , the second term by x+4 2x −1 1 12 ( 2 − x ) 12 ( 2 − x ) 3x x−4 3x x + 4 x − 4 2 x − 1 1 − + = ⋅ − ⋅ + ⋅ 2 x − 1 x + 4 ( 2 x − 1)( x + 4 ) 2 x − 1 x + 4 x + 4 2 x − 1 ( 2 x − 1)( x + 4 ) 1 3x 2 + 12 x − 2 x 2 + 9 x − 4 + 24 − 12 x = ( 2 x − 1)( x + 4 )
ANSWER:
x+5 2x −1
=
x 2 + 9 x + 20 ( 2 x − 1)( x + 4 )
=
( x + 5)( x + 4 ) ( 2 x − 1)( x + 4 )
=
x+5 2x −1
41.
PROBLEM: 1 1 − 2 2 x + 8 x − 9 x + 11x + 18 SOLUTION: 1 1 1 1 − 2 = − 2 x + 8 x − 9 x + 11x + 18 ( x + 9 )( x − 1) ( x + 9 )( x + 2 )
LCD = ( x + 9 )( x + 2 )( x − 1) To obtain equivalent terms with this common denominator, we will multiply the x+2 x −1 first term by . and the second term by x+2 x −1 1 1 1 x+2 1 x −1 − = ⋅ − ⋅ ( x + 9 )( x − 1) ( x + 9 )( x + 2 ) ( x + 9 )( x − 1) x + 2 ( x + 9 )( x + 2 ) x − 1
ANSWER:
=
x + 2 − x +1 ( x + 9 )( x + 2 )( x − 1)
=
3 ( x + 9 )( x + 2 )( x − 1)
3 ( x + 9 )( x + 2 )( x − 1)
42.
PROBLEM: 4 3 + 2 2 x + 13x + 36 x + 6 x − 27 SOLUTION: 4 3 4 3 + 2 = + 2 x + 13x + 36 x + 6 x − 27 ( x + 4 )( x + 9 ) ( x + 9 )( x − 3)
LCD = ( x + 9 )( x − 3)( x + 4 ) To obtain equivalent terms with this common denominator, we will multiply the x−3 x+4 first term by . and the second term by x−3 x+4 4 3 4 3 x−3 x+4 + = ⋅ + ⋅ ( x + 4 )( x + 9 ) ( x + 9 )( x − 3) ( x + 4 )( x + 9 ) x − 3 ( x + 9 )( x − 3) x + 4
ANSWER:
=
4 x − 12 + 3x + 12 ( x + 9 )( x − 3)( x + 4 )
=
7x ( x + 9 )( x − 3)( x + 4 )
7x ( x + 9 )( x − 3)( x + 4 )
43.
PROBLEM: y +1 1 2y − + 2 y+2 2− y y −4 SOLUTION: y +1 1 2y y +1 1 2y − + 2 = + + y + 2 2 − y y − 4 y + 2 y − 2 ( y + 2 )( y − 2 )
LCD = ( y + 2 )( y − 2 ) To obtain equivalent terms with this common denominator, we will multiply the 1 y−2 y+2 first term by and the third term by . , the second term by 1 y−2 y+2 y +1 1 2y y +1 y − 2 1 y+2 2y 1 + + = ⋅ + ⋅ + ⋅ y + 2 y − 2 ( y + 2 )( y − 2 ) y + 2 y − 2 y − 2 y + 2 ( y + 2 )( y − 2 ) 1 =
y2 − y − 2 + y + 2 + 2 y ( y + 2 )( y − 2 )
y2 + 2 y = ( y + 2 )( y − 2 ) = =
ANSWER:
y y−2
y ( y + 2)
( y + 2 )( y − 2 ) y y−2
44.
PROBLEM: 1 1 2 − − 2 y 1 − y y −1 SOLUTION: 1 1 2 1 1 2 − − 2 = + − y 1 − y y − 1 y y − 1 ( y + 1)( y − 1)
LCD = y ( y + 1)( y − 1) To obtain equivalent terms with this common denominator, we will multiply the ( y + 1)( y − 1) , the second term by y ( y + 1) and the third term by first term by y ( y + 1) ( y + 1)( y − 1) y . y 1 1 2 1 ( y + 1)( y − 1) 1 y ( y + 1) 2 y + − = ⋅ + ⋅ − ⋅ y y − 1 ( y + 1)( y − 1) y ( y + 1)( y − 1) y − 1 y ( y + 1) ( y + 1)( y − 1) y y2 −1 + y2 + y − 2 y = y ( y + 1)( y − 1)
ANSWER:
2 y +1 y ( y + 1)
=
2 y2 − y −1 y ( y + 1)( y − 1)
=
( 2 y + 1)( y − 1) y ( y + 1)( y − 1)
=
2 y +1 y ( y + 1)
45.
PROBLEM:
Given f ( x ) =
x +1 x and g ( x ) = , calculate ( f + g )( x ) and state the x +1 2x − 5
restrictions. SOLUTION:
x x +1 + 2x − 5 x + 1 LCD = ( 2 x − 5 )( x + 1) To obtain equivalent terms with this common denominator, we will multiply the x +1 2x − 5 first term by . and the second term by x +1 2x − 5 x +1 x x +1 x +1 x 2x − 5 + + = ⋅ ⋅ 2x − 5 x +1 2x − 5 x +1 x +1 2x − 5 x2 + 2 x + 1 + 2 x2 − 5x = ( 2 x − 5)( x + 1)
( f + g ) ( x) =
=
3x 2 − 3x + 1 ( 2 x − 5)( x + 1)
5 In this case, the domain of f ( x) consists of all real numbers except , and the 2 g ( x ) consists all real numbers except –1. domain of ANSWER: ( f + g ) ( x) =
3x 2 − 3x + 1 ( 2 x − 5)( x + 1)
The domain consists of any real number x where x ≠
5 and x ≠ −1. 2
46.
PROBLEM:
Given f ( x ) =
x +1 2 and g ( x ) = , calculate ( f − g )( x ) and state the x −8 3x
restrictions. SOLUTION: 2 x +1 − ( f − g ) ( x) = 3x x − 8 LCD = 3x ( x − 8)
To obtain equivalent terms with this common denominator, we will multiply the x+8 3x first term by and the second term by . x+8 3x x +1 2 x +1 2 − = ⋅ 3x ( x − 8) − ⋅ 3x ( x − 8) 3x x − 8 3x x −8 x2 − 7 x − 8 − 6 x = 3x ( x − 8) x 2 − 13 x − 8 3x ( x − 8) In this case, the domain of f ( x) consists of all real numbers except 0 , and the domain of g ( x) consists all real numbers except 8. =
x 2 − 13x − 8 3x ( x − 8 ) The domain consists of any real number x where x ≠ 0 and x ≠ 8. ANSWER: ( f − g ) ( x) =
7.4 Complex Fractions Simplify.
47.
PROBLEM: 2 4− x 2x −1 3x SOLUTION: 2 4x − 2 4− 4 x − 2 3x x = x = ⋅ 2x −1 2x −1 2x −1 x 3x 3x 6 ( 2 x − 1) = 2x −1 =6 ANSWER: 6
48.
PROBLEM: 1 1 − 3 3y 1 1 − 5 5y SOLUTION: 1 1 y −1 − y −1 5 y 5 3 3y 3y = = ⋅ = 1 1 y −1 3 y y −1 3 − 5 5y 5y ANSWER:
5 3
49.
PROBLEM: 1 1 + 6 x 1 1 − 36 x 2 SOLUTION: 1 1 x+6 + 2 6 x = 6 x = x + 6 ⋅ 36 x 2 1 1 6 x x 2 − 36 − 2 x − 36 36 x 36 x 2 36 x 2 x+6 = ⋅ 6 x ( x + 6 )( x − 6 ) =
6x x−6
ANSWER:
50.
6x x−6
PROBLEM: 1 1 − 2 100 x 1 1 − 10 x SOLUTION: 1 1 x 2 − 100 − 2 2 100 x = 100 x 2 = x − 100 ⋅ 10 x 1 1 x − 10 100 x 2 x − 10 − 10 x 10 x ( x − 10 )( x + 10 ) ⋅ 10 x = 100 x 2 x − 10 x + 10 = 10 x ANSWER:
x + 10 10 x
51.
PROBLEM: x 2 − x + 3 x +1 x 1 + x+4 x+3 SOLUTION:
x2 + x − 2x − 6 x 2 x x +1 2 x+3 − ⋅ − x + 3 x + 1 = x + 3 x + 1 x + 1 x + 3 = ( x + 3)( x + 1) x 1 x x+3 1 x + 4 x 2 + 3x + x + 4 + + ⋅ x+4 x+3 x + 4 x + 3 x + 3 x + 4 ( x + 4 )( x + 3) x2 − x − 6 ( x + 3)( x + 1) = x2 + 4 x + 4 ( x + 4 )( x + 3) =
x 2 − x − 6 ( x + 4 )( x + 3) ⋅ ( x + 3)( x + 1) x 2 + 4 x + 4
( x − 3)( x + 2 ) ⋅ ( x + 4 )( x + 3) ( x + 3)( x + 1) ( x + 2 )2 ( x + 4 )( x − 3) = ( x + 1)( x + 2 ) =
ANSWER:
( x + 4 )( x − 3) ( x + 1)( x + 2 )
52.
PROBLEM: 3 1 − x x −5 5 2 − x+2 x SOLUTION:
3 ( x − 5) 1 x 3 x − 15 − x 3 1 ⋅ − ⋅ − x x − 5 = x ( x − 5) x − 5 x = x ( x − 5) 5 2 5 x 2 x+2 5x − 2 x − 4 − ⋅ − ⋅ x+2 x x+2 x x x+2 x ( x + 2) 2 x − 15 x ( x − 5) = 3x − 4 x ( x + 2) =
2 x − 15 x ( x + 2 ) ⋅ x ( x − 5) 3x − 4
=
( 2 x − 15)( x + 2 ) ( x − 5)( 3x − 4 )
ANSWER:
( 2 x − 15)( x + 2 ) ( x − 5)( 3x − 4 )
53.
PROBLEM: 12 35 1− + 2 x x 25 1− 2 x SOLUTION: ⎛ 12 35 ⎞ 2 12 35 1− + 2 ⎜1 − + 2 ⎟ ⋅ x x 2 − 12 x + 35 x x ⎠ x x = ⎝ = 25 x 2 − 25 ⎛ 25 ⎞ 2 1− 2 1 − ⋅ x ⎜ 2 ⎟ x ⎝ x ⎠ ( x − 7 )( x − 5) = ( x + 5)( x − 5) =
ANSWER:
54.
x−7 x+5
x−7 x+5
PROBLEM: 15 25 2− + 2 x x 2x − 5 SOLUTION: 15 25 ⎛ 15 25 ⎞ 2 2 − + 2 ⎜2 − + 2 ⎟⋅ x 2 x 2 − 15 x + 25 x x ⎠ x x =⎝ = 2x − 5 x 2 ( 2 x − 5) ( 2 x − 5) ⋅ x2
=
( 2 x − 5)( x − 5) x 2 ( 2 x − 5)
=
x −5 x2
ANSWER:
x−5 x2
7.5 Solving Rational Equations Solve.
55.
PROBLEM: 6 2 = x − 6 2x −1
1 . 2 Multiply both sides with the LCD, ( x − 6 )( 2 x − 1) .
SOLUTION: We first note that x ≠ 6 and x ≠
6 2 ⋅ ( x − 6 )( 2 x − 1) = ⋅ ( x − 6 )( 2 x − 1) x−6 2x −1 12 x − 6 = 2 x − 12 10 x = −6 6 x=− 10 3 x=− 5 3 We can check our answer by substituting − in for x to see if we obtain a true 5 statement. 6 2 = 3 3 − − 6 2 ⋅ ⎛⎜ − ⎞⎟ − 1 5 ⎝ 5⎠ 10 10 − =− 11 11 3 ANSWER: The solution is − . 5
56.
PROBLEM: x x+2 = x−6 x−2 SOLUTION: We first note that x ≠ 6 and x ≠ 2 . Multiply both sides with the LCD, ( x − 6 )( x − 2 ) .
x x+2 ⋅ ( x − 6 )( x − 2 ) = ⋅ ( x − 6 )( x − 2 ) x−6 x−2 x 2 − 2 x = x 2 − 4 x − 12 2 x = −12 x = −6 We can check our answer by substituting −6 in for x to see if we obtain a true statement. −6 −6 + 2 = −6 − 6 −6 − 2 −6 −4 = −12 −8 1 1 = 2 2 ANSWER: The solution is −6 .
57.
PROBLEM: 1 2 1 − = 3x 9 x SOLUTION: We first note that x ≠ 0 . Multiply both sides with the LCD, 9x . 1 ⎛ 1 2⎞ ⎜ − ⎟ 9x = ⋅ 9x x ⎝ 3x 9 ⎠ 3 − 2x = 9
2 x = −6 x = −3 We can check our answer by substituting −3 in for x to see if we obtain a true statement. 1 2 1 − =− 3 ( −3) 9 3 3 1 − =− 9 3 1 1 − =− 3 3 ANSWER: The solution is −3 .
58.
PROBLEM: 2 3 1 + = x −5 5 x −5 SOLUTION: We first note that x ≠ 5 . Multiply both sides with the LCD, 5 ( x − 5 ) .
3⎞ 1 ⎛ 2 + ⎟ 5 ( x − 5) = ⋅ 5 ( x − 5) ⎜ x −5 ⎝ x−5 5⎠ 10 + 3x − 15 = 5 3x = 10 10 x= 3 We can check our answer by substituting statement. 2 3 1 + = 10 10 −5 5 −5 3 3 6 3 3 − + =− 5 5 5 3 3 − =− 5 5 ANSWER: The solution is
10 . 3
10 in for x to see if we obtain a true 3
59.
PROBLEM: x 4 10 + =− 2 x −5 x +5 x − 25 SOLUTION: x 4 10 + =− 2 x −5 x +5 x − 25 x 4 10 + =− x −5 x +5 ( x − 5)( x + 5)
We first note that x ≠ ±5 . Multiply both sides with the LCD, ( x − 5 )( x + 5 ) .
4 ⎞ 10 ⎛ x + ⋅ ( x − 5)( x + 5) ⎜ ⎟ ( x − 5)( x + 5) = − ( x − 5)( x + 5) ⎝ x−5 x +5⎠ x 2 + 5 x + 4 x − 20 = −10 x 2 + 9 x − 10 = 0
( x − 1)( x + 10) = 0 x = 1 or x = −10 Check x = 1 1 4 10 + =− 1− 5 1+ 5 (1 − 5)(1 + 5) 4 1 5 − = 6 4 12 8−3 5 = 12 12 5 5 = 12 12 ANSWER: The solution is 1 or –10.
Check x = −10 −10 4 10 + =− −10 − 5 −10 + 5 ( −10 − 5 )( −10 + 5 ) 2 4 2 − =− 3 5 15 2 2 − =− 15 15
60.
PROBLEM: 2 12 2 − 3x 2 − = 2 x 2 x + 3 2 x + 3x SOLUTION: 2 12 2 − 3x 2 − = 2 x 2 x + 3 2 x + 3x 2 12 2 − 3x 2 − = x 2 x + 3 x ( 2 x + 3)
3 We first note that x ≠ 0 and x ≠ − . 2 Multiply both sides with the LCD, x ( 2 x + 3) .
12 ⎞ 2 − 3x 2 ⎛2 ⋅ x ( 2 x + 3) ⎜ − ⎟ x ( 2 x + 3) = x ( 2 x + 3) ⎝ x 2x + 3 ⎠ 4 x + 6 − 12 x = 2 − 3x 2 3x 2 − 8 x + 4 = 0
( 3x − 2 )( x − 2) = 0 x=
2 or x = 2 3 Check x =
2 3 2
⎛2⎞ 2 − 3⎜ ⎟ 2 12 ⎝3⎠ − = 2 2 2⎛ 2 ⎞ 2⋅ + 3 2 ⋅ + 3⎟ ⎜ 3 3 3⎝ 3 ⎠ 36 1 3− = 13 13 3 3 3 = 13 13
ANSWER: The solution is
2 or 2. 3
Check x = 2 2 12 2 − 3 ⋅ 22 − = 2 2 ⋅ 2 + 3 2 ( 2 ⋅ 2 + 3) 12 10 =− 7 14 5 5 − =− 7 7
1−
61.
PROBLEM: x +1 x−6 + =1 2 ( x − 2) x SOLUTION: We first note that x ≠ 0 and x ≠ 2 . Multiply both sides with the LCD, 2 x ( x − 2 ) .
⎛ x +1 x−6⎞ + ⎜⎜ ⎟ 2 x ( x − 2 ) = 1⋅ 2 x ( x − 2 ) x ⎟⎠ ⎝ 2 ( x − 2) x 2 + x + 2 x 2 − 16 x + 24 = 2 x 2 − 4 x x 2 − 11x + 24 = 0
( x − 8)( x − 3) = 0 x = 8 or x = 3 Check x = 8 8 +1 8−6 + =1 2 (8 − 2 ) 8 9 1 + =1 12 4 1=1
ANSWER: The solution is 8 or 3.
Check x = 3 3 +1 3−6 + =1 2 (3 − 2) 3 2 −1 = 1 1=1
62.
PROBLEM: x 5x + 2 − =4 x +1 x + 4 SOLUTION: We first note that x ≠ −1 and x ≠ −4 . Multiply both sides with the LCD, ( x + 1)( x + 4 ) .
x ⎞ ⎛ 5x + 2 − ⎜ ⎟ ( x + 1)( x + 4 ) = 4 ⋅ ( x + 1)( x + 4 ) ⎝ x +1 x + 4 ⎠ 5 x 2 + 22 x + 8 − x 2 − x = 4 x 2 + 20 x + 16 x =8 We can check our answer by substituting 8 in for x to see if we obtain a true statement. 5⋅8 + 2 8 − =4 8 +1 8 + 4 14 2 − =4 3 3 4=4 ANSWER: The solution is 8 .
63.
PROBLEM: x 1 4x − 7 + = 2 x + 5 x − 4 x + x − 20 SOLUTION: x 1 4x − 7 + = x + 5 x − 4 ( x + 5)( x − 4 ) We first note that x ≠ −5 and x ≠ 4 . Multiply both sides with the LCD, ( x + 5 )( x − 4 ) .
1 ⎞ 4x − 7 ⎛ x + ⋅ ( x + 5)( x − 4 ) ⎜ ⎟ ⋅ ( x + 5)( x − 4 ) = ( x + 5)( x − 4) ⎝ x+5 x−4⎠ x2 − 4x + x + 5 = 4 x − 7 x 2 − 7 x + 12 = 0
( x − 3)( x + 4) = 0 x = 3 or x = −4 x ≠ 4. ANSWER: Hence the only solution is 3.
We can check our answer by substituting 3 in for x to see if we obtain a true statement. 3 1 4⋅3 − 7 + = 3 + 5 3 − 4 ( 3 + 5 )( 3 − 4 )
3 5 −1 = − 8 8 5 5 − =− 8 8
64.
PROBLEM: 2 (3 − 4x ) x 2 + = 2 3x − 1 2 x + 1 6 x + x − 1 SOLUTION: 2 (3 − 4x ) 2 x + = 2 3x − 1 2 x + 1 6 x + x − 1 2 (3 − 4x ) 2 x + = 3x − 1 2 x + 1 ( 3 x − 1)( 2 x + 1)
1 1 and x ≠ − . 3 2 Multiply both sides with the LCD, ( 3 x − 1)( 2 x + 1) . We first note that x ≠
2 (3 − 4x ) x ⎞ ⎛ 2 + ( 3x − 1)( 2 x + 1) ⎜ ⎟ ( 3x − 1)( 2 x + 1) = ( 3x − 1)( 2 x + 1) ⎝ 3x − 1 2 x + 1 ⎠ 4 x + 2 + 3x 2 − x = 6 − 8 x
3x 2 + 11x − 4 = 0
( x + 4)( 3x − 1) = 0 x = −4 or x =
1 3
1 x≠ . 3 ANSWER: Hence the only solution is –4. We can check our answer by substituting −4 in for x to see if we obtain a true statement. 2 ( 3 − 4 ( −4 ) ) ( −4 ) = 2 + 3 ( −4 ) − 1 2 ( −4 ) + 1 ( 3 ( −4 ) − 1) ( 2 ( −4 ) + 1)
2 4 2 ⋅19 + = 13 7 13 ⋅ 7 38 38 = 91 91
−
65.
PROBLEM: x 1 2x + = 2 x −1 x + 1 x −1 SOLUTION: x 1 2x + = 2 x −1 x + 1 x −1 1 2x x + = x − 1 x + 1 ( x + 1)( x − 1)
We first note that x ≠ ±1 . Multiply both sides with the LCD, ( x + 1)( x − 1) .
1 ⎞ 2x ⎛ x + ( x + 1)( x − 1) ⎜ ⎟ ( x + 1)( x − 1) = ( x + 1)( x − 1) ⎝ x −1 x +1 ⎠ x2 + x + x −1 = 2x x2 −1 = 0
( x + 1)( x − 1) = 0 x = −1 or x = 1 x ≠ ±1 . ANSWER: Hence, the solution is∅.
66.
PROBLEM: 2x 1 4 − 7x − = 2 x + 5 2 x − 3 2 x + 7 x − 15 SOLUTION: 2x 1 4 − 7x − = 2 x + 5 2 x − 3 2 x + 7 x − 15 2x 1 4 − 7x − = x + 5 2 x − 3 ( x + 5 )( 2 x − 3)
3 . 2 Multiply both sides with the LCD, ( x + 5 )( 2 x − 3) . We first note that x ≠ −5 and x ≠
1 ⎞ 4 − 7x ⎛ 2x − ⋅ ( x + 5)( 2 x − 3) ⎜ ⎟ ( x + 5)( 2 x − 3) = ( x + 5)( 2 x − 3) ⎝ x + 5 2x − 3 ⎠ 4x2 − 6x − x − 5 = 4 − 7 x 4x2 − 9 = 0
( 2 x + 3)( 2 x − 3) = 0 x=± x≠
3 2
3 . 2
3 ANSWER: Hence the only solution is − . 2 We can check our answer by substituting −
3 in for x to see if we obtain a true 2
statement. ⎛ 3⎞ ⎛ 3⎞ 2⎜ − ⎟ 4 − 7⎜− ⎟ 1 ⎝ 2⎠ − ⎝ 2⎠ = 3 3 − + 5 2 ⎛⎜ − ⎞⎟ − 3 ⎜⎛ ⎛⎜ − 3 ⎞⎟ + 5 ⎟⎞ ⎜⎛ 2 ⎛⎜ − 3 ⎞⎟ − 3 ⎟⎞ 2 ⎝ 2⎠ ⎝⎝ 2 ⎠ ⎠⎝ ⎝ 2 ⎠ ⎠ 29 6 1 − + =− 2 7 6 21 29 29 − =− 42 42
67.
PROBLEM: 1 1 1 Solve for a: = + a b c SOLUTION: Multiply both the sides with the LCD, abc. 1 ⎛1 1⎞ ⋅ abc = ⎜ + ⎟ ⋅ abc a ⎝b c⎠ bc = ac + ab
bc = a ( c + b ) a=
bc c+b
ANSWER: a =
68.
PROBLEM:
Solve for y: x =
bc c+b 2 y −1 3y
SOLUTION: 2 y −1 x= 3y 3xy = 2 y − 1 2 y − 3xy = 1 y ( 2 − 3x ) = 1 y=
1 2 − 3x
ANSWER: y =
1 2 − 3x
7.6 Applications of Rational Equations Use algebra to solve the following applications.
69.
PROBLEM: A positive integer is twice that of another. The sum of the reciprocals of the two positive integers is 1/4. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: 2n 1 1 1 + = n 2n 4 We first note that n ≠ 0. Multiply both sides with the LCD, 2n. 1 ⎛1 1 ⎞ ⎜ + ⎟ 2n = ⋅ 2n 4 ⎝ n 2n ⎠ n 2 +1 = 2 n=6 Larger integer: 2n = 2 ⋅ 6 = 12 ANSWER: n = 6 , 2n = 12
70.
PROBLEM: If the reciprocal of the smaller of two consecutive integers is subtracted from three times the reciprocal of the larger, the result is 3/10. Find the integers. SOLUTION: Smaller integer is represented by n. Larger integer: n + 1 3 1 3 − = n + 1 n 10 We first note that n ≠ 0 and n ≠ 1 . Multiply both the sides with the LCD, 10n ( n + 1) .
1⎞ 3 ⎛ 3 − ⎟10n ( n + 1) = ⋅10n ( n + 1) ⎜ 10 ⎝ n +1 n ⎠ 2 30n − 10n − 10 = 3n + 3n 3n 2 − 17 n + 10 = 0
( 3n − 2 )( n − 5) = 0 2 or n = 5 3 The problem speaks about integers. Hence, the solution is 5. Larger integer: n + 1 = 5 +1 = 6 n=
ANSWER: The two integers are 5 and 6.
71.
PROBLEM: Mary can jog, on average, 2 mph faster than her husband James. James can jog 6.6 miles in the same amount of time it takes Mary to jog 9 miles. How fast, on average, can Mary jog? SOLUTION: James’ speed is represented by x. Mary’ speed: x + 2 D 9 Time taken by Mary to cover 9 miles: t = = r x+2 D 6.6 Time taken by James to cover 6.6 miles: t = = r x
Time taken by Mary to cover 9 miles = Time taken by James to cover 6.6 miles 9 6.6 = x+2 x We first note that x ≠ 0 and x ≠ −2 . Multiply both the sides with the LCD, x ( x + 2 ) . 9 x = 6.6 ( x + 2 ) 9 x = 6.6 x + 13.2 90 x = 66 x + 132 24 x = 132 x = 5.5
ANSWER: Mary’s speed: 5.5 + 2 = 7.5 miles/hour
72.
PROBLEM: Billy traveled 140 miles to visit his grandmother on the bus and then drove the 140 miles back in a rental car. On average, the bus was 14 mph slower than the car. If the total time spent traveling was 4.5 hours, then what was the average speed of the bus? SOLUTION: Speed of the rental car is represented by x. Speed of the bus: x – 14 D 140 Time taken by the bus to cover 140 miles: t = = r x − 14 D 140 Time taken by the rental car to cover 140 miles: t = = r x
Time taken by the bus + Time taken by the rental car to to cover 140 miles cover 140 miles 140 140 + = 4.5 x − 14 x We first note that x ≠ 14 and x ≠ 0 . Multiply both the sides with the LCD, x ( x − 14 ) .
= 4.5 hours
⎛ 140 140 ⎞ + ⎜ ⎟ x ( x − 14 ) = 4.5 ⋅ x ( x − 14 ) x ⎠ ⎝ x − 14 140 x + 140 x − 196 = 4.5 x 2 − 63 x 4.5 x 2 − 343 x + 1960 = 0 45 x 2 − 3430 x + 19600 = 0
( x − 70 )( 9 x − 56 ) = 0 56 9 Consider the whole number to be the solution. x = 70 or x =
ANSWER: Speed of the bus: x – 14 = 70 – 14 = 56 miles/hour
73.
PROBLEM: Jerry takes twice as long as Manny to assemble a skateboard. If they work together, they can assemble a skateboard in 6 minutes. How long would it take Manny to assemble the skateboard without Jerry’s help? SOLUTION: If Manny can assemble a skateboard in x minutes, Manny’s work 1 rate is . x 1 Jerry can assemble the skateboard in 2x minutes. Hence, Jerry’s work rate is . 2x If they work together, they can assemble the skateboard in 6 minutes. Multiply the individual work rates by 6. The sum of these two products is equal to 1. 1 1 ⋅6 + ⋅6 =1 x 2x We first note that x ≠ 0. 1 1 ⋅6 + ⋅6 =1 x 2x 6 3 + =1 x x 9 =1 x x=9 ANSWER: Manny can assemble the skateboard in 9 minutes.
74.
PROBLEM: Working alone, Joe can complete the yard work in 30 minutes. It takes Mike 45 minutes to complete the same yard. How long would it take them working together? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 30 minutes; t2 = 45 minutes 1 1 ⋅t + ⋅t = 1 30 45 t t + =1 30 45 t ⎞ ⎛ t ⎜ + ⎟ ⋅ 90 = 1⋅ 90 ⎝ 30 45 ⎠ 3t + 2t = 90 5t = 90 t = 18
ANSWER: It would take Joe and Manny to build the yard together in 18 minutes. 7.7 Variation Construct a mathematical model given the following:
75.
PROBLEM: y varies directly x and y = 12 when x = 4. SOLUTION: y = kx Substitute the values of x and y in the expression above. y = kx 12 = k 4 k =3 ANSWER: The expression becomes y = 3x.
76.
PROBLEM: y varies inversely as x and y = 2 when x = 5.
k x Substitute the values of x and y in the expression above. k y= x k 2= 5 k = 10 SOLUTION: y =
ANSWER: The expression becomes y =
77.
10 . x
PROBLEM: y is jointly proportional to x and z where y = 36 when x = 3 and z = 4. SOLUTION: y = kxz Substitute the values of x, y, and z in the expression above. y = kxz 36 = k ⋅ 3 ⋅ 4 k =3 ANSWER: The expression becomes y = 3xz .
78.
PROBLEM: y is directly proportional to the square of x and inversely proportional to z, where y = 20 when x = 2 and z = 5.
x2 z Substitute the values of x, y, and z in the expression above. x2 y=k z 22 20 = k 5 20 ⋅ 5 k= = 25 4 25x 2 ANSWER: The expression becomes y = . z SOLUTION: y = k
79.
PROBLEM: The distance an object in freefall falls varies directly with the time that it has been falling. It is observed that an object falls 16 feet in 1 second, find an equation that models the distance an object will fall and use it to determine how far it will fall in 2 seconds. SOLUTION: Distance an object falls in freefall is represented by d. Time of fall is represented by t. d = kt It is observed that an object falls 16 feet in 1 second. d = kt 16 = k ⋅1 k = 16 The formula is d = 16t . ANSWER: In 2 seconds the object falls at a distance of , d = 16 ⋅ 2 = 32 feet.
80.
PROBLEM: The weight of an object varies inversely as the square of its distance from the center of the Earth. If an object weighs 180 lbs on the surface of the Earth (approximately 4000 miles from the center), then how much will it weigh at 2000 miles above the Earth’s surface? SOLUTION: The weight of the object is represented by w. The distance from the center of the earth is represented by d. k w= 2 d An object weighs 180 lbs on the surface of the Earth (approximately 4000 miles from the center). k w= 2 d k 180 = 40002 k = 288 × 107 Weight of the object 2000 miles above the Earth’s surface, i.e. 4000 + 2000 = k 6000 miles: w = 2 d 7 288 ×10 w= = 80 60002 ANSWER: Weight of the object 2000 miles above the Earth’s surface is 80 lb.
Chapter 7 Sample Exam Simplify and state the restrictions.
1.
PROBLEM: 2 15 x3 ( 3x − 1)
3x ( 3x − 1)
SOLUTION: 2 15 x3 ( 3x − 1)
= 5 x 2 ( 3x − 1)
3x ( 3x − 1) To find the restrictions to the domain, set the denominator equal to zero and solve: 3x ( 3x − 1) = 0 x = 0 or 3x − 1 = 0 1 x = 0 or x = 3
ANSWER: 5 x 2 ( 3 x − 1)
The domain consists of any real number x where x ≠ 0 and x ≠ 2.
1 . 3
PROBLEM: x 2 − 144 x 2 + 12 x SOLUTION: x 2 − 144 ( x + 12 )( x − 12 ) x − 12 = = x 2 + 12 x x ( x + 12 ) x To find the restrictions to the domain, set the denominator equal to zero and solve: x ( x + 12 ) = 0 x = 0 or x + 12 = 0 x = 0 or x = −12
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ −12 .
3.
PROBLEM: x 2 + x − 12 2x2 + 7 x − 4 SOLUTION: ( x + 4 )( x − 3) = x − 3 x 2 + x − 12 = 2 2 x + 7 x − 4 ( 2 x − 1)( x + 4 ) 2 x − 1 To find the restrictions to the domain, set the denominator equal to zero and solve: ( 2 x − 1)( x + 4 ) = 0
2 x − 1 = 0 or x + 4 = 0 1 x = or x = −4 2 ANSWER:
x −3 2x −1
The domain consists of any real number x where x ≠ 4.
1 and x ≠ −4 . 2
PROBLEM: 9 − x2
( x − 3)
2
SOLUTION: ( x − 3)( x + 3) = − x + 3 9 − x2 =− 2 2 x−3 ( x − 3) ( x − 3) To find the restrictions to the domain, set the denominator equal to zero and solve:
( x − 3)
2
=0
x −3 = 0 x=3
ANSWER: The domain consists of any real number x where x ≠ 3 .
Simplify. (Assume all variables in the denominator are positive.)
5.
PROBLEM: 5x x −5 ⋅ 2 x − 25 25 x 2 SOLUTION: 5x x −5 5x x−5 1 ⋅ = ⋅ = 2 2 2 x − 25 25 x ( x − 5)( x + 5) 25 x 5 x ( x + 5) ANSWER:
6.
1 5x ( x + 5)
PROBLEM: x 2 + x − 6 3x 2 − 5 x − 2 ⋅ x2 − 4x + 4 x2 − 9 SOLUTION: x 2 + x − 6 3x 2 − 5 x − 2 ( x + 3)( x − 2 ) ( 3 x + 1)( x − 2 ) 3 x + 1 ⋅ = ⋅ = 2 x2 − 4x + 4 x2 − 9 ( x + 3)( x − 3) x − 3 ( x − 2) ANSWER:
7.
3x + 1 x −3
PROBLEM: x 2 − 4 x − 12 x − 6 ÷ 12 x 2 6x SOLUTION: ( x − 6 )( x + 2 ) ⋅ 6 x x 2 − 4 x − 12 x − 6 x 2 − 4 x − 12 6 x ÷ = ⋅ = 2 2 12 x 6x 12 x 12 x 2 x−6 x−6 x+2 = 2x x+2 ANSWER: 2x
8.
PROBLEM: 2 x2 − 7 x − 4 2 x2 + 7 x + 3 ÷ 6 x 2 − 24 x 10 x 2 + 30 x SOLUTION: 2 x 2 − 7 x − 4 2 x 2 + 7 x + 3 2 x 2 − 7 x − 4 10 x 2 + 30 x ÷ = ⋅ 6 x 2 − 24 x 10 x 2 + 30 x 6 x 2 − 24 x 2 x 2 + 7 x + 3 ( 2 x + 1)( x − 4 ) ⋅ 10 x ( x + 3) = 6x ( x − 4) ( 2 x + 1)( x + 3)
= ANSWER:
9.
5 3
5 3
PROBLEM: 1 1 + x −5 x +5 SOLUTION: LCD = ( x − 5 )( x + 5 )
To obtain equivalent terms with this common denominator, we will multiply the x+5 x−5 and the second term by first term by . x+5 x−5 1 1 1 x+5 1 x −5 + = ⋅ + ⋅ x −5 x +5 x −5 x +5 x +5 x −5 x +5+ x −5 = ( x − 5)( x + 5)
=
ANSWER:
2x ( x − 5)( x + 5)
2x ( x − 5)( x + 5)
10.
PROBLEM: x 8 12 x − − 2 x +1 2 − x x − x − 2 SOLUTION: x 8 12 x x 8 12 x − − 2 = + − x + 1 2 − x x − x − 2 x + 1 x − 2 ( x − 2 )( x + 1) LCD = ( x − 2 )( x + 1)
To obtain equivalent terms with this common denominator, we will multiply the x−2 x +1 1 first term by , and the third term by . , the second term by x−2 1 x +1 x 8 12 x x x−2 8 x +1 12 x 1 + − = ⋅ + ⋅ − ⋅ x + 1 x − 2 ( x − 2 )( x + 1) x + 1 x − 2 x − 2 x + 1 ( x − 2 )( x + 1) 1
=
x 2 − 2 x + 8 x + 8 − 12 x ( x − 2 )( x + 1)
x2 − 6 x + 8 = ( x − 2 )( x + 1)
ANSWER:
x−4 x +1
=
( x − 2 )( x − 4 ) ( x − 2 )( x + 1)
=
x−4 x +1
11.
PROBLEM: 1 1 + y x 1 1 − 2 2 y x SOLUTION: 1 1 1 x 1 y x+ y + ⋅ + ⋅ y x y x x y xy = = 2 2 2 1 1 1 x 1 y x − y2 − ⋅ − ⋅ y 2 x2 y 2 x2 x2 y 2 x2 y2 x+ y xy = ( x + y )( x − y ) x2 y2 =
x+ y x2 y2 ⋅ xy ( x + y )( x − y )
=
xy x− y
ANSWER:
xy x− y
12.
PROBLEM: 6 9 1− + 2 x x 5 3 2− − 2 x x SOLUTION: ⎛ 6 9 ⎞ 2 6 9 1 − + 2 ⎜1 − + 2 ⎟ x 2 x x = ⎝ x x ⎠ = x − 6x + 9 2 5 3 5 3 2 − − 2 ⎜⎛ 2 − − 2 ⎟⎞ x 2 2 x − 5 x − 3 x x x x ⎠ ⎝
( x − 3) = ( 2 x + 1)( x − 3) 2
=
x −3 2x +1
ANSWER:
x −3 2x +1
13.
PROBLEM:
Given f ( x ) =
x 2 − 81
( 4 x − 3)
2
and g ( x ) =
4x − 3 , calculate ( f ⋅ g )( x ) and state the x −9
restrictions. SOLUTION: x 2 − 81 4 x − 3 ( x − 9 )( x + 9 ) 4 x − 3 ( ) ⋅ f g x ⋅ = ⋅ = ( ) 2 2 x −9 ( 4 x − 3) x − 9 ( 4 x − 3)
=
x+9 4x − 3
3 In this case, the domain of f ( x) consists of all real numbers except , and the 4 domain of g ( x) consists all real numbers except 9. ANSWER: ( f ⋅ g ) ( x) =
x+9 4x − 3
The domain consists of any real number x where x ≠
3 and x ≠ 9. 4
14.
PROBLEM:
Given f ( x ) =
x 1 and g ( x ) = , calculate ( f − g )( x ) and state the x −5 3x − 5
restrictions. SOLUTION:
x 1 − x − 5 3x − 5 LCD = ( x − 5)( 3x − 5) To obtain equivalent terms with this common denominator, we will multiply the 3x − 5 x−5 first term by . and the second term by 3x − 5 x−5 x x 3x − 5 x−5 1 1 − − = ⋅ ⋅ x − 5 3x − 5 x − 5 3x − 5 3x − 5 x − 5 3x 2 − 5 x − x + 5 = ( x − 5 )( 3x − 5)
( f − g ) ( x) =
3x 2 − 6 x + 5 ( x − 5)( 3x − 5) In this case, the domain of f ( x) consists of all real numbers except 5 , and the 5 domain of g ( x) consists all real numbers except . 3 5 ANSWER: The domain consists of any real number x where x ≠ 5 and x ≠ . 3 =
Solve.
15.
PROBLEM: 1 1 + =2 3 x SOLUTION: We first note that x ≠ 0. Multiplying both the side with the LCD, 3x. ⎛1 1⎞ ⎜ + ⎟ 3x = 2 ⋅ 3x ⎝3 x⎠ x + 3 = 6x
5x = 3 3 x= 5 We can check our answer by substituting statement. 1 1 + =2 3 3 5 1 5 + =2 3 3 1+ 5 =2 3 2=2 3 ANSWER: The solution is . 5
3 in for x to see if we obtain a true 5
16.
PROBLEM: 1 3 = x − 5 2x − 3
3 SOLUTION: We first note that x ≠ 5 and x ≠ . 2 Multiplying both the sides with the LCD, ( x − 5 )( 2 x − 3) .
⎛ 1 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ( x − 5 )( 2 x − 3) = ⎜ ⎟ ( x − 5 )( 2 x − 3) ⎝ x−5⎠ ⎝ 2x − 3 ⎠ 2 x − 3 = 3x − 15 x = 12 We can check our answer by substituting 12 in for x to see if we obtain a true statement. 1 3 = 12 − 5 2 ⋅12 − 3 1 3 = 7 21 1 1 = 7 7 ANSWER: The solution is 12 .
17.
PROBLEM: 9 20 1− + 2 = 0 x x SOLUTION: We first note that x ≠ 0 . Multiply both the sides with the LCD, x2. ⎛ 9 20 ⎞ 2 2 ⎜1 − + 2 ⎟ x = 0 ⋅ x ⎝ x x ⎠ x 2 − 9 x + 20 = 0
( x − 4)( x − 5) = 0 x = 4 or x = 5 Check x = 4 9 20 1− + 2 = 0 4 4 9 20 1− + =0 4 16 16 − 36 + 20 =0 16 0=0 ANSWER: The solution is 4 or 5.
Check x = 5 9 20 1− + 2 = 0 5 5 9 20 =0 1− + 5 25 25 − 45 + 20 =0 25 0=0
18.
PROBLEM: 4 ( x + 1) x+2 1 + = 2 x−2 x+2 x −4 SOLUTION: 4 ( x + 1) x+2 1 + = 2 x−2 x+2 x −4 4 ( x + 1) x+2 1 + = x − 2 x + 2 ( x + 2 )( x − 2 ) We first note that x ≠ ±2 . Multiply both the sides with the LCD, ( x − 2 )( x + 2 ) .
4 ( x + 1) 1 ⎞ ⎛ x+2 + ⋅ ( x + 2 )( x − 2 ) ⎜ ⎟ ( x + 2 )( x − 2 ) = ( x + 2 )( x − 2 ) ⎝ x−2 x+2⎠ x2 + 4x + 4 + x − 2 = 4 x + 4 x2 + x − 2 = 0
( x + 2 )( x − 1) = 0 x = −2 or x = 1 x ≠ ±2 . ANSWER: Hence, the only solution is 1. We can check our answer by substituting 1 in for x to see if we obtain a true statement. 4 (1 + 1) 1+ 2 1 + = 1 − 2 1 + 2 (1 + 2 )(1 − 2 )
1 8 −3 + = − 3 3 8 8 − =− 3 3
19.
PROBLEM: x 1 3x − 10 − = 2 x − 2 x − 3 x − 5x + 6 SOLUTION: x 1 3 x − 10 − = 2 x − 2 x − 3 x − 5x + 6 x 1 3x − 10 − = x − 2 x − 3 ( x − 2 )( x − 3)
We first note that x ≠ 2 and x ≠ 3 . Multiply both the sides with the LCD, ( x − 2 )( x − 3) .
1 ⎞ 3x − 10 ⎛ x − ⋅ ( x − 2 )( x − 3) ⎜ ⎟ ( x − 2 )( x − 3) = ( x − 2)( x − 3) ⎝ x −2 x −3⎠ x 2 − 3x − x + 2 = 3x − 10 x 2 − 7 x + 12 = 0
( x − 3)( x − 4) = 0 x = 3 or x = 4 x ≠ 3. ANSWER: Hence, the only solution is 4. We can check our answer by substituting 4 in for x to see if we obtain a true statement. 4 1 3 ⋅ 4 − 10 − = 4 − 2 4 − 3 ( 4 − 2 )( 4 − 3)
2 −1 = 1 1=1
20.
PROBLEM: 5 x 9x − 4 − = 2 x + 4 4 − x x − 16 SOLUTION: x 5 9x − 4 − = 2 x + 4 4 − x x − 16 x 5 9x − 4 + = x + 4 x − 4 ( x + 4 )( x − 4 )
We first note that x ≠ ±4 . Multiply both the sides with the LCD, ( x − 4 )( x + 4 ) .
x ⎞ 9x − 4 ⎛ 5 + ⋅ ( x − 4 )( x + 4 ) ⎜ ⎟ ( x − 4 )( x + 4 ) = ( x + 4)( x − 4) ⎝ x+4 x−4⎠
5 x − 20 + x 2 + 4 x = 9 x − 4 x 2 − 16 = 0
( x − 4)( x + 4 ) = 0 x = 4 or x = −4 x ≠ ±4 . ANSWER: Hence, the solution is∅.
21.
PROBLEM:
Solve for r: P =
120 1 + 3r
SOLUTION:
1 We first note that r ≠ − . 3 120 P= 1 + 3r 120 1 + 3r = P 120 3r = −1 P 120 1 r= − 3P 3 40 1 r= − P 3 ANSWER: r =
40 1 − P 3
Set up an algebraic equation and then solve:
22.
PROBLEM: An integer is three times that of another. The sum of the reciprocals of the two integers is 1/3. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: 3n 1 1 1 + = n 3n 3 We first note that n ≠ 0. Multiply both sides by the LCD, 3n. 1 ⎛1 1 ⎞ ⎜ + ⎟ 3n = ⋅ 3n 3 ⎝ n 3n ⎠ 3 +1 = n n=4 Larger integer: 3n = 3 ⋅ 4 = 12 ANSWER: The two integers are 4 and 12.
23.
PROBLEM: Working alone, Joe can paint the room in 6 hours. If Manny helps, then together they can paint the room in 2 hours. How long would it take Manny to paint the room by himself? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 6 hours; t = 2 hours 1 1 ⋅t + ⋅t =1 t1 t2 1 1 ⋅2+ ⋅2 =1 6 t2 2 2 + =1 6 t2
Multiplying both sides by the LCD, 6t2. ⎛2 2⎞ ⎜ + ⎟ 6t2 =1 ⋅ 6t2 ⎝ 6 t2 ⎠ 2t2 + 12 = 6t2 4t2 = 12 t2 = 3
ANSWER: It would take Manny to paint the room in 3 hours.
24.
PROBLEM: A river tour boat can average 6 mph in still water. With the current, the boat can travel 17 miles in the same time it can travel 7 miles against it. What is the speed of the current? SOLUTION: The speed of the current is represented by x. Speed of the boat downstream: 6 + x Speed of the boat upstream: 6 – x
D 17 = t 6+ x D 7 Time taken by the boat to cover 7 miles upstream: t = = t 6− x Time taken by the boat to cover = Time taken by the boat to cover 17 miles downstream 7 miles upstream 17 7 = 6+ x 6− x Multiply both the sides by the LCD, (6 + x)(6 – x) 17 7 ⋅ ( 6 + x )( 6 − x ) = ⋅ ( 6 + x )( 6 − x ) 6+ x 6− x 102 − 17 x = 42 + 7 x 24 x = 60 5 1 x= =2 2 2 1 ANSWER: The speed of the current is 2 miles/hour. 2
Time taken by the boat to cover 17 miles downstream: t =
25.
PROBLEM: The breaking distance of an automobile is directly proportional to the square of its speed. Under optimal conditions, a certain automobile moving at 35 mph can break in 25 feet. Find an equation that models the breaking distance under optimal conditions and use it to determine the breaking distance if the automobile is moving 28 mph. SOLUTION: Breaking distance of the automobile is represented by d. Speed of the automobile is represented by s. d = ks 2 Under optimal conditions, the automobile moving at 35 mph can break in 25 feet. d = ks 2
25 = k 352 25 1 k= 2 = 35 49 1 2 d= s 49 If the automobile is moving at 28 mph: d = ks 2 25 d = 2 282 = 16 35 ANSWER: If the automobile is moving at 28 mph, the breaking distance is 16 feet.
ANSWER: Substitute the values in for x. x = −1
x=0
5 5 = = undefined x 0
5 5 = = −5 x ( −1)
2.
x =1
5 5 = =5 x 1
PROBLEM: 4x ; {−1, 0, 1} 3x 2 ANSWER: Substitute the values in for x. x = −1
4 ( −1) 4x 4 = =− 2 2 3x 3 3 ( −1) 3.
x=0
4x 4 (0) = = undefined 3x 2 3 ( 0 )
x =1
4 (1) 4 4x = = 2 2 3x 3 3 (1)
PROBLEM: 1 ; {−10, −9, 0} x+9 ANSWER: Substitute the values in for x. x = −10
1 1 = = −1 x + 9 −10 + 9
x = −9
5 5 = = undefined x 0
x=0
5 5 = =5 x 1
4.
PROBLEM: x+6 ; {−6, 0, 5} x −5 ANSWER: Substitute the values in for x. x = −6
x=0
x + 6 −6 + 6 = =0 x − 5 −6 − 5
5.
x+6 0+6 6 = =− x −5 0−5 5
x=5
x + 6 5 + 6 11 = = = undefined x−5 5−5 0
PROBLEM: 3x ( x − 2 ) ; {0, 1/2, 2} 2x −1 ANSWER: Substitute the values in for x. x=0
x=
3x ( x − 2 ) =
2x −1 3 ( 0 )( 0 − 2 ) 2 ( 0) −1
=0
x=2
1 2
⎛ 1 ⎞⎛ 1 ⎞ 3⎜ ⎟ ⎜ − 2 ⎟ 3x ( x − 2 ) 2 2 ⎠ = ⎝ ⎠⎝ 2x −1 ⎛1⎞ 2 ⎜ ⎟ −1 ⎝2⎠ 3⎛ 3⎞ ⎜− ⎟ 2⎝ 2⎠ = = undefined 0
3x ( x − 2 ) 2x −1
=
3 ( 2 )( 2 − 2 ) 2 ( 2) −1
=
0 =0 3
6.
PROBLEM: 9 x2 − 1 ; {0, 1/3, 7} x−7 ANSWER: Substitute the values in for x. x=0
9 x 2 − 1 9 ( 0 ) − 1 −1 1 = = = −7 7 x−7 0−7
7.
x=
x=7
1 3
2
⎛1⎞ 9 ⎜ ⎟ −1 2 9x −1 1−1 3 = ⎝ ⎠ = 1 20 x−7 −7 − 3 3 0 = 20 − 3 =0
9 x 2 − 1 9 ( 7 ) − 1 440 = = x−7 7−7 0 = undefined 2
PROBLEM: 5 ; {−3, 0, 3} 2 x −9 ANSWER: Substitute the values in for x. x = −3
5 5 5 = = 2 2 x − 9 ( −3) − 9 0 = undefined
x=0
5 5 5 = =− x −9 0−9 9 2
x=3
5 5 5 = = 2 x − 9 ( 3) − 9 0 2
= undefined
8.
PROBLEM: x 2 − 25 ; {−5, −4, 5} x 2 − 3x − 10 ANSWER: Substitute the values in for x. x = −5
x = −4
x=5
2 2 x 2 − 25 −5 ) − 25 −4 ) − 25 ( ( x 2 − 25 x 2 − 25 = = 2 x 2 − 3x − 10 ( −5 )2 − 3 ( −5) − 10 x 2 − 3 x − 10 ( −4 )2 − 3 ( −4 ) − 10 x − 3 x − 10 2 5 − 25 ( ) 0 1 0 −9 = = 2 = = =− ( 5) − 3 ( 5) − 10 0 25 + 15 − 10 18 2 =0 = undefined
9.
PROBLEM: Fill in the following chart:
1 x
x −3 −2 0 2 3
ANSWER:
1 x
x −3 −2 0 2 3
1 x 1 x 1 x 1 x 1 x
1 1 =− −3 3 1 1 = =− −2 2 1 = = undefined 0 1 = 2 1 = 3 =
10.
PROBLEM: Fill in the following chart:
x
1 x−2
−1 0 1 2 3 ANSWER:
x −1 0
1 2 3
1 x−2 1 1 1 = =− 3 x − 2 −1 − 2 1 1 1 = =− 2 x−2 0−2 1 1 1 = = = −1 x − 2 1 − 2 −1 1 1 1 = = = undefined x−2 2−2 0 1 1 1 = = =1 x − 2 3− 2 3− 2
11.
PROBLEM: Fill in the following chart:
x
1 x+2
−4 −3 −2 −1 0 ANSWER:
x −4 −3 −2 −1 0
1 x+2 1 1 1 1 = = =− x + 2 −4 + 2 −2 2 1 1 1 = = = −1 x + 2 −3 + 2 −1 1 1 1 = = = undefined x + 2 −2 + 2 0 1 1 1 = = =1 x + 2 −1 + 2 1 1 1 1 = = x+2 0+2 2
An objects weight depends on the height above the surface of the earth. If an object weighs 120 pounds on the surface of the Earth, then its weight W, in pounds, x miles above the surface is approximated by the formula: 120* 40002 W= (4000 + x) 2 Approximate the weight of a 120 lb object at the given height above the surface of the Earth. (1 mile = 5280 feet)
12.
PROBLEM: Fill in the following chart:
x
1 −2 x
−2 −1 0 1 2 ANSWER:
x −2 −1 0 1 2
1 −2 x 1 5 − −2 = − 2 2 1 1 − 2 = − − 2 = −3 1 x 1 − 2 = undefined 0 1 − 2 = −1 1 1 3 −2 = − 2 2
An objects weight depends on the height above the surface of the earth. If an object weighs 120 pounds on the surface of the Earth, then its weight W, in pounds, x miles above the surface is approximated by the formula:
13.
PROBLEM: 100 miles SOLUTION: 120* 40002 120* 40002 W= = = 114.2 ~ 114 lbs (4000 + x) 2 (4000 + 100) 2 ANSWER: ~ 114 lbs
14.
PROBLEM: 1000 miles SOLUTION: 120* 40002 120* 40002 W= = = 76.8 lbs (4000 + x) 2 (4000 + 1000) 2 ANSWER: 76.8 lbs
15.
PROBLEM: 44, 350 feet SOLUTION:
44,350 feet × W=
1 mile 4435 = miles 5280 feet 528
120* 40002 120* 40002 = = 119.5 lbs 2 (4000 + x) 2 ⎛ 4435 ⎞ ⎜ 4000 + ⎟ 528 ⎠ ⎝
ANSWER: 119.5 lbs
The price to earnings ratio (P/E) is a metric used to compare the valuations of similar publically traded companies. The P/E ratio is calculated using the stock price and the earnings per share (EPS) over the previous 12 month period as follows: Price per Share P/E = Earnings per Share If each share of a company stock is priced at $22.40, then calculate the P/E ratio given the following values for the earnings per share.
16.
PROBLEM: 90,000 feet SOLUTION:
90, 000 feet ×
1 mile 375 = miles 5280 feet 22
120* 40002 120* 40002 W= = = 118.9 lb ~ 119 lbs 2 (4000 + x) 2 ⎛ 375 ⎞ ⎜ 4000 + ⎟ 22 ⎠ ⎝ ANSWER: ~ 119 lbs The price to earnings ratio (P/E) is a metric used to compare the valuations of similar publically traded companies. The P/E ratio is calculated using the stock price and the earnings per share (EPS) over the previous 12 month period as follows: Price per Share P/E = Earnings per Share If each share of a company stock is priced at $22.40, then calculate the P/E ratio given the following values for the earnings per share.
17.
PROBLEM: $1.40 SOLUTION: Price per Share $22.40 P/E = = = 16 Earnings per Share $1.40 ANSWER: 16
18.
PROBLEM: $1.21 SOLUTION: Price per Share $22.40 P/E = = = 18.5 Earnings per Share $1.21 ANSWER: 18.5
19.
PROBLEM: What happens to the P/E ratio when earnings decrease? ANSWER: If the denominator increases, i.e. the earnings per share decreases the P/E ratio increases.
State the restrictions to the domain.
20.
PROBLEM: What happens to the P/E ratio when earnings increase? ANSWER: If the denominator increases, i.e. the earnings per share increases, the P/E ratio decreases.
State the restrictions to the domain.
21.
PROBLEM: 1 3x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 3x = 0 x=0
ANSWER: The domain consists of any real number x where x ≠ 0 .
22.
PROBLEM: 3x 2 7 x5 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 7 x5 = 0
x=0 ANSWER: The domain consists of any real number x where x ≠ 0 .
23.
PROBLEM: 3x ( x + 1)
x+4 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x+4=0 x + 4 − 4 = −4 x = −4 ANSWER: The domain consists of any real number x where x ≠ −4 .
24.
PROBLEM: 2 x 2 ( x − 3)
x −1 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x −1 = 0
x −1+1 = 0 +1 x =1 ANSWER: The domain consists of any real number x where x ≠ 1 .
25.
PROBLEM: 1 5x − 1 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 5x − 1 = 0 5x − 1 + 1 = 0 + 1 5x = 1 5 1 x= 5 5 1 x= 5 1 ANSWER: The domain consists of any real number x where x ≠ . 5
26.
PROBLEM: x−2 3x − 2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 3x − 2 = 0
3x − 2 + 2 = 0 + 2 3x = 2 3 2 x= 3 3 2 x= 3 ANSWER: The domain consists of any real number x where x ≠
27.
2 . 3
PROBLEM: x −9 5x ( x − 2) SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 5x ( x − 2) = 0
5x = 0 x=0
x−2=0 x−2+2 = 0+2
or
x=2 ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ 2 . 28.
PROBLEM: 1 ( x − 3)( x + 6 ) SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: ( x − 3)( x + 6) = 0
x −3 = 0
or
x − 3 + 3 = 0 + 3 or x=3 or
x+6 =0 x+6−6 = 0−6 x = −6
ANSWER: The domain consists of any real number x where x ≠ 3 and x ≠ −6 .
29.
PROBLEM: x 1 − x2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 1 − x2 = 0
(1 − x )(1 + x ) = 0 1− x = 0 1 − 1 − x = −1 x =1
or or or
1+ x = 0 1 − 1 + x = −1 x = −1
ANSWER: The domain consists of any real number x where x ≠ ±1 .
30.
PROBLEM: x2 − 9 x 2 − 36 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x 2 − 36
( x − 6)( x + 6 ) = 0 x−6 = 0 or x − 6 + 6 = 0 + 6 or x=6 or
x+6=0 x+6−6 = 0−6 x = −6
ANSWER: The domain consists of any real number x where x ≠ ±6 .
31.
PROBLEM: 1 2 x ( x + 3)( 2 x − 1) SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 x ( x + 3)( 2 x − 1) = 0
2x = 0 x=0 x=0
x + 3 = 0 or 2x −1 = 0 x + 3 − 3 = 0 − 3 or 2 x − 1 + 1 = 0 + 1 x = −3 or 2 x = 1 2 1 x = 0 or x = −3 or x = 2 2 1 x = 0 or x = −3 or x = 2 ANSWER: The domain consists of any real number x where x ≠ 0 , x ≠ −3 , and 1 x≠ . 2 32.
or or or
PROBLEM: x−3 ( 3x − 1)( 2 x + 3) SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: ( 3x − 1)( 2 x + 3) = 0
3x − 1 = 0 3x − 1 + 1 = 0 + 1 3x = 1 3 1 x= 3 3 1 x= 3
or or or or or
2x + 3 = 0 2x + 3 − 3 = 0 − 3 2 x = −3 2 3 x=− 2 2 3 x=− 2
ANSWER: The domain consists of any real number x where x ≠
1 3 and x ≠ − . 3 2
33
PROBLEM: 4 x ( 2 x + 1)
12 x 2 + x − 1 SOLUTION: 4 x(2 x + 1) 2 12 x + x − 1
4 x(2 x + 1) (4 x − 1)(3 x + 1) Therefore, 4 x − 1 = 0and 3 x + 1 = 0 1 x = and 4 1 x=− 3 =
ANSWER: The domain consists of any real number, where 34.
x≠
1 1 x≠− 4 and 3
PROBLEM: x −5 3x 2 − 15 x SOLUTION: x−5 3x 2 − 15 x x −5 = 3x( x − 5) 1 = 3x ANSWER: The domain consists of any real number where x ≠ 0
Part B: Simplifying Rational Expressions State the restrictions and then simplify.
35.
PROBLEM: 5x2 20 x3 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 20 x 3 = 0
x=0 ANSWER: The domain consists of any real number x where x ≠ 0 . 5x2 1 1 = 3− 2 = 3 20 x 4x 4x
36.
PROBLEM: 12 x 6 60 x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 60 x = 0 x=0 ANSWER: The domain consists of any real number x where x ≠ 0 . 12 x 6 1 6−1 x5 = x = 60 x 5 5
37.
PROBLEM: 3x 2 ( x − 2 ) 9 x ( x − 2)
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 9x ( x − 2) = 0
9x = 0 x=0
or or
x−2=0 x−2+2 = 0+2
x=0
or
x=2
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ 2 . 3x 2 ( x − 2 ) 3x 2 x 2−1 x = = = 9 x ( x − 2) 9x 9 9
38.
PROBLEM: 20 ( x − 3)( x − 5 ) 6 ( x − 3)( x + 1) SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 6 ( x − 3)( x + 1) = 0
( x − 3)( x + 1) = 0 x −3 = 0 x −3+3 = 0+3
or or
x +1 = 0 x +1 −1 = 0 −1
x=3
or
x = −1
ANSWER: The domain consists of any real number x where x ≠ 3 and x ≠ −1 . 20 ( x − 3)( x − 5 ) 10 ( x − 5) = 6 ( x − 3)( x + 1) 3 ( x + 1)
39.
PROBLEM: 6 x2 ( x − 8)
36 x ( x + 9 )( x − 8 )
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 36 x ( x + 9 )( x − 8 ) = 0
x=0 x=0
or or
x+9 = 0 or x + 9 − 9 = 0 − 9 or
x −8 = 0 x −8+8 = 0+8
x=0
or
x = −9
x =8
or
ANSWER: The domain consists of any real number x where x ≠ 0 , x ≠ −9 and x≠8 . 6 x2 ( x − 8) x 2−1 x = = 36 x ( x + 9 )( x − 8 ) 6 ( x + 9 ) 6 ( x + 9 )
40.
PROBLEM: 16 x 2 − 1
( 4 x + 1)
2
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 ( 4 x + 1) = 0 4x +1 = 0 4x +1−1 = 0 −1 4 x = −1 4 1 x=− 4 4 1 x=− 4
1 ANSWER: The domain consists of any real number x where x ≠ − . 4 2 16 x − 1 ( 4 x − 1)( 4 x + 1) 4 x − 1 = = 2 2 4x +1 ( 4 x + 1) ( 4 x + 1)
41.
PROBLEM: 9x2 − 6x + 1
( 3 x − 1)
2
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 ( 3x − 1) = 0 3x − 1 = 0 3x − 1 + 1 = 0 + 1 3x = 1 3 1 x= 3 3 1 x= 3
ANSWER: The domain consists of any real number x where x ≠
( 3x − 1) = 2 2 ( 3x − 1) ( 3x − 1)
9 x2 − 6 x + 1
42.
1 3
2
=1
PROBLEM: x−7 x 2 − 49 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x 2 − 49 = 0
( x + 7 )( x − 7 ) = 0 x+7 = 0
or
x+7−7 = 0−7 x = −7
or or
x−7 = 0 x−7+7 = 0+7 x=7
ANSWER: The domain consists of any real number x where x ≠ ±7 . x−7 x−7 1 = = 2 x − 49 ( x − 7 )( x + 7 ) x + 7
43.
PROBLEM: x 2 − 64 x2 + 8x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x2 + 8x = 0
x ( x + 8) = 0 x=0 x=0 x=0 x=0
x+8 = 0 x+8 = 0 x +8−8 = 0−8 x = −8
or or or or
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ −8 . x 2 − 64 ( x + 8)( x − 8 ) x − 8 = = x2 + 8x x ( x + 8) x
44.
PROBLEM: x + 10 x 2 − 100 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x 2 − 100 = 0
( x + 10 )( x − 10 ) = 0 x + 10 = 0 x + 10 − 10 = 0 − 10 x = −10
or or or
x − 10 = 0 x − 10 + 10 = 0 + 10 x = 10
ANSWER: The domain consists of any real number x where x ≠ ±10 . x + 10 x + 10 1 = = 2 x − 100 ( x + 10 )( x − 10 ) x − 10
45.
PROBLEM: 2 x3 − 12 x 2 5 x 2 − 30 x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 5 x 2 − 30 x = 0
5x ( x − 6) = 0 5x = 0 x=0 x=0
x−6 = 0 x−6+6 = 0+6 x=6
or or or
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ 6 . 2 2 x3 − 12 x 2 2 x ( x − 6 ) 2 x 2−1 2 x = = = 5 x 2 − 30 x 5 x ( x − 6 ) 5 5
46.
PROBLEM: 30 x 5 + 60 x 4 2 x3 − 8 x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 x3 − 8 x = 0
(
)
2 x x2 − 4 = 0 2 x ( x + 2 )( x − 2 ) = 0 2x = 0 x=0 x=0
x+2=0 or x + 2 − 2 = 0 − 2 or x = −2 or
or or or
x−2 = 0 x−2+2 = 0+2 x=2
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ ±2 . 4 30 x 4 ( x + 2 ) 30 x5 + 60 x 4 30 x ( x + 2 ) 15 x 4−1 15 x3 = = = = 2 x3 − 8 x 2 x ( x − 2 )( x + 2 ) ( x − 2 ) x − 2 2x x2 − 4
(
)
47.
PROBLEM: 2x −1 2 x2 + x − 6 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 x2 + x − 6 = 0 a = 2; b = 1; c = −6 2 −b ± b 2 − 4ac −1 ± 1 − 4 ( 2 )( −6 ) −1 ± 49 −1 ± 7 x= = = = 2a 2 ( 2) 2 ( 2) 4
−1 + 7 −1 − 7 or 4 4 3 = or − 2 2 =
ANSWER: The domain consists of any real number x where x ≠
The fraction cannot be simplified. 2x −1 2 x2 + x − 6
3 and x ≠ −2 . 2
48.
PROBLEM: x2 − x − 6 3x 2 − 8 x − 3 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 3x 2 − 8 x − 3 = 0
( x − 3)( 3x + 1) = 0 x −3 = 0
or
3x + 1 = 0
x −3+3 = 0+3
or
3x + 1 − 1 = 0 − 1
x=3
or
3 x = −1
x=3
or
x=3
or
3 1 x=− 3 3 1 x=− 3
ANSWER: The domain consists of any real number x where x ≠ 3 and x ≠ −
( x − 3)( x + 2 ) = x + 2 x2 − x − 6 = 2 3x − 8 x − 3 ( x − 3)( 3x + 1) 3 x + 1
1 . 3
49.
PROBLEM: 6 x 2 − 25 x + 25 3x 2 + 16 x − 35 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 3 x 2 + 16 x − 35 = 0
( 3x − 5)( x + 7 ) = 0 3x − 5 = 0
or
x+7 =0
3x − 5 + 5 = 0 + 5
or
x+7−7 = 0−7
3x = 5
or
x = −7
or
x = −7
or
x = −7
3 5 x= 3 3 5 x= 3
ANSWER: The domain consists of any real number x where x ≠
6 x 2 − 25 x + 25 ( 2 x − 5)( 3x − 5 ) 2 x − 5 = = 3x 2 + 16 x − 35 ( 3x − 5)( x + 7 ) x+7 50.
5 and x ≠ −7 . 3
PROBLEM: 3x 2 + 4 x − 15 x2 − 9 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x2 − 9 = 0
( x + 3)( x − 3) = 0 x+3= 0 x +3−3 = 0−3 x = −3
or or or
x −3 = 0 x −3+3 = 0+3 x=3
ANSWER: The domain consists of any real number x where x ≠ ±3 . 3x 2 + 4 x − 15 ( 3x − 5)( x + 3) 3x − 5 = = x2 − 9 ( x − 3)( x + 3) x − 3
51.
PROBLEM: x 2 − 10 x + 21 x 2 − 4 x − 21 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x 2 − 4 x − 21 = 0
x 2 − 7 x + 3 x − 21 = 0 x ( x − 7) + 3( x − 7) = 0
( x + 3)( x − 7 ) = 0 x+3= 0 x +3−3 = 0−3 x = −3
or or or
x−7 = 0 x−7+7 = 0+7 x=7
ANSWER: The domain consists of any real number x where x ≠ −3 and x ≠ 7 . x 2 − 10 x + 21 ( x − 3)( x − 7 ) x − 3 = = x 2 − 4 x − 21 ( x + 3)( x − 7 ) x + 3
52.
PROBLEM: x3 − 1 x2 − 1 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x2 − 1 = 0
( x − 1)( x + 1) = 0 x −1 x −1 + 1 = 0 + 1 x =1
or or or
x +1 = 0 x +1−1 = 0 −1 x = −1
ANSWER: The domain consists of any real number x where x ≠ ±1 . 2 x3 − 1 ( x − 1) x + x + 1 x 2 + x + 1 = = x2 − 1 x +1 ( x − 1)( x + 1)
(
)
53.
PROBLEM: x3 + 8 x2 − 4 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x2 − 4 = 0
( x + 2 )( x − 2 ) = 0 x+2=0 x+2−2 = 0−2 x = −2
x−2=0 x−2+2 = 0+2 x=2
or or or
ANSWER: The domain consists of any real number x where x ≠ ±2 . x3 + 8 ( x + 2 )( x − 2 x + 4 ) x − 2 x + 4 = = x2 − 4 x−2 ( x + 2 )( x − 2 )
54.
PROBLEM: x 4 − 16 x2 − 4 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x2 − 4 = 0
( x + 2 )( x − 2 ) = 0 x+2=0 x+2−2 = 0−2 x = −2
x−2=0 x−2+2 = 0+2 x=2
or or or
ANSWER: The domain consists of any real number x where x ≠ ±2 . x2 − 4 x2 + 4 x 4 − 16 = = x2 + 4 2 2 x −4 x −4
(
)(
)
Part C: Simplifying Rational Expressions State the restrictions and then simplify.
55.
PROBLEM: x −9 9− x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 9− x = 0 9 − 9 x = −9 − x = −9 x=9
ANSWER: The domain consists of any real number x where x ≠ 9 . x −9 x −9 = = −1 9 − x − ( x − 9)
56.
PROBLEM: 3x − 2 2 − 3x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 − 3x = 0 2 − 2 − 3x = 0 − 2 − 3 x = −2 −3 −2 x= −3 −3 2 x= 3 2 ANSWER: The domain consists of any real number x where x ≠ . 3 3x − 2 3x − 2 = = −1 2 − 3x − ( 3x − 2 )
57.
PROBLEM: x+6 6+ x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 6+ x = 0 6−6+ x = 0−6 x = −6 ANSWER: The domain consists of any real number x where x ≠ −6 . x+6 x+6 = =1 6+ x x+6
58.
PROBLEM: 3x + 1 1 + 3x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 1 + 3x = 0 1 − 1 + 3x = 0 − 1 3x = −1 3 1 x=− 3 3 1 x=− 3 1 ANSWER: The domain consists of any real number x where x ≠ − . 3 3x + 1 3x + 1 = =1 1 + 3x 3x + 1
59.
PROBLEM: ( 2 x − 5)( x − 7 )
( 7 − x )( 2 x − 1)
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: ( 7 − x )( 2 x − 1) = 0
7−x =0 7−7− x = 0−7 x = −7
or or or
x=7
or
x=7
or
2x −1 = 0 2x −1+1 = 0 +1 2x = 1 2 1 x= 2 2 1 x= 2
ANSWER: The domain consists of any real number x where x ≠
( 2 x − 5)( x − 7 ) = ( 2 x − 5)( x − 7 ) = − 2 x − 5 ( 7 − x )( 2 x − 1) − ( x − 7 )( 2 x − 1) 2 x − 1 60.
1 and x ≠ 7 . 2
PROBLEM: ( 3x + 2 )( x + 5 )
( x − 5 )( 2 + 3x )
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: ( x − 5)( 2 + 3x ) = 0
x −5 = 0 x −5+5 = 0+5 x=5
or or or
x=5
or
x=5
or
2 + 3x = 0 2 − 2 + 3x = 0 − 2 3 x = −2 3 2 x=− 3 3 2 x=− 3
2 ANSWER: The domain consists of any real number x where x ≠ − and x ≠ 5 . 3 3 x + 2 x + 5 3 x + 2 x + 5 ( )( ) = ( )( ) = x + 5 ( x − 5)( 2 + 3x ) ( x − 5)( 3x + 2 ) x − 5
61.
PROBLEM: x2 − 4
(2 − x)
2
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve:
(2 − x)
2
=0
2− x = 0 2−2− x = 0−2 x=2
ANSWER: The domain consists of any real number x where x ≠ 2 . x 2 − 4 ( x − 2 )( x + 2 ) x + 2 = = 2 2 x−2 (2 − x) ( x − 2)
62.
PROBLEM: 16 − 9 x 2
( 3x + 4 )
2
SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve:
( 3x + 4 )
2
=0
3x + 4 = 0 3x + 4 − 4 = 0 − 4 3 4 x=− 3 3 4 x=− 3 4 ANSWER: The domain consists of any real number x where x ≠ − . 3 2 ( 3x + 4 )( 4 − 3x ) = 4 − 3x 16 − 9 x = 2 2 3x + 4 ( 3x + 4 ) ( 3x + 4 )
63.
PROBLEM: 4 x 2 (10 − x ) 3 x 3 − 300 x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 3 x 3 − 300 x = 0
(
)
3 x x 2 − 100 = 0 3x = 0
or
x 2 − 100 = 0
x=0
or
( x + 10 )( x − 10 ) = 0
x=0 x=0
or or
x + 10 = 0 x + 10 − 10 = 0 − 10
or x − 10 = 0 or x − 10 + 10 = 0 + 10
x=0
or
x = −10
or
x = 10
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ ±10 . 4 x 2 (10 − x ) 4 x 2 (10 − x ) 4x = =− 3 3x − 300 x −3 x ( x + 10 )(10 − x ) 3 ( x + 10 )
64.
PROBLEM: −2 x + 14 x3 − 49 x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x3 − 49 x = 0
(
)
x x 2 − 49 = 0 x ( x − 7 )( x + 7 ) = 0 x=0 x=0 x=0
or or or
x−7 = 0 or x − 7 + 7 = 0 + 7 or x=7 or
x+7 =0 x+7−7 = 0−7 x = −7
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ ±7 . −2 ( x − 7 ) −2 x + 14 2 = =− 3 x − 49 x x ( x − 7 )( x + 7 ) x ( x + 7)
65.
PROBLEM: 2x2 − 7 x − 4 1 − 4x2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 1 − 4x2 = 0
(1 + 2 x )(1 − 2 x ) = 0 1 + 2x = 0
or
1− 2x = 0
1 −1+ 2x = 0 −1
or
1 − 1 − 2 x = −1
2 x = −1
or
− 2 x = −1
2 1 x=− 2 2 1 x=− 2
−2 −1 x= −2 −2 1 x= 2
or or
ANSWER: The domain consists of any real number x where x ≠ ±
2 x 2 − 7 x − 4 ( 2 x + 1)( x − 4 ) x−4 = = 2 1 − 4x ( 2 x + 1)(1 − 2 x ) 1 − 2 x 66.
1 . 2
PROBLEM: 9 x2 − 4 4 x − 6x2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 4 x − 6 x2 = 0 2 x ( 2 − 3x ) = 0 2x = 0
or
2 − 3x = 0
x=0
or
2 − 2 − 3x = 0 − 2
x=0
or
x=0
or
x=0
or
− 3 x = −2 −3 −2 x= −3 −3 2 x= 3
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠
9 x 2 − 4 ( 3x − 2 )( 3x + 2 ) 3x + 2 = =− 2 −2 x ( 3x − 2 ) 4x − 6x 2x
2 . 3
67.
PROBLEM: x 2 − 5 x − 14 7 − 15 x + 2 x 2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 7 − 15 x + 2 x 2 = 0
( 2 x − 1)( x − 7 ) = 0 2x −1 = 0
or
x−7 = 0
2x −1+1 = 0 +1
or
x−7+7 = 0+7
2x = 1
or
x=7
or
x=7
or
x=7
2 1 x= 2 2 1 x= 2
ANSWER: The domain consists of any real number x where x ≠
( x + 2 )( x − 7 ) = x + 2 x 2 − 5 x − 14 = 2 7 − 15 x + 2 x ( 2 x − 1)( x − 7 ) 2 x − 1 68.
1 and x ≠ 7 . 2
PROBLEM: 2 x3 + x 2 − 2 x − 1 1 + x − 2x2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 1 + x − 2x2 = 0
( 2 x + 1)( − x + 1) = 0 2x +1 = 0 2x +1−1 = 0 −1 2 x = −1 1 x=− 2
or or or or
− x +1 = 0 − x + 1 −1 = 0 −1 − x = −1 x =1
1 ANSWER: The domain consists of any real number x where x ≠ − and x ≠ 1 . 2 3 2 ( x + 1)( 2 x + 1)(1 − x ) = − x − 1 2x + x − 2x −1 =− 2 1+ x − 2x ( 2 x + 1)(1 − x )
69.
PROBLEM: x3 + 2 x − 3x 2 − 6 2 + x2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 + x2 = 0
x2 + 2 − 2 = 0 − 2 x 2 = −2 x = ± −2 ANSWER: The roots of x are complex numbers. Hence, there are no restrictions to the domain. The fraction cannot be simplified further. 2 + x 2 ( x − 3) x3 + 2 x − 3x 2 − 6 = = x −3 2 + x2 2 + x2
(
70.
(
)
)
PROBLEM: 27 + x3 x2 + 6 x + 9 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x2 + 6 x + 9 = 0
( x + 3)
2
=0
x+3= 0 x +3−3 = 0−3 x = −3 ANSWER: The domain consists of any real number x where x ≠ −3 . ( x + 3) x 2 − 3 x + 9 x 2 − 3 x + 9 27 + x3 = = 2 x2 + 6 x + 9 x+3 ( x + 3)
(
)
71.
PROBLEM: 64 − x3 x 2 − 8 x + 16 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x 2 − 8 x + 16 = 0
( x − 4)
2
=0
x−4 = 0 x−4+4 = 0+4 x=4 ANSWER: The domain consists of any real number x where x ≠ 4 . − ( x − 4 ) x 2 + 4 x + 16 x 2 + 4 x + 16 64 − x3 = = − 2 x 2 − 8 x + 16 x−4 ( x − 4)
(
72.
)
PROBLEM: x2 + 4 4 − x2 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 4 − x2 = 0 2− x = 0 or 2+ x = 0 2−2− x = 0−2 or 2−2+ x = 0−2 − x = −2 or x = −2 or x = −2 x=2 ANSWER: The domain consists of any real number x where x ≠ ±2 . The fraction cannot be simplified. x2 + 4 4 − x2
Simplify. (Assume all denominators are nonzero.)
73.
PROBLEM: −15 x 3 y 2 5 xy 2 ( x + y ) SOLUTION: 3x 2 −15 x 3 y 2 = − 5 xy 2 ( x + y ) x+ y ANSWER: −
74.
3x 2 x+ y
PROBLEM: 4 14 x 7 y 2 ( x − 2 y ) 7 x8 y ( x − 2 y )
2
SOLUTION: 4 14 x 7 y 2 ( x − 2 y ) 7 x8 y ( x − 2 y )
ANSWER:
75.
2
=
2 y 2−1 ( x − 2 y )
2y ( x − 2y)
x8−7 2
x
PROBLEM: y+x x2 − y 2 SOLUTION: ( x + y) = 1 y+x = x 2 − y 2 ( x + y )( x − y ) x − y ANSWER:
1 x− y
4−2
=
2y ( x − 2y) x
2
76.
PROBLEM: y−x x2 − y2 SOLUTION: −( x − y) y−x 1 = =− 2 2 x −y ( x − y )( x + y ) x + y ANSWER: −
77.
1 x+ y
PROBLEM: x2 − y 2
( x − y)
2
SOLUTION: x 2 − y 2 ( x − y )( x + y ) x + y = = 2 2 x− y ( x − y) ( x − y)
ANSWER:
78.
x+ y x− y
PROBLEM: a 2 − ab − 6b 2 a 2 − 6ab + 9b 2 SOLUTION: a 2 − ab − 6b 2 ( a + 2b )( a − 3b ) a + 2b = = 2 a 2 − 6ab + 9b 2 a − 3b ( a − 3b ) ANSWER:
a + 2b a − 3b
79.
PROBLEM: 2a 2 − 11a + 12 −32 + 2a 2 SOLUTION: 2a 2 − 11a + 12 ( a − 4 )( 2a − 3) 2a − 3 = = 2 −32 + 2a 2 ( a − 4 )( a + 4 ) 2 ( a + 4 ) ANSWER:
80.
2a − 3 2 ( a + 4)
PROBLEM: a 2b − 3a 2 3a 2 − 3ab SOLUTION: 2 a 2b − 3a 2 a ( b − 3) a ( b − 3) = = 3a 2 − 3ab 3a ( a − b ) 3 ( a − b ) ANSWER:
81.
a ( b − 3)
3( a − b)
PROBLEM: xy 2 − x + y 3 − y x − xy 2 SOLUTION:
(
)
2 xy 2 − x + y 3 − y ( x + y ) y − 1 x+ y = =− 2 2 x − xy x −x y −1
(
ANSWER: −
x+ y x
)
82.
PROBLEM: x3 − xy 2 − x 2 y + y 3 x 2 − 2 xy + y 2 SOLUTION: x3 − xy 2 − x 2 y + y 3 ( x − y )( x + y )( x − y ) = = x+ y 2 x 2 − 2 xy + y 2 ( x − y) ANSWER: x + y
83.
PROBLEM: x3 − 27 x 2 + 3x + 9 SOLUTION:
(
)
( x − 3) x 2 + 3 x + 9 x3 − 27 = = x −3 x 2 + 3x + 9 x 2 + 3x + 9
ANSWER: x − 3
84.
PROBLEM: x2 − x + 1 x3 + 1 SOLUTION:
(
)
x3 + 1 x2 − x + 1 1 = = 3 3 x +1 ( x + 1) x + 1 x + 1
ANSWER:
1 x +1
(
)
Part D: Rational Functions Calculate the following.
85.
PROBLEM: 5x f ( x) = ; f ( 0) , f ( 2) , f ( 4) x −3 ANSWER: Substitute the values in for x. f ( x ) = f (0)
f ( x) = 86.
5 ( 0) 0−3
f ( x ) = f ( 2)
f ( x) =
=0
5 ( 2) 2−3
= −10
f ( x ) = f ( 4)
f ( x) =
5 ( 4) 4−3
= 20
PROBLEM: x+7 f ( x) = 2 ; f ( −1) , f ( 0 ) , f (1) x +1 ANSWER: Substitute the values in for x. f ( x ) = f ( −1)
f ( x) = 87.
( −1) + 7 ( −1)
2
+1
=
f ( x ) = f (0)
f ( x) =
6 =3 2
0+7 =7 0 +1
f ( x ) = f (1)
f ( x) =
1+ 7 =4 12 + 1
PROBLEM: x3 ; g ( 0 ) , g ( 2 ) , g ( −2 ) g ( x) = 2 ( x − 2) ANSWER: Substitute the values in for x. g ( x ) = g (0)
g ( x) =
0
( 0 − 2)
2
g ( x ) = g ( 2)
=0
g ( x) =
23
( 2 − 2)
2
= undefined
g ( x ) = g ( −2 )
g ( x) =
( −2 )
3
( −2 − 2 )
2
=
−8 1 =− 16 2
88.
PROBLEM: x2 − 9 g ( x) = ; g ( −2 ) , g ( 0 ) , g ( 2 ) 9 − x2 ANSWER: Substitute the values in for x. g ( x ) = g ( −2 )
g ( x) = 89.
( −2 ) − 9 = −1 2 9 − ( −2 ) 2
g ( x ) = g (0)
g ( x) =
0−9 = −1 9−0
g ( x ) = g ( 2)
g ( x) =
22 − 9 = −1 9 − 22
PROBLEM: x3 g ( x) = 2 ; g ( −1) , g ( 0 ) , g (1) x +1 ANSWER: Substitute the values in for x. g ( x ) = g ( −1)
g ( x) = 90.
( −1) = − 1 2 ( −1) + 1 2 3
g ( x ) = g (0)
g ( x) =
0 =0 0 +1
g ( x ) = g (1)
g ( x) =
13 1 = 2 1 +1 2
PROBLEM: 5x + 1 g ( x) = 2 ; g ( −1/ 5 ) , g ( −1) , g ( −5 ) x − 25 ANSWER: Substitute the values in for x. ⎛ 1⎞ g ( x) = g ⎜ − ⎟ ⎝ 5⎠
⎛ 1⎞ 5⎜ − ⎟ +1 5 g ( x) = ⎝ 2 ⎠ ⎛ 1⎞ ⎜ − ⎟ − 25 ⎝ 5⎠ 0 = =0 2 ⎛ 1⎞ ⎜ − ⎟ − 25 ⎝ 5⎠
g ( x ) = g ( −1)
g ( x) =
5 ( −1) + 1
( −1)
2
− 25
=
g ( x ) = g ( −5 )
−4 1 = −24 6
g ( x) =
5 ( −5 ) + 1
( −5 )
= undefined
2
− 25
=
−24 0
State the restrictions to the domain and then simplify.
91.
PROBLEM: −3x 2 − 6 x f ( x) = 2 x + 4x + 4 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x2 + 4 x + 4 = 0
( x + 2)
2
=0
x+2=0 x + 2 − 2 = −2 x = −2 ANSWER: The domain consists of any real number x where x ≠ −2 . −3x 2 − 6 x −3x ( x + 2 ) 3x = =− 2 2 x + 4x + 4 x+2 ( x + 2)
f ( x) = − 92.
3x x+2
PROBLEM: x2 + 6 x + 9 f ( x) = 2 2 x + 5x − 3 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 2 x2 + 5x − 3 = 0
( 2 x − 1)( x + 3) = 0 2x −1 = 0 2x −1+1 = 0 +1 x=
1 2
or or
x+3= 0 x + 3 − 3 = −3
or
x = −3
ANSWER: The domain consists of any real number x where x ≠
( x + 3) x2 + 6x + 9 x+3 = = 2 2 x + 5 x − 3 ( 2 x − 1)( x + 3) 2 x − 1 2
f ( x) =
x+3 2x −1
1 and x = −3 . 2
93.
PROBLEM: 9− x g ( x) = 2 x − 81 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x 2 − 81 = 0
( x + 9 )( x − 9 ) = 0 x+9 = 0
or
x+9−9 = 0−9 x = −9
or or
x −9 = 0 x −9+9 = 0+9 x=9
ANSWER: The domain consists of any real number x where x ≠ ±9 . − ( x − 9) 9− x 1 = =− 2 x − 81 ( x + 9 )( x − 9 ) x−9
g ( x) = − 94.
1 x −9
PROBLEM: x3 − 27 g ( x) = 3− x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 3− x = 0 3−3− x = 0−3 − x = −3 x=3
ANSWER: The domain consists of any real number x where x ≠ 3 . 2 x3 − 27 ( x − 3) x + 3x + 9 = = − x 2 + 3x + 9 3− x − ( x − 3)
(
(
g ( x ) = − x 2 + 3x + 9
)
)
(
)
95.
PROBLEM: 3x − 15 g ( x) = 10 − 2 x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 10 − 2 x = 0 10 + 10 − 2 x = 0 − 10 −2 x = −10 −2 −10 x= −2 −2 x=5 ANSWER: The domain consists of any real number x where x ≠ 5 . 3x − 15 3 ( x − 5) 3 = =− 10 − 2 x −2 ( x − 5) 2
96.
PROBLEM: 25 − 5 x g ( x) = 4 x − 20 SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: 4 x − 20 = 0 4 x − 20 + 20 = 0 + 20 4 x = 20 4 20 x= 4 4 x=5 ANSWER: The domain consists of any real number x where x = 5 . 25 − 5 x −5 ( x − 5) 5 = =− 4 x − 20 4 ( x − 5) 4
g ( x) = −
5 4
97.
PROBLEM: The cost, in dollars, of producing coffee mugs with a company logo is given by C ( x ) = x + 40 where x represents the number of mugs produced. Calculate the
average cost of producing 100 mugs and the average cost of producing 500 mugs. SOLUTION: C ( x ) = x + 40
C (100 ) = 100 + 40 = 140 C ( 500 ) = 500 + 40 = 540 Total cost Average cost = Number of mugs 140 = $1.40 . For 100 mugs, average cost = 100 540 = $1.08 . For 500 mugs, average cost = 500 ANSWER: The average cost of producing 100 mugs is $1.40 per mug. The average cost of producing 500 mugs is $1.08 per mug.
98.
PROBLEM: The cost, in dollars, of renting a moving truck for the day is given by C ( x ) = 0.45 x + 90 where x represents the number of miles driven. Calculate the
average cost per mile if the truck is driven 250 miles in one day. SOLUTION: C ( x ) = 0.45 x + 90
C ( 250 ) = 0.45 ( 250 ) + 90 = 202.5 Average cost =
Total cost Number of miles
ANSWER: For 250 miles, average cost =
202.5 = $0.81 . 250
99.
PROBLEM: The cost, in dollars, of producing sweatshirts with a custom design on the back is given by C ( x) = 1200 + (12 − 0.05 x) x where x represents the number of sweatshirts produced. Calculate the average cost of producing 150 custom sweatshirts. SOLUTION: C ( x) = 1200 + (12 − 0.05 x) x
C (150) = 1200 + (12 − 0.05 (150 ) )150 = 1875 Average cost =
Total cost Number of sweatshirts
ANSWER: For 150 sweat shirts, average cost = 100.
1875 = $12.5 150
PROBLEM: The cost, in dollars, of producing a custom injected molded part is given by C ( x) = 500 + (3 − 0.001x) x where x represents the number of parts produced. Calculate the average cost of producing 1000 custom parts. SOLUTION: C ( x) = 500 + (3 − 0.001x) x C (1000) = 500 + ( 3 − 0.001(1000 ) )1000 = 2500
Average cost =
Total cost Number of parts
ANSWER: For 1000 parts, average cost =
2500 = $2.5 1000
Part E: Discussion Board
101.
PROBLEM:
Explain why
b−a = −1 and illustrate this fact by substituting in some numbers. a −b
ANSWER: b−a b−a = = −1 a − b − (b − a )
(Students answers may vary.) Put a = 1; b = 2 b − a 2 −1 1 = = = −1 a − b 1 − 2 −1
102.
PROBLEM:
Explain why
b+a = 1 and illustrate this fact by substituting in some numbers. a+b
ANSWER: b+a a+b = =1 a+b a+b (Students answers may vary.) Put a = 1; b = 2 b + a 2 +1 3 = = =1 a + b 1+ 2 3
103.
PROBLEM:
Explain why we cannot cancel x in the expression
x . x +1
ANSWER: The denominator cannot be factored into a multiple of x. Hence, we cannot cancel x in the expression. 7.2 Multiplying and Dividing Rational Expressions Part A: Multiplying Rational Expressions Multiply. (Assume all denominators are nonzero)
1.
PROBLEM: 2x 9 ⋅ 3 4x2 SOLUTION: 2x 9 3 3 ⋅ 2 = 2−1 = 3 4x 2x 2x ANSWER:
3 2x
2.
PROBLEM: −5 x 3 y 2 ⋅ y 25 x SOLUTION: −5 x3 y 2 x 3−1 y 2−1 x2 y ⋅ =− =− y 25 x 5 5 ANSWER: −
3.
x2 y 5
PROBLEM: 5x2 4 y 2 ⋅ 2 y 15 x3 SOLUTION: 5 x 2 4 y 2 2 y 2−1 2 y ⋅ = = 2 y 15 x3 3x3−2 3x
4.
ANSWER:
2y 3x
PROBLEM: 16a 4 49b ⋅ 7b 2 32a3 SOLUTION: 16a 4 49b 7a 4−3 7a ⋅ = = 7b2 32a3 2b 2−1 2b ANSWER:
7a 2b
5.
PROBLEM: x − 6 24 x 2 ⋅ 12 x 3 x − 6 SOLUTION: x − 6 24 x 2 2 2 ⋅ = 3− 2 = 3 12 x x − 6 x x ANSWER:
6.
2 x
PROBLEM: x + 10 x − 2 ⋅ 2 x − 1 x + 10 SOLUTION: x + 10 x − 2 x−2 ⋅ = 2 x − 1 x + 10 2 x − 1 ANSWER:
7.
x−2 2x −1
PROBLEM:
( y − 1) y +1
2
⋅
1 y −1
SOLUTION: 2 ( y − 1) ⋅ 1 = y − 1 y +1 y −1 y +1 ANSWER:
y −1 y +1
8.
PROBLEM: y2 − 9 2 y − 3 ⋅ y +3 y −3 SOLUTION: y 2 − 9 2 y − 3 ( y − 3)( y + 3)( 2 y − 3) ⋅ = = 2y −3 y +3 y −3 ( y + 3)( y − 3) ANSWER: 2 y − 3
9.
PROBLEM: 2a − 5 2a + 5 ⋅ a − 5 4a 2 − 25 SOLUTION: ( 2a − 5 ) ⋅ ( 2 a + 5 ) = 1 2a − 5 2 a + 5 ⋅ 2 = a − 5 4a − 25 ( a − 5 ) ( 2a + 5 )( 2a − 5 ) a − 5 ANSWER:
10.
1 a −5
PROBLEM: 2a 2 − 9a + 4 2 ⋅ ( a + 4a ) a 2 − 16 SOLUTION: ( 2a − 1)( a − 4 ) a ( a + 4 ) = a 2a − 1 2a 2 − 9 a + 4 2 ⋅ ( a + 4a ) = ( ) 2 a − 16 ( a + 4 )( a − 4 ) ANSWER: a ( 2a − 1)
11.
PROBLEM: 2 x 2 + 3x − 2 2 x ⋅ 2 ( 2 x − 1) x + 2 SOLUTION: ( 2 x − 1)( x + 2 ) 2 x = 2 x 2 x 2 + 3x − 2 2 x ⋅ = 2 2 ( 2 x − 1) x + 2 ( 2 x − 1) ( x + 2 ) 2 x − 1 ANSWER:
12.
2x 2x −1
PROBLEM: 9 x 2 + 19 x + 2 x 2 − 4 x + 4 ⋅ 2 4 − x2 9x − 8x −1 SOLUTION:
( x + 2 )( 9 x + 1)( x − 2 ) 9 x 2 + 19 x + 2 x 2 − 4 x + 4 x−2 ⋅ 2 = =− 2 4− x 9 x − 8 x − 1 − ( x − 2 )( x + 2 )( x − 1)( 9 x + 1) x −1 x−2 ANSWER: − x −1 2
13.
PROBLEM: x 2 + 8 x + 16 x 2 − 3x − 4 ⋅ 2 16 − x 2 x + 5x + 4 SOLUTION:
( x + 4 ) ( x − 4 )( x + 1) = −1 x 2 + 8 x + 16 x 2 − 3x − 4 ⋅ = 16 − x 2 x 2 + 5 x + 4 − ( x − 4 )( x + 4 )( x + 1)( x + 4 ) 2
ANSWER: −1
14.
PROBLEM: x 2 − x − 2 x 2 + 2 x − 15 ⋅ x2 + 8x + 7 x2 − 5x + 6 SOLUTION: x 2 − x − 2 x 2 + 2 x − 15 ( x + 1)( x − 2 )( x − 3)( x + 5 ) x + 5 ⋅ = = x 2 + 8 x + 7 x 2 − 5 x + 6 ( x + 1)( x + 7 )( x − 2 )( x − 3) x + 7 ANSWER:
15.
x+5 x+7
PROBLEM: x +1 3 − x ⋅ x −3 x +5 SOLUTION: x + 1 3 − x x + 1 − ( x − 3) x +1 ⋅ = ⋅ =− x −3 x +5 x −3 x +5 x+5 ANSWER: −
16.
x +1 x+5
PROBLEM: 2x −1 x + 6 ⋅ x −1 1 − 2x SOLUTION: 2x −1 x + 6 2x −1 x + 6 x+6 ⋅ = ⋅ =− x − 1 1 − 2 x x − 1 − ( 2 x − 1) x −1 ANSWER: −
x+6 x −1
17.
PROBLEM: 9+ x 3 ⋅ 3x + 1 x + 9 SOLUTION: 9+ x 3 x+9 3 3 ⋅ = ⋅ = 3x + 1 x + 9 3x + 1 x + 9 3x + 1 ANSWER:
18.
3 3x + 1
PROBLEM: 1 5x + 2 ⋅ 2 + 5x 5x SOLUTION: 1 5x + 2 1 2 + 5x 1 ⋅ = ⋅ = 2 + 5x 5x 2 + 5x 5x 5x ANSWER:
19.
1 5x
PROBLEM: 100 − y 2 25 y 2 ⋅ y − 10 y + 10 SOLUTION: 2 100 − y 2 25 y 2 − ( y + 10 )( y − 10 ) 25 y ⋅ = = −25 y 2 y − 10 y + 10 ( y + 10 )( y − 10 ) ANSWER: −25y 2
20.
PROBLEM: 3 y 3 36 y 2 − 25 ⋅ 6y −5 5+ 6y SOLUTION: 3 3 y 3 36 y 2 − 25 3 y ( 6 y − 5 )( 6 y + 5 ) ⋅ = = 3 y3 6y − 5 5 + 6y ( 6 y − 5)( 6 y + 5) ANSWER: 3y 3
21.
PROBLEM: 3a 2 + 14a − 5 3a + 1 ⋅ a2 + 1 1 − 9a 2 SOLUTION: ( 3a − 1)( a + 5)( 3a + 1) = − a + 5 3a 2 + 14a − 5 3a + 1 ⋅ = 2 2 a +1 1 − 9a a2 + 1 − ( a 2 + 1) ( 3a + 1)( 3a − 1) ANSWER: −
22.
a+5 a2 + 1
PROBLEM: 4a 2 − 16a 1 − 16a 2 ⋅ 2 4a − 1 4a − 15a − 4 SOLUTION: −4a ( a − 4 )( 4a + 1)( 4a − 1) 4a 2 − 16a 1 − 16a 2 ⋅ 2 = 4a − 1 4a − 15a − 4 ( 4a − 1)( 4a + 1)( a − 4 )
= −4a
ANSWER: −4a
23.
PROBLEM: x+9 ⋅ x 2 − 81 2 − x + 14 x − 45
(
)
SOLUTION:
( x + 9 )( x + 9 )( x − 9 ) = − ( x + 9 ) x+9 ⋅ ( x 2 − 81) = 2 − x + 14 x − 45 − ( x − 5)( x − 9 ) x −5 ( x + 9) ANSWER: − x−5
2
2
24.
PROBLEM: 1 ⋅ 25 x 2 + 20 x + 4 2 + 5x
(
)
SOLUTION:
( 2 + 5x ) = 5x + 2 1 ⋅ ( 25 x 2 + 20 x + 4 ) = 2 + 5x ( 2 + 5x ) 2
ANSWER: 5 x + 2
25.
PROBLEM: x2 + x − 6 2x2 − 8 ⋅ 3x 2 + 15 x + 18 x 2 − 4 x + 4 SOLUTION: ( x + 3)( x − 2 ) 2 ( x − 2 )( x + 2 ) = 2 x2 + x − 6 2x2 − 8 ⋅ = 2 2 2 3x + 15 x + 18 x − 4 x + 4 3 3 ( x + 2 )( x + 3)( x − 2 ) ANSWER:
26.
2 3
PROBLEM: 5 x 2 − 4 x − 1 25 x 2 − 10 x + 1 ⋅ 5x2 − 6 x + 1 3 − 75 x 2 SOLUTION:
( 5x + 1)( x − 1)( 5x − 1) 5 x 2 − 4 x − 1 25 x 2 − 10 x + 1 1 ⋅ = =− 2 2 5x − 6x + 1 3 − 75 x − ( x − 1)( 5 x − 1) 3 ( 5 x + 1)( 5 x − 1) 3 2
ANSWER: −
1 3
Part B: Dividing Rational Expressions Divide. (Assume all denominators are nonzero)
27.
PROBLEM: 5 x 15 x 2 ÷ 8 4 SOLUTION: 5 x 15 x 2 5 x 4 1 ÷ = ⋅ = 2 8 4 8 15 x 6x ANSWER:
28.
1 6x
PROBLEM: 3 −15 ÷ 8 y 2 y2 SOLUTION: 3 −15 3 ⎛ 2 y 2 ⎞ y ÷ 2 = ⋅⎜ − ⎟=− 8y 2y 8 y ⎝ 15 ⎠ 20 ANSWER: −
29.
y 20
PROBLEM: 5 x9 3 y3 25 x10 9 y5 SOLUTION: 5 x9 3 y2 5 x9 9 y 5 3 y3 = ⋅ = 25 x10 3 y 3 25 x10 5 x 9 y5 ANSWER:
3y2 5x
30.
PROBLEM: 12 x 4 y 2 21z 5 6 x3 y 2 7 z3 SOLUTION: 12 x 4 y 2 4 2 3 21z 5 = 12 x y ⋅ 7 z = 2 x 6 x3 y 2 21z 5 6 x3 y 2 3z 2 7 z3 ANSWER:
31.
2x 3z 2
PROBLEM: 2 ( x − 4) ÷ x − 4 30 x 4 15 x SOLUTION: 2 2 ( x − 4 ) ÷ x − 4 = ( x − 4 ) ⋅ 15 x = x − 4 30 x 4 15 x 30 x 4 x − 4 2 x3 ANSWER:
32.
x−4 2 x3
PROBLEM: 5 y4 10 y 5 ÷ 2 3 10 ( 3 y − 5) 2 ( 3 y − 5 ) SOLUTION:
5 y4 10 ( 3 y − 5)
2
ANSWER:
÷
10 y 5 2 ( 3 y − 5)
3y − 5 10 y
3
=
5 y4 10 ( 3 y − 5)
2
⋅
2 ( 3 y − 5) 10 y
5
3
=
3y − 5 10 y
33.
PROBLEM: x2 − 9 ÷ ( x − 3) 5x SOLUTION: ( x + 3)( x − 3) = x + 3 x2 − 9 x2 − 9 1 ÷ ( x − 3) = ⋅ = 5x 5 x ( x − 3) 5 x ( x − 3) 5x ANSWER:
34.
x+3 5x
PROBLEM: y 2 − 64 ÷ (8 + y ) 8y SOLUTION: ( y + 8)( y − 8) = y − 8 y 2 − 64 y 2 − 64 1 ÷ (8 + y ) = ⋅ = 8y 8y 8+ y 8 y ( y + 8) 8y ANSWER:
35.
y −8 8y
PROBLEM: 2 ( a − 8) ÷ a − 8 2a 2 + 10a a SOLUTION: 2 2 2 ( a − 8) ÷ a − 8 = ( a − 8) ⋅ a = ( a − 8) ⋅ a = a − 8 2a 2 + 10a a 2a 2 + 10a a − 8 2a ( a + 5) a − 8 2 ( a + 5 ) ANSWER:
a −8 2 ( a + 5)
36.
PROBLEM:
24a 2b3 (a − 2b) ÷ SOLUTION:
24a 2b3 (a − 2b) ÷
12ab ( a − 2b ) 5 12ab ( a − 2b ) 5
= 24a 2b3 (a − 2b) ⋅
5 = 10ab 2 12ab ( a − 2b )
ANSWER: 10ab 2
37.
PROBLEM: x 2 + 7 x + 10 1 ÷ 2 2 x + 4x + 4 x − 4 SOLUTION: ( x + 2 )( x + 5)( x + 2 )( x − 2 ) = x + 5 x − 2 x 2 + 7 x + 10 1 x 2 + 7 x + 10 2 ÷ = ⋅x −4= ( )( ) 2 2 2 2 x + 4x + 4 x − 4 x + 4x + 4 ( x + 2) ANSWER: ( x + 5 )( x − 2 )
38.
PROBLEM: 2x2 − x −1 1 ÷ 2 2 2 x − 3x + 1 4 x − 1 SOLUTION: ( x − 1)( 2 x + 1)( 2 x + 1)( 2 x − 1) 2x2 − x −1 1 2 x2 − x −1 ÷ = ⋅ 4 x2 − 1 = 2 2 2 2 x − 3x + 1 4 x − 1 2 x − 3x + 1 ( 2 x − 1)( x − 1)
= ( 2 x + 1) ANSWER: ( 2 x + 1)
2
2
39.
PROBLEM: y +1 y2 −1 ÷ y2 − 3y y2 − 6 y + 9 SOLUTION:
( y + 1)( y − 3) y +1 y2 −1 y +1 y2 − 6 y + 9 ÷ = ⋅ = 2 2 2 2 y − 3y y − 6 y + 9 y − 3y y −1 y ( y − 3)( y + 1)( y − 1) 2
= ANSWER:
40.
y −3 y ( y − 1)
y −3 y ( y − 1)
PROBLEM: 9 − a2 2a 2 − 10a ÷ a 2 − 8a + 15 a 2 − 10a + 25 SOLUTION:
a 2 − 10a + 25 − ( a − 3)( a + 3)( a − 5 ) 9 − a2 2a 2 − 10a 9 − a2 ÷ = ⋅ = a 2 − 8a + 15 a 2 − 10a + 25 a 2 − 8a + 15 2a 2 − 10a ( a − 3)( a − 5) 2a ( a − 5) 2
=−
ANSWER: −
41.
a+3 2a
a+3 2a
PROBLEM: a 2 − 3a − 18 a 2 + a − 6 ÷ 2a 2 − 11a − 6 2a 2 − a − 1 SOLUTION: a 2 − 3a − 18 a 2 + a − 6 a 2 − 3a − 18 2a 2 − a − 1 ÷ = ⋅ 2a 2 − 11a − 6 2a 2 − a − 1 2a 2 − 11a − 6 a 2 + a − 6 ( a + 3)( a − 6 )( 2a + 1)( a − 1) = ( 2a + 1)( a − 6 )( a − 2 )( a + 3)
= ANSWER:
a −1 a−2
a −1 a−2
42.
PROBLEM: y 2 − 7 y + 10 y 2 + 5 y − 14 2 y2 − 9 y − 5 y 2 + 14 y + 49 SOLUTION: y 2 − 7 y + 10 2 y − 2 )( y − 5)( y + 7 ) ( y 2 − 7 y + 10 y 2 + 14 y + 49 y 2 + 5 y − 14 = 2 ⋅ = 2 y2 − 9 y − 5 y + 5 y − 14 2 y 2 − 9 y − 5 ( y − 2 )( y + 7 )( 2 y + 1)( y − 5) y 2 + 14 y + 49 y+7 = 2 y +1 y+7 ANSWER: 2 y +1
43.
PROBLEM: 6 y2 + y −1 4 y2 + 4 y +1 3 y2 + 2 y −1 2 y2 − 7 y − 4 SOLUTION: 6 y2 + y −1 6 y2 + y −1 2 y2 − 7 y − 4 4 y2 + 4 y +1 = ⋅ 3 y2 + 2 y −1 4 y2 + 4 y +1 3 y2 + 2 y −1 2 y2 − 7 y − 4 =
( 3 y − 1)( 2 y + 1)( 2 y + 1)( y − 4 ) 2 ( 2 y + 1) ( 3 y − 1)( y + 1)
y−4 y +1 y−4 ANSWER: y +1 =
44.
PROBLEM: x 2 − 7 x − 18 x 2 − 81 ÷ x 2 + 8 x + 12 x 2 + 12 x + 36 SOLUTION: x 2 − 7 x − 18 x 2 − 81 x 2 − 7 x − 18 x 2 + 12 x + 36 ÷ = ⋅ x 2 + 8 x + 12 x 2 + 12 x + 36 x 2 + 8 x + 12 x 2 − 81
( x − 9 )( x + 2 )( x + 6 ) = ( x + 2 )( x + 6 )( x − 9 )( x + 9 ) 2
= ANSWER:
45.
x+6 x+9
x+6 x+9
PROBLEM: 4a 2 − b 2 2 ÷ ( b − 2a ) b + 2a SOLUTION: − ( b + 2a )( b − 2a ) 4a 2 − b 2 4a 2 − b 2 1 2 ÷ ( b − 2a ) = ⋅ = b + 2a b + 2a ( b − 2a )2 ( b + 2a )( b − 2a )2
=− ANSWER:
46.
1 1 = b − 2a 2a − b
1 2a − b
PROBLEM: x2 − y2 2 ÷ ( y − x) y+x SOLUTION: ( x + y )( x − y ) x2 − y2 x2 − y2 1 2 ÷ ( y − x) = ⋅ = 2 2 y+x y + x ( y − x) ( x + y )( x − y ) =
ANSWER:
1 x− y
1 x− y
47.
PROBLEM: 5 y 2 ( y − 3) 25 y ( 3 − y ) ÷ 4 x3 2 x2 SOLUTION: 5 y 2 ( y − 3) 25 y ( 3 − y ) 5 y 2 ( y − 3) 5 y 2 ( y − 3) ⋅ 2 x 2 2 x2 ÷ = ⋅ = 4 x3 2x2 4 x3 25 y ( 3 − y ) −4 x 3 ⋅ 25 y ( y − 3) =−
ANSWER: −
48.
y 10 x
y 10 x
PROBLEM: 15 x 3 25 x 6 ÷ 3 ( y + 7 ) 9 ( 7 + y )2 SOLUTION:
15 x 3 25 x 6 15 x3 9 ( 7 + y ) 15 x 3 9 ( y + 7 ) ÷ = ⋅ = ⋅ 3 ( y + 7 ) 9 ( 7 + y )2 3 ( y + 7 ) 25 x 6 3( y + 7) 25 x 6 2
=
ANSWER:
49.
9 ( y + 7)
9 ( y + 7) 5 x3
5 x3
PROBLEM: 3x + 4 7 x ÷ x −8 8− x SOLUTION: 3x + 4 7 x 3x + 4 8 − x 3x + 4 x − 8 ÷ = ⋅ =− ⋅ x − 8 8 − x x − 8 7x x − 8 7x 3x + 4 =− 7x ANSWER: −
3x + 4 7x
2
50.
PROBLEM: 3x − 2 2 − 3x ÷ 2x +1 3x SOLUTION: 3x − 2 2 − 3x 3x − 2 3x 3x − 2 3x ÷ = ⋅ = ⋅ 2x +1 3x 2 x + 1 2 − 3x 2 x + 1 − ( 3x − 2 )
=− ANSWER: −
51.
3x 2x + 1
3x 2x +1
PROBLEM:
( 7 x − 1) 4x +1
2
÷
28 x 2 − 11x + 1 1 − 4x
SOLUTION: 2 2 ( 7 x − 1) ÷ 28 x 2 − 11x + 1 = ( 7 x − 1) ⋅ 1 − 4 x 4x +1 1− 4x 4 x + 1 28 x 2 − 11x + 1
− ( 7 x − 1) ( 4 x − 1) 2
=
( 4 x + 1)( 4 x − 1)( 7 x − 1)
=− ANSWER: −
52.
7 x −1 4x +1
7x −1 4x +1
PROBLEM: 4x 2− x ÷ 2 2 ( x + 2) x − 4 SOLUTION: 4x 2− x 4x x 2 − 4 4 x ( x − 2 )( x + 2 ) ÷ = ⋅ = 2 2 2 2 ( x + 2) x − 4 ( x + 2) 2 − x − ( x + 2) ( x − 2) =−
ANSWER: −
4x x+2
4x x+2
53.
PROBLEM: a 2 − b2 2 ÷ (b − a ) a SOLUTION: ( a + b )( a − b ) a 2 − b2 a 2 − b2 1 2 ÷ (b − a ) = ⋅ = 2 2 a a a ( a − b) (b − a )
= ANSWER:
54.
a+b a ( a − b)
a+b a (a − b)
PROBLEM: (a − 2b) 2 ÷ 2b 2 + ab − a 2 2b
(
)
SOLUTION: (a − 2b) 2 (a − 2b) 2 (a − 2b) 2 1 1 ÷ 2b 2 + ab − a 2 = ⋅ = ⋅ 2 2 2b 2b 2b ( 2b − a )( a + b ) 2b + ab − a
(
)
(
( 2b − a ) = 2b
=
ANSWER:
2b − a 2b ( a + b )
2
⋅
2b − a 2b ( a + b )
)
1 ( 2b − a )( a + b )
55.
PROBLEM: x2 − 6 x + 9 9 − x2 ÷ x 2 + 7 x + 12 x 2 + 8 x + 16 SOLUTION: x2 − 6 x + 9 9 − x2 x 2 − 6 x + 9 x 2 + 8 x + 16 ÷ = ⋅ x 2 + 7 x + 12 x 2 + 8 x + 16 x 2 + 7 x + 12 9 − x2
( x − 3) ( x + 4 ) = − ( x + 4 )( x + 3)( x + 3)( x − 3) ( x − 3)( x + 4 ) =− 2 ( x + 3) ( x − 3)( x + 4 ) ANSWER: − 2 ( x + 3) 2
56.
2
PROBLEM: 2x2 − 9x − 5 1 − 4x + 4x2 ÷ 25 − x 2 −2 x 2 − 9 x + 5 SOLUTION: 2 x2 − 9 x − 5 1 − 4 x + 4 x2 2 x 2 − 9 x − 5 −2 x 2 − 9 x + 5 ÷ = ⋅ 25 − x 2 25 − x 2 −2 x 2 − 9 x + 5 1 − 4x + 4x2 − ( 2 x + 1)( x − 5 )( 2 x − 1)( x + 5 ) = 2 − ( x − 5 )( x + 5 )( 2 x − 1)
= ANSWER:
2x +1 2x −1
2x +1 2x −1
57.
PROBLEM: 3 x 2 − 16 x + 5 100 − 4 x 2 9x2 − 6x + 1 3 x 2 + 14 x − 5 SOLUTION: 3 x 2 − 16 x + 5 3x 2 − 16 x + 5 3x 2 + 14 x − 5 100 − 4 x 2 = ⋅ 2 9x2 − 6x + 1 100 − 4 x 2 9x − 6x +1 2 3 x + 14 x − 5 ( 3x − 1)( x − 5)( 3x − 1)( x + 5) = 2 −4 ( x + 5 )( x − 5 )( 3x − 1)
=− ANSWER: −
58.
1 4
1 4
PROBLEM: 10 x 2 − 25 x − 15 x2 − 6x + 9 9 − x2 x2 + 6 x + 9 SOLUTION: 10 x 2 − 25 x − 15 2 2 2 x 2 − 6 x + 9 = 10 x − 25 x − 15 ⋅ x + 6 x + 9 = 5 ( 2 x + 1)( x − 3)( x + 3) 2 9 − x2 x2 − 6x + 9 9 − x2 − ( x − 3) ( x − 3)( x + 3) x2 + 6 x + 9 5 ( 2 x + 1)( x + 3) =− 2 ( x − 3) ANSWER: −
5 ( 2 x + 1)( x + 3)
( x − 3)
2
Recall that multiplication and division is to be performed in the order as they appear from left to right. Simplify the following:
59.
PROBLEM: 1 x −1 x −1 ⋅ ÷ x 2 x + 3 x3 SOLUTION: 1 x − 1 x − 1 1 x − 1 x3 x ⋅ ÷ 3 = 2⋅ ⋅ = 2 x x+3 x x x + 3 x − 1 ( x + 3) ANSWER:
60.
x ( x + 3)
PROBLEM: x−7 1 x−7 ⋅ ÷ x + 9 x3 x SOLUTION: x−7 1 x−7 x−7 1 x 1 ⋅ 3÷ = ⋅ 3⋅ = 2 x+9 x x x + 9 x x − 7 x ( x + 9) ANSWER:
61.
1 x ( x + 9) 2
PROBLEM: x +1 x x2 ÷ ⋅ x − 2 x − 5 x +1 SOLUTION: x ( x − 5) x +1 x x2 x + 1 x − 5 x2 ÷ ⋅ = ⋅ ⋅ = x − 2 x − 5 x +1 x − 2 x x +1 x−2 ANSWER:
x ( x − 5) x−2
62.
PROBLEM: x+4 x −3 x + 4 ÷ ⋅ 2x + 5 2x + 5 x − 3 SOLUTION:
x+4 x − 3 x + 4 x + 4 2x + 5 x + 4 ⎛ x + 4 ⎞ ÷ ⋅ = ⋅ ⋅ =⎜ ⎟ 2x + 5 2x + 5 x − 3 2x + 5 x − 3 x − 3 ⎝ x − 3 ⎠ ⎛ x+4⎞ ANSWER: ⎜ ⎟ ⎝ x−3 ⎠ 63.
2
2
PROBLEM: 2x −1 x − 4 x − 4 ÷ ⋅ x + 1 x2 + 1 2x −1 SOLUTION: 2 x − 1 x − 4 x − 4 2 x − 1 x2 + 1 x − 4 x2 + 1 ÷ ⋅ = ⋅ ⋅ = x + 1 x2 + 1 2x −1 x + 1 x − 4 2x −1 x + 1 ANSWER:
64.
x2 + 1 x +1
PROBLEM: 4 x 2 − 1 2 x − 1 3x + 2 ÷ ⋅ 3x + 2 x + 5 2 x + 1 SOLUTION: 4 x 2 − 1 2 x − 1 3x + 2 4 x 2 − 1 x + 5 3x + 2 ÷ ⋅ = ⋅ ⋅ 3x + 2 x + 5 2 x + 1 3x + 2 2 x − 1 2 x + 1 ( 2 x − 1)( 2 x + 1)( x + 5)( 3x + 2 ) = ( 3x + 2 )( 2 x − 1)( 2 x + 1) ANSWER: x + 5
= x+5
Part C: Multiplying and Dividing Rational Functions Calculate ( f ⋅ g ) ( x) and determine the restrictions to the domain.
65.
PROBLEM: 1 1 f ( x) = and g ( x) = x x −1 SOLUTION: 1 1 f ( x) = and g ( x) = x x −1 1 1 1 = ( f ⋅ g ) ( x) = ⋅ x x − 1 x ( x − 1) In this case, the domain of f ( x) consists of all real numbers except 0, and the domain of g ( x) consists all real numbers except 1. 1 x ( x − 1) The domain consists of any real number x where x ≠ 0 and x ≠ 1 .
ANSWER: ( f ⋅ g ) ( x) =
66.
PROBLEM: x +1 f ( x) = and g ( x) = x 2 − 1 x −1 SOLUTION: x +1 f ( x) = and g ( x) = x 2 − 1 x −1 ( x + 1)( x + 1)( x − 1) = x + 1 2 x +1 2 ⋅ x −1 = ( f ⋅ g ) ( x) = ( ) x −1 ( x − 1) In this case, the domain of f ( x) consists of all real numbers except 1, and the domain of g ( x) consists all real numbers. ANSWER: ( f ⋅ g ) ( x) = ( x + 1)
2
The domain consists of any real number x where x ≠ 1 .
67.
PROBLEM: 3x + 2 x2 − 4 f ( x) = and g ( x) = 2 x+2 ( 3x + 2 ) SOLUTION: 3x + 2 x2 − 4 f ( x) = and g ( x) = 2 x+2 ( 3x + 2 )
( f ⋅ g ) ( x) =
( 3x + 2 )( x − 2 )( x + 2 ) = x − 2 3x + 2 x 2 − 4 ⋅ = 2 2 x + 2 ( 3x + 2 ) 3x + 2 ( x + 2 )( 3x + 2 )
In this case, the domain of f ( x) consists of all real numbers except –2, and the 2 domain of g ( x) consists all real numbers except − . 3 ANSWER: ( f ⋅ g ) ( x) =
x−2 3x + 2
2 The domain consists of any real number x where x ≠ −2 and x ≠ − . 3
68.
PROBLEM: 2 2 1 − 3x ) x − 6) ( ( f ( x) = and g ( x) = 9x2 −1 x−6 SOLUTION:
(1 − 3x ) f ( x) =
2
( x − 6) g ( x) =
2
and 9x2 −1 x−6 2 2 2 2 1 − 3x ) ( x − 6 ) 3x − 1) ( x − 6 ) ( ( 3x − 1)( x − 6 ) ( ⋅ 2 = = ( f ⋅ g ) ( x) = x−6 9 x − 1 ( x − 6 )( 3 x − 1)( 3 x + 1) ( 3x + 1) In this case, the domain of f ( x) consists of all real numbers except 6, and the 1 domain of g ( x) consists all real numbers except ± . 3 3 x − 1 x − 6 ( )( ) ANSWER: ( f ⋅ g ) ( x) = ( 3x + 1) 1 The domain consists of any real number x where x ≠ 6 and x ≠ ± . 3
69.
PROBLEM: 25 x 2 − 1 x2 − 9 f ( x) = 2 and g ( x) = x + 6x + 9 5x + 1 SOLUTION: 25 x 2 − 1 x2 − 9 f ( x) = 2 and g ( x ) = x + 6x + 9 5x + 1 2 2 25 x − 1 x − 9 ( 5 x + 1)( 5 x − 1)( x − 3)( x + 3) ( 5 x − 1)( x − 3) ⋅ = = ( f ⋅ g ) ( x) = 2 2 x + 6 x + 9 5x + 1 ( x + 3) ( x + 3) ( 5 x + 1)
In this case, the domain of f ( x) consists of all real numbers except –3, and the 1 domain of g ( x) consists all real numbers except − . 5 ANSWER: ( f ⋅ g ) ( x) =
( 5 x − 1)( x − 3) ( x + 3)
1 The domain consists of any real number x where x ≠ −3 and x ≠ − . 5
70.
PROBLEM: x 2 − 49 4 x2 − 4 x + 1 f ( x) = 2 and g ( x) = 2 x + 13x − 7 7−x SOLUTION: x 2 − 49 4 x2 − 4 x + 1 f ( x) = 2 and g ( x) = 7−x 2 x + 13x − 7 2 x 2 − 49 4 x 2 − 4 x + 1 ( x + 7 )( x − 7 )( 2 x − 1) ⋅ = = 1− 2x ( f ⋅ g ) ( x) = 2 2 x + 13x − 7 7−x − ( 2 x − 1)( x + 7 )( x − 7 )
In this case, the domain of f ( x) consists of all real numbers except –7 and and the domain of g ( x) consists all real numbers except 7 . ANSWER: ( f ⋅ g ) ( x) = 1 − 2x
The domain consists of any real number x where x ≠ ±7 and x ≠
1 . 2
1 , 2
Calculate ( f / g ) ( x) and state the restrictions.
71.
PROBLEM: 1 x−2 f ( x) = and g ( x) = x x −1 SOLUTION: 1 x−2 f ( x) = and g ( x ) = x x −1 1 1 x −1 x −1 = ( f / g ) ( x) = x −x 2 = ⋅ x x − 2 x ( x − 2) x −1 In this case, the domain of f ( x) consists of all real numbers except 0, and the domain of g ( x) consists all real numbers except 1. In addition, the reciprocal of g ( x) has a restriction of 2. x −1 x ( x − 2) The domain consists of any real number x where x ≠ 0 , x ≠ 1 , and x ≠ 2 .
ANSWER: ( f / g ) ( x) =
72.
PROBLEM: 2 5 x + 3) ( 5x + 3 f ( x) = and g ( x) = 2 x 6− x SOLUTION: 2 5 x + 3) ( 5x + 3 f ( x) = and g ( x ) = 2 x 6− x
( 5 x + 3)
2
2 ( 5 x + 3) 6 − x ( 5 x + 3)( 6 − x ) ⋅ = ( f / g ) ( x) = 5 xx+ 3 = 5x + 3 x2 x2 6− x In this case, the domain of f ( x) consists of all real numbers except 0, and the domain of g ( x) consists all real numbers except 6. In addition, the reciprocal of 3 g ( x) has a restriction of − . 5 ( 5 x + 3)( 6 − x ) ANSWER: ( f / g ) ( x) = x2 3 The domain consists of any real number x where x ≠ 0 , x ≠ 6 , and x ≠ − . 5 2
73.
PROBLEM: x 2 − 25 5− x g ( x ) and = f ( x) = 2 x −8 ( x − 8) SOLUTION: 5− x x 2 − 25 f ( x) = and = g ( x ) 2 x −8 ( x − 8)
5− x
(f
/ g)
( x − 8) ( x) =
2
x − 25 x −8 2
=
5− x
( x − 8)
2
⋅
− ( x − 5 )( x − 8 ) 1 x −8 = =− 2 2 x − 25 ( x − 8 ) ( x − 5 )( x + 5 ) ( x + 5)( x − 8)
In this case, the domain of f ( x) consists of all real numbers except 8, and the domain of g ( x) consists all real numbers except 8. In addition, the reciprocal of g ( x) has a restriction of ±5 . ANSWER: ( f / g ) ( x) = −
1
( x + 5 )( x − 8)
The domain consists of any real number x where x ≠ 8 and x ≠ ±5 .
74.
PROBLEM: x 2 − 2 x − 15 2 x2 − 5x − 3 f ( x) = 2 and g ( x) = 2 x − 3 x − 10 x − 7 x + 12 SOLUTION: x 2 − 2 x − 15 2 x2 − 5x − 3 f ( x) = 2 and g ( x ) = 2 x − 3 x − 10 x − 7 x + 12 2 x − 2 x − 15 2 x 2 − 2 x − 15 x 2 − 7 x + 12 ⋅ ( f / g ) ( x) = x 2− 3x − 10 = 2 2 x − 5 x − 3 x − 3 x − 10 2 x 2 − 5 x − 3 x 2 − 7 x + 12 ( x + 3)( x − 5)( x − 3)( x + 4 ) = ( x − 5 )( x + 2 )( 2 x + 1)( x − 3) =
( x + 3)( x + 4 ) ( x + 2 )( 2 x + 1)
In this case, the domain of f ( x) consists of all real numbers except 5 and –2 , and 1 the domain of g ( x) consists all real numbers except 3 and − . In addition, the 2 reciprocal of g ( x) has a restriction of 3 and 4 . ANSWER: ( f / g ) ( x) =
( x + 3)( x + 4 ) ( x + 2 )( 2 x + 1)
The domain consists of any real number x where x ≠ 5 , x ≠ −2 , x ≠ 3 , and x ≠ − .
1 2
75.
PROBLEM: 3x 2 + 11x − 4 x2 − 2 x + 1 f ( x) = and g ( x ) = 9x2 − 6x + 1 3x 2 − 4 x + 1 SOLUTION: 3x 2 + 11x − 4 x2 − 2 x + 1 f ( x) = and g ( x) = 2 9 x2 − 6 x + 1 3x − 4 x + 1 2 3x + 11x − 4 2 3x 2 + 11x − 4 3 x 2 − 4 x + 1 ⋅ ( f / g ) ( x) = 9 x2 − 6 x + 1 = 2 x − 2x +1 9x − 6x + 1 x2 − 2x + 1 3x 2 − 4 x + 1 ( 3x − 1)( x + 4 )( 3x − 1)( x − 1) = 2 2 ( 3x − 1) ( x − 1)
=
x+4 x −1
1 , and the 3 domain of g ( x) consists all real numbers except 1. In addition, the reciprocal of 1 g ( x) has a restriction of and 1 . 3
In this case, the domain of f ( x) consists of all real numbers except
ANSWER: ( f / g ) ( x) =
x+4 x −1
1 The domain consists of any real number x where x ≠ 1 and x ≠ . 3
76.
PROBLEM: 36 − x 2 x 2 − 12 x + 36 f ( x) = 2 and g ( x) = 2 x + 12 x + 36 x + 4 x − 12 SOLUTION: 36 − x 2 x 2 − 12 x + 36 f ( x) = 2 and g ( x ) = 2 x + 12 x + 36 x + 4 x − 12 2 36 − x 2 36 − x 2 x 2 + 4 x − 12 ⋅ 2 ( f / g ) ( x) = x 2 + 12 x + 36 = 2 x − 12 x + 36 x + 12 x + 36 x − 12 x + 36 x 2 + 4 x − 12 − ( x − 6 )( x + 6 )( x + 6 )( x − 2 ) = 2 2 ( x + 6) ( x − 6)
x−2 x−6 In this case, the domain of f ( x) consists of all real numbers except 6, and the domain of g ( x) consists all real numbers except –6. In addition, the reciprocal of g ( x) has a restriction of –6 and 2 . =−
x−2 x−6 The domain consists of any real number x where x ≠ ±6 and x ≠ 2 . ANSWER: ( f / g ) ( x) = −
Part D: Discussion Board Topics
77.
PROBLEM: History of fractions, who is credited for the first use of the fraction bar? ANSWER: The earliest known use of fractions is ca. 2800 BC as Ancient Indus Valley units of measurement. The Egyptians used Egyptian fractions ca. 1000 BC. The Greeks used unit fractions and later continued fractions and followers of the Greek philosopher Pythagoras, ca. 530 BC, discovered that the square root of two cannot be expressed as a fraction. In 150 BC Jain mathematicians in India wrote the “Sthananga Sutra”, which contains work on the theory of numbers, arithmetical operations, operations with fractions. The fraction bar is called the vinculum, which was developed by in the 12th century by the Moroccan mathematician Abu Bakr al-Hassar. (Source: http://en.wikipedia.org/wiki/Fraction_%28mathematics%29)
78.
PROBLEM: How did the ancient Egyptians use fractions? ANSWER: An Egyptian fraction is the sum of distinct unit fractions. 1 1 1 For e.g.: + + 2 3 16 Each fraction in the expression has a numerator equal to 1 and a denominator that is a positive integer, and all the denominators differ from each other. The sum of an expression of this type is a positive rational number a/b; for instance the Egyptian fraction above sums to 43/48. Every positive rational number can be represented by an Egyptian fraction. Sums of this type, and similar sums also including 2/3 and 3/4 as summands, were used as a serious notation for rational numbers by the ancient Egyptians, and continued to be used by other civilizations into medieval times. In modern mathematical notation, Egyptian fractions have been superseded by vulgar fractions and decimal notation. (Source: http://en.wikipedia.org/wiki/Egyptian_fraction)
79.
PROBLEM:
Explain why x = 7 is a restriction to
1 x−7 ÷ . x x−2
ANSWER: 1 x−7 1 x−2 x−2 ÷ = ⋅ = x x − 2 x x − 7 x ( x − 7)
We can find the restrictions but equating the denominator to zero. x ( x − 7) = 0
x = 0 or x = 7
7.3 Adding and Subtracting Rational Expressions
Part A: Adding and Subtracting with Common Denominators Simplify. (Assume all denominators are nonzero.)
1.
PROBLEM: 3 7 + x x SOLUTION: 3 7 3 + 7 10 + = = x x x x ANSWER:
2.
10 x
PROBLEM: 9 10 − x x SOLUTION: 9 10 9 − 10 1 − = =− x x x x ANSWER: −
3.
1 x
PROBLEM: x 3 − y y SOLUTION: x 3 x −3 − = y y y ANSWER:
x−3 y
4.
PROBLEM: 4 6 + x −3 x −3 SOLUTION: 4 6 4+6 10 + = = x −3 x −3 x −3 x −3 ANSWER:
5.
10 x−3
PROBLEM: 7 x − 2x −1 2x −1 SOLUTION: 7 x 7−x − = 2x −1 2x −1 2x −1 ANSWER:
6.
7−x 2x −1
PROBLEM: 8 3x − 3x − 8 3x − 8 SOLUTION: 8 3x 8 − 3x − ( 3x − 8 ) − = = = −1 3x − 8 3x − 8 3x − 8 3x − 8 ANSWER: −1
7.
PROBLEM: 2 x − 11 + x −9 x −9 SOLUTION: 2 x − 11 2 + x − 11 x − 9 + = = =1 x −9 x −9 x −9 x −9 ANSWER: 1
8.
PROBLEM: y+2 y +3 − 2y + 3 2y + 3 SOLUTION: y+2 y +3 y + 2− y −3 1 − = =− 2y + 3 2y + 3 2y + 3 2y + 3 ANSWER: −
9.
1 2y + 3
PROBLEM: 2x − 3 x − 4 − 4x −1 4x −1 SOLUTION: 2x − 3 x − 4 2x − 3 − x + 4 x +1 − = = 4x −1 4x −1 4x −1 4x −1 ANSWER:
10.
x +1 4x −1
PROBLEM: 2 x 3x + 4 x − 2 − + x −1 x −1 x −1 SOLUTION: 2 x 3x + 4 x − 2 2 x − 3x − 4 + x − 2 6 − + = =− x −1 x −1 x −1 x −1 x −1 ANSWER: −
6 x −1
11.
PROBLEM: 1 2 y − 9 13 − 5 y − − 3y 3y 3y SOLUTION: 1 2 y − 9 13 − 5 y 1 − 2 y + 9 − 13 + 5 y 3 y − 3 − − = = 3y 3y 3y 3y 3y
=
ANSWER:
12.
3 ( y − 1) 3y
=
y −1 y
y −1 y
PROBLEM: −3 y + 2 y+7 3y + 4 + − 5 y − 10 5 y − 10 5 y − 10 SOLUTION: −3 y + 2 y+7 3 y + 4 −3 y + 2 + y + 7 − 3 y − 4 −5 y + 5 + − = = 5 y − 10 5 y − 10 5 y − 10 5 y − 10 5 y − 10
=
ANSWER:
13.
5 (1 − y )
5 ( y − 2)
=
1− y y−2
1− y y−2
PROBLEM: x 3 − ( x + 1)( x − 3) ( x + 1)( x − 3) SOLUTION: x 3 x −3 1 − = = ( x + 1)( x − 3) ( x + 1)( x − 3) ( x + 1)( x − 3) x + 1 ANSWER:
1 x +1
14.
PROBLEM: 3x + 5 x+6 − ( 2 x − 1)( x − 6 ) ( 2 x − 1)( x − 6 ) SOLUTION: 3x + 5 x+6 3x + 5 − x − 6 2x −1 − = = ( 2 x − 1)( x − 6 ) ( 2 x − 1)( x − 6 ) ( 2 x − 1)( x − 6 ) ( 2 x − 1)( x − 6 ) =
ANSWER:
15.
1 x−6
1 x−6
PROBLEM: x 6 + 2 2 x − 36 x − 36 SOLUTION: ( x + 6) = 1 x x+6 6 + = = x 2 − 36 x 2 − 36 x 2 − 36 ( x + 6 )( x − 6 ) x − 6 ANSWER:
16.
1 x−6
PROBLEM: x 9 − 2 2 x − 81 x − 81 SOLUTION: x 9 x −9 x−9 1 − 2 = 2 = = 2 x − 81 x − 81 x − 81 ( x − 9 )( x + 9 ) x + 9 ANSWER:
1 x+9
17.
PROBLEM: x2 + 2 x − 22 + 2 2 x + 3 x − 28 x + 3 x − 28 SOLUTION: x2 + 2 x − 22 x 2 + x − 20 ( x + 5 )( x − 4 ) + = = x 2 + 3 x − 28 x 2 + 3x − 28 x 2 + 3 x − 28 ( x − 4 )( x + 7 )
=
ANSWER:
18.
x+5 x+7
x+5 x+7
PROBLEM: x 3− x − 2 2 2x − x − 3 2x − x − 3 SOLUTION: x 3− x x −3+ x 2x − 3 1 − 2 = 2 = = 2 2 x − x − 3 2 x − x − 3 2 x − x − 3 ( 2 x − 3)( x + 1) x + 1
ANSWER:
1 x +1
Part B: Adding and Subtracting with Unlike Denominators Simplify. (Assume all denominators are nonzero.)
19.
PROBLEM: 1 1 + 2 3x SOLUTION: LCD = 6 x To obtain equivalent terms with this common denominator, we will multiply the 3x 2 first term by and the second term by . 3x 2 1 1 1 3x 1 2 3x 2 3x + 2 + = ⋅ + ⋅ = + = 2 3x 2 3x 3x 2 6 x 6 x 6x ANSWER:
3x + 2 6x
20.
PROBLEM: 1 1 − 5x 2 x SOLUTION: LCD = 5 x 2 To obtain equivalent terms with this common denominator, we will multiply the 1 5x2 first term by and the second term by 2 . 1 5x 2 1 1 1 1 1 5x 1 5x 1 − 5x − = 2⋅ − ⋅ 2 = 2− 2 = 2 x 5x 1 x 5x 5x 5x 5x 5x2 ANSWER:
21.
1 − 5x 5x2
PROBLEM: 1 3 + 2 12 y 10 y 3 SOLUTION: LCD = 60 y 3 To obtain equivalent terms with this common denominator, we will multiply the 5y 6 first term by and the second term by . 5y 6 1 3 1 5y 3 6 5y 18 5 y + 18 + = ⋅ + ⋅ = + = 2 3 2 3 3 3 12 y 10 y 12 y 5 y 10 y 6 60 y 60 y 60 y 3 ANSWER:
22.
5 y + 18 60 y 3
PROBLEM: 1 1 − x 2y SOLUTION: LCD = 2 xy To obtain equivalent terms with this common denominator, we will multiply the 2y x first term by and the second term by . 2y x 1 1 1 2y 1 x 2y x 2y − x − = ⋅ − ⋅ = − = x 2 y x 2 y 2 y x 2 xy 2 xy 2 xy ANSWER:
2y − x 2 xy
23.
PROBLEM: 1 −2 y SOLUTION: LCD = y To obtain equivalent terms with this common denominator, we will multiply the 1 y first term by and the second term by . y 1 1 1 1 y 1 2 y 1− 2 y − 2 = ⋅ − 2⋅ = − = y y 1 y y y y ANSWER:
24.
1− 2 y y
PROBLEM: 3 −4 y+2 SOLUTION: LCD = y + 2 To obtain equivalent terms with this common denominator, we will multiply the 1 y+2 first term by and the second term by . y+2 1 3 3 1 3 4y +8 y+2 −4 = ⋅ − 4⋅ = − y+2 y+2 1 y+2 y+2 y+2 3− 4y −8 4y + 5 = =− y+2 y+2 ANSWER: −
4y + 5 y+2
25.
PROBLEM: 2 +2 x+4 SOLUTION: LCD = x + 4 To obtain equivalent terms with this common denominator, we will multiply the 1 x+4 first term by and the second term by . 1 x+4 2 2 1 x + 4 2 + 2x + 8 = +2= ⋅ + 2⋅ x+4 1 x+4 x+4 x+4 2 x + 10 = x+4 2 ( x + 5) = x+4 ANSWER:
26.
2 ( x + 5) x+4
PROBLEM: 2 1 − y y2 SOLUTION: LCD = y 2 To obtain equivalent terms with this common denominator, we will multiply the 1 y first term by and the second term by . y 1 2 1 2 y 1 1 2y 1 − = ⋅ − ⋅ = − y y2 y y y2 1 y2 y2 2 y −1 = y2 ANSWER:
2 y −1 y2
27.
PROBLEM: 3 1 + x +1 x SOLUTION: LCD = x ( x + 1)
To obtain equivalent terms with this common denominator, we will multiply the x x +1 first term by and the second term by . x x +1 3 1 3 x 1 x +1 3x x +1 + = ⋅ + ⋅ = + x + 1 x x + 1 x x x + 1 x ( x + 1) x ( x + 1)
3x + x + 1 4x +1 = x ( x + 1) x ( x + 1)
=
ANSWER:
28.
4x +1 x ( x + 1)
PROBLEM: 1 2 − x −1 x SOLUTION: LCD = x ( x − 1)
To obtain equivalent terms with this common denominator, we will multiply the x x −1 first term by and the second term by . x x −1 1 2 1 x 2 x −1 x 2x − 2 = − − = ⋅ − ⋅ x − 1 x x − 1 x x x − 1 x ( x − 1) x ( x − 1)
=
x − 2x + 2 2− x = x ( x − 1) x ( x − 1)
ANSWER:
2− x x ( x − 1)
29.
PROBLEM: 1 1 + x −3 x +5 SOLUTION: LCD = ( x − 3)( x + 5)
To obtain equivalent terms with this common denominator, we will multiply the x+5 x−3 first term by . and the second term by x+5 x−3 1 1 1 x+5 1 x −3 x+5 x −3 + = + + = ⋅ ⋅ x − 3 x + 5 x − 3 x + 5 x + 5 x − 3 ( x − 3)( x + 5) ( x − 3)( x + 5 )
= =
ANSWER:
30.
x +5+ x −3 2x + 2 = ( x − 3)( x + 5) ( x − 3)( x + 5) 2 ( x + 1)
( x − 3)( x + 5) 2 ( x + 1)
( x − 3)( x + 5)
PROBLEM: 1 1 − x + 2 x −3 SOLUTION: LCD = ( x + 2 )( x − 3)
To obtain equivalent terms with this common denominator, we will multiply the x−3 x+2 first term by . and the second term by x−3 x+2 1 1 1 x −3 1 x+2 x −3 x+2 = − − = ⋅ − ⋅ x + 2 x − 3 x + 2 x − 3 x − 3 x + 2 ( x + 2 )( x − 3) ( x + 2 )( x − 3)
=
ANSWER: −
5 x −3− x − 2 =− ( x + 2 )( x − 3) ( x + 2 )( x − 3) 5
( x + 2 )( x − 3)
31.
PROBLEM: x 2 − x +1 x − 2 SOLUTION: LCD = ( x + 1)( x − 2 )
To obtain equivalent terms with this common denominator, we will multiply the x−2 x +1 first term by . and the second term by x +1 x−2 x 2 x x−2 2 x +1 x2 − 2 x 2x + 2 − = ⋅ − ⋅ = − x + 1 x − 2 x + 1 x − 2 x − 2 x + 1 ( x + 1)( x − 2 ) ( x + 1)( x − 2 )
=
ANSWER:
32.
x2 − 2 x − 2 x − 2 x2 − 4x − 2 = ( x + 1)( x − 2 ) ( x + 1)( x − 2 )
x2 − 4x − 2 ( x + 1)( x − 2 )
PROBLEM: 2x − 3 x − x +5 x −3 SOLUTION: LCD = ( x + 5 )( x − 3)
To obtain equivalent terms with this common denominator, we will multiply the x−3 x+5 first term by . and the second term by x−3 x+5 2x − 3 x 2x − 3 x − 3 x x + 5 2x2 − 9 x + 9 x2 + 5x = − − = ⋅ − ⋅ x + 5 x − 3 x + 5 x − 3 x − 3 x + 5 ( x + 5)( x − 3) ( x + 5)( x − 3)
2 x2 − 9 x + 9 − x2 − 5x x 2 − 14 x + 9 = = ( x + 5)( x − 3) ( x + 5)( x − 3) ANSWER:
x 2 − 14 x + 9 ( x + 5)( x − 3)
33.
PROBLEM: y +1 y −1 + y −1 y +1 SOLUTION: LCD = ( y + 1)( y − 1)
To obtain equivalent terms with this common denominator, we will multiply the y +1 y −1 first term by . and the second term by y +1 y −1 y + 1 y −1 y +1 y + 1 y −1 y −1 y2 + 2 y +1 y2 − 2 y +1 + = + + = ⋅ ⋅ y − 1 y + 1 y − 1 y + 1 y + 1 y − 1 ( y − 1)( y + 1) ( y − 1)( y + 1) = =
ANSWER:
34.
y2 + 2 y +1+ y2 − 2 y +1 2 y2 + 2 = ( y − 1)( y + 1) ( y − 1)( y + 1)
(
)
2 y2 +1
( y − 1)( y + 1)
(
)
2 y2 +1
( y − 1)( y + 1)
PROBLEM: 3y −1 y + 4 − 3y y−2 SOLUTION: LCD = ( 3 y )( y − 2 )
To obtain equivalent terms with this common denominator, we will multiply the y−2 3y first term by and the second term by . y−2 3y 3 y − 1 y + 4 3 y − 1 y − 2 y + 4 3 y 3 y 2 − 7 y + 2 3 y 2 + 12 y − = ⋅ − ⋅ = − 3y y−2 3y y − 2 y − 2 3y 3 y ( y − 2) 3 y ( y − 2) =
ANSWER:
3 y 2 − 7 y + 2 − 3 y 2 − 12 y 2 − 19 y = 3 y ( y − 2) 3 y ( y − 2)
2 − 19 y 3 y ( y − 2)
35.
PROBLEM: 2x − 5 2x + 5 − 2x + 5 2x − 5 SOLUTION: LCD = ( 2 x + 5 )( 2 x − 5 )
To obtain equivalent terms with this common denominator, we will multiply the 2x − 5 2x + 5 first term by . and the second term by 2x − 5 2x + 5 2x − 5 2x + 5 2x − 5 2x − 5 2x + 5 2x + 5 − = ⋅ − ⋅ 2x + 5 2x − 5 2x + 5 2x − 5 2x − 5 2x + 5 4 x 2 − 20 x + 25 4 x 2 + 20 x + 25 = − ( 2 x + 5)( 2 x − 5) ( 2 x + 5)( 2 x − 5)
4 x 2 − 20 x + 25 − 4 x 2 − 20 x − 25 = ( 2 x + 5)( 2 x − 5) =−
ANSWER: −
40 x ( 2 x + 5) ( 2 x − 5)
40 x ( 2 x + 5)( 2 x − 5)
36.
PROBLEM: 2 2x +1 − 2x −1 1− 2x SOLUTION: LCD = ( 2 x − 1)(1 − 2 x )
To obtain equivalent terms with this common denominator, we will multiply the 1− 2x 2x −1 first term by . and the second term by 1− 2x 2x −1 2 2x +1 2 1 − 2x 2x +1 2x −1 − = ⋅ − ⋅ 2x −1 1 − 2x 2x −1 1 − 2x 1 − 2x 2x −1 2 − 4x 4 x2 − 1 = − 2 2 − ( 2 x − 1) − ( 2 x − 1)
= = =
2 − 4x − 4x2 + 1 − ( 2 x − 1)
2
(
− 4 x2 + 4 x − 3 − ( 2 x − 1)
)
2
4 x2 + 4 x − 3
( 2 x − 1) ( 2 x − 1)( 2 x + 3) = 2 ( 2 x − 1) =
ANSWER:
2
2x + 3 2x −1
2x + 3 2x −1
37.
PROBLEM: 3x + 4 2 − x −8 8− x SOLUTION: LCD = ( x − 8 )( 8 − x )
To obtain equivalent terms with this common denominator, we will multiply the 8− x x −8 first term by . and the second term by 8− x x −8 3x + 4 2 3x + 4 8 − x 2 x −8 − = ⋅ − ⋅ x −8 8− x x −8 8− x 8− x x −8 −3x 2 + 20 x + 32 2 x − 16 = − 2 2 − ( x − 8) − ( x − 8)
= = = =
ANSWER:
−3x 2 + 20 x + 32 − 2 x + 16 − ( x − 8)
2
−3x 2 + 18 x + 48 − ( x − 8)
2
−3 ( x + 2 )( x − 8) − ( x − 8) 3( x + 2)
( x − 8)
3( x + 2)
( x − 8)
2
38.
PROBLEM: 1 1 + y −1 1 − y SOLUTION: LCD = ( y − 1)(1 − y )
To obtain equivalent terms with this common denominator, we will multiply the 1− y y −1 first term by . and the second term by y −1 1− y 1 1 1 1− y 1 y −1 1− y y −1 + = ⋅ + ⋅ = + 2 y − 1 1 − y y − 1 1 − y 1 − y y − 1 − ( y − 1) − ( y − 1)2 =
1 − y + y −1 − ( y − 1)
=0 ANSWER: 0
2
=
0 − ( y − 1)
2
39.
PROBLEM: x + 15 2 x2 + 2 x − 9 9 − x2
(
)(
SOLUTION: LCD = x 2 − 9 9 − x 2
)
To obtain equivalent terms with this common denominator, we will multiply the 9 − x2 x2 − 9 first term by . and the second term by 9 − x2 x2 − 9 2 x2 x + 15 2 x 2 9 − x 2 x + 15 x 2 − 9 + = ⋅ + ⋅ x2 − 9 9 − x2 x2 − 9 9 − x2 9 − x2 x2 − 9 9 − x 2 2 x 2 − x − 15 = 2 − x2 − 9
(
)( ) ( ) ( 9 − x ) ( 2 x + 5)( x − 3) = − ( x − 9) 2
2
=
ANSWER:
( 2 x + 5)( x − 3)
(x
2
−9
)
=
( 2 x + 5)( x − 3) ( x − 3)( x + 3)
=
2x + 5 x+3
2x + 5 x+3
2
40.
PROBLEM: x 1 15 − x + − x + 3 x − 3 ( x + 3)( x − 3) SOLUTION: LCD = ( x + 3)( x − 3)
To obtain equivalent terms with this common denominator, we will multiply the x−3 x+3 1 and the third term by . , the second term by first term by x−3 x+3 1 x 1 15 − x x x−3 1 x+3 15 − x 1 + − = ⋅ + ⋅ − ⋅ x + 3 x − 3 ( x + 3)( x − 3) x + 3 x − 3 x − 3 x + 3 ( x + 3)( x − 3) 1
x 2 − 3x x+3 15 − x = + − ( x + 3)( x − 3) ( x + 3)( x − 3) ( x + 3)( x − 3) =
x 2 − 3 x + x + 3 − 15 + x ( x + 3) ( x − 3)
x 2 − x − 12 = ( x + 3)( x − 3)
ANSWER:
x−4 x −3
=
( x − 4 )( x + 3) ( x + 3)( x − 3)
=
x−4 x−3
41.
PROBLEM: 2x 1 2( x − 1) − + 3 x − 1 3 x + 1 ( 3 x − 1)( 3 x + 1) SOLUTION: LCD = ( 3x − 1)( 3x + 1)
To obtain equivalent terms with this common denominator, we will multiply the 3x + 1 3x − 1 1 , and the third term by . , the second term by first term by 3x + 1 3x − 1 1 2x 1 2( x − 1) 2 x 3x + 1 1 3x − 1 2( x − 1) 1 − + = ⋅ − ⋅ + ⋅ 3x − 1 3 x + 1 ( 3x − 1)( 3x + 1) 3 x − 1 3x + 1 3x + 1 3x − 1 ( 3 x − 1)( 3 x + 1) 1 6 x2 + 2 x 3x − 1 2x − 2 = − + ( 3x − 1)( 3x + 1) ( 3x − 1)( 3x + 1) ( 3x − 1)( 3x + 1) =
6 x 2 + 2 x − 3x + 1 + 2 x − 2 ( 3x − 1)( 3x + 1)
6 x2 + x − 1 = ( 3x − 1)( 3x + 1)
ANSWER:
2x +1 3x + 1
=
( 3x − 1)( 2 x + 1) ( 3x − 1)( 3x + 1)
=
2x +1 3x + 1
42.
PROBLEM: 4x x 16 x − 3 − + 2 x + 1 x − 5 ( 2 x + 1)( x − 5 ) SOLUTION: LCD = ( 2 x + 1)( x − 5 )
To obtain equivalent terms with this common denominator, we will multiply the x−5 2x +1 1 and the third term by . , the second term by first term by x−5 2x +1 1 4x x 16 x − 3 4x x − 5 x 2x +1 16 x − 3 1 − + = ⋅ − ⋅ + ⋅ 2 x + 1 x − 5 ( 2 x + 1)( x − 5 ) 2 x + 1 x − 5 x − 5 2 x + 1 ( 2 x + 1)( x − 5 ) 1
ANSWER:
x−3 x−5
=
4 x 2 − 20 x 2x2 + x 16 x − 3 − + ( 2 x + 1)( x − 5) ( 2 x + 1)( x − 5) ( 2 x + 1)( x − 5)
=
4 x 2 − 20 x − 2 x 2 − x + 16 x − 3 ( 2 x + 1)( x − 5 )
=
2x2 − 5x − 3 ( 2 x + 1)( x − 5 )
=
( 2 x + 1)( x − 3) ( 2 x + 1)( x − 5 )
=
x−3 x −5
43.
PROBLEM: x 2 4 + + 3x x − 2 3x ( x − 2 ) SOLUTION: LCD = 3x ( x − 2 )
To obtain equivalent terms with this common denominator, we will multiply the x−2 3x 1 , the second term by , and the third term by . first term by x−2 3x 1 x 2 4 x x−2 2 3x 4 1 + + = ⋅ + ⋅ + ⋅ 3x x − 2 3x ( x − 2 ) 3x x − 2 x − 2 3x 3x ( x − 2 ) 1 =
x2 − 2x 6x 4 + + 3x ( x − 2 ) 3x ( x − 2 ) 3x ( x − 2 )
=
x2 − 2 x + 6 x + 4 3x ( x − 2 )
=
x2 + 4 x + 4 3x ( x − 2 )
x2 + 4x + 4 ANSWER: 3x ( x − 2 )
44.
PROBLEM: −2 x 3x 18( x − 2) − − x + 6 6 − x ( x + 6 )( x − 6 ) SOLUTION: LCD = ( x + 6 )( x − 6 )
To obtain equivalent terms with this common denominator, we will multiply the x−6 x+6 1 and the third term by . , the second term by first term by x−6 x+6 1 −2 x 3x 18( x − 2) −2 x x − 6 3x x + 6 18( x − 2) 1 − − = ⋅ − ⋅ − ⋅ x + 6 6 − x ( x + 6 )( x − 6 ) x + 6 x − 6 6 − x x + 6 ( x + 6 )( x − 6 ) 1 =
−2 x 2 + 12 x 3 x 2 + 18 x 18 x − 36 + − ( x + 6 )( x − 6 ) ( x + 6 )( x − 6 ) ( x + 6 )( x − 6 )
=
−2 x 2 + 12 x + 3x 2 + 18 x − 18 x + 36 ( x + 6 )( x − 6 )
=
x 2 + 12 x + 36 ( x + 6 )( x − 6 )
( x + 6) = ( x + 6 )( x − 6 ) 2
= ANSWER:
x+6 x−6
x+6 x−6
45.
PROBLEM: x 1 25 − 7 x − − x + 5 x − 7 ( x + 5 )( x − 7 ) SOLUTION: LCD = ( x + 5 )( x − 7 )
To obtain equivalent terms with this common denominator, we will multiply the x−7 x+5 1 , and the third term by . , the second term by first term by x−7 x+5 1 x 1 25 − 7 x x x−7 1 x+5 25 − 7 x 1 − − = ⋅ − ⋅ − ⋅ x + 5 x − 7 ( x + 5)( x − 7 ) x + 5 x − 7 x − 7 x + 5 ( x + 5)( x − 7 ) 1 x2 − 7 x 25 − 7 x x+5 = − − ( x + 5)( x − 7 ) ( x + 5)( x − 7 ) ( x + 5)( x − 7 ) =
x 2 − 7 x − x − 5 − 25 + 7 x ( x + 5)( x − 7 )
x 2 − x − 30 = ( x + 5)( x − 7 )
ANSWER:
x−6 x−7
=
( x + 5)( x − 6 ) ( x + 5)( x − 7 )
=
x−6 x−7
46.
PROBLEM: x 2 + 2 x − 2x − 3 x − 3 SOLUTION: x 2 x 2 + = + 2 x − 2 x − 3 x − 3 ( x − 3)( x + 1) x − 3
LCD = ( x − 3)( x + 1) To obtain equivalent terms with this common denominator, we will multiply the 1 x +1 . first term by and the second term by 1 x +1 2 1 2 x +1 x x + = ⋅ + ⋅ ( x − 3)( x + 1) x − 3 ( x − 3)( x + 1) 1 x − 3 x + 1
=
ANSWER:
x
( x − 3)( x + 1)
=
x + 2x + 2 ( x − 3)( x + 1)
=
3x + 2 ( x − 3)( x + 1)
3x + 2 ( x − 3)( x + 1)
+
2x + 2 ( x − 3)( x + 1)
47.
PROBLEM: 1 x2 − 2 x + 5 x − 25 SOLUTION: 1 x2 1 x2 − 2 = − x + 5 x − 25 x + 5 ( x + 5)( x − 5)
LCD = ( x + 5)( x − 5) To obtain equivalent terms with this common denominator, we will multiply the x−5 1 and the second term by . first term by x−5 1 2 1 1 x−5 1 x x2 − = ⋅ − ⋅ x + 5 ( x + 5 )( x − 5 ) x + 5 x − 5 ( x + 5 )( x − 5 ) 1 =
x −5 x2 − ( x + 5)( x − 5) ( x + 5)( x − 5)
=
− x2 + x − 5 ( x + 5)( x − 5)
− x2 + x − 5 ANSWER: ( x + 5)( x − 5)
48.
PROBLEM: 5x − 2 2 − 2 x −4 x−2 SOLUTION: 5x − 2 2 5x − 2 2 − = − 2 x − 4 x − 2 ( x + 2 )( x − 2 ) x − 2
LCD = ( x − 3)( x + 1) To obtain equivalent terms with this common denominator, we will multiply the 1 x+2 . first term by and the second term by 1 x+2 5x − 2 2 5x − 2 1 2 x+2 − = ⋅ − ⋅ ( x + 2 )( x − 2 ) x − 2 ( x + 2 )( x − 2 ) 1 x − 2 x + 2 =
5x − 2 2x + 4 − ( x + 2 )( x − 2 ) ( x + 2 )( x − 2 )
=
5x − 2 − 2 x − 4 ( x + 2 )( x − 2 )
=
3x − 6 ( x + 2 )( x − 2 )
= = ANSWER:
3 x+2
3 ( x − 2)
( x + 2 )( x − 2 ) 3 x+2
49.
PROBLEM: 1 6x − 3 − 2 x +1 x − 7x − 8 SOLUTION: 1 6x − 3 1 6x − 3 − 2 = − x + 1 x − 7 x − 8 x + 1 ( x − 8 )( x + 1)
LCD = ( x − 8)( x + 1) To obtain equivalent terms with this common denominator, we will multiply the x −8 1 and the second term by . first term by x −8 1 1 6x − 3 1 x −8 6x − 3 1 − = ⋅ − ⋅ x + 1 ( x − 8 )( x + 1) x + 1 x − 8 ( x − 8 )( x + 1) 1 =
x −8 6x − 3 − ( x − 8)( x + 1) ( x − 8)( x + 1)
=
x − 8 − 6x + 3 ( x − 8)( x + 1)
=
−5 x − 5 ( x − 8)( x + 1)
=
( x − 8)( x + 1)
=−
ANSWER: −
5 x −8
−5 ( x + 1)
5 x −8
50.
PROBLEM: 3x 1 − 2 9 x − 16 3x + 4 SOLUTION: 3x 1 3x 1 − = − 2 9 x − 16 3x + 4 ( 3x + 4 )( 3x − 4 ) 3x + 4
LCD = ( 3x + 4 )( 3x − 4 ) To obtain equivalent terms with this common denominator, we will multiply the 1 3x − 4 . first term by and the second term by 1 3x − 4 3x 1 3x 1 1 3x − 4 − = ⋅ − ⋅ ( 3x + 4 )( 3x − 4 ) 3x + 4 ( 3x + 4 )( 3x − 4 ) 1 3x + 4 3x − 4 =
3x 3x − 4 − ( 3x + 4 )( 3x − 4 ) ( 3x + 4 )( 3x − 4 )
=
3x − 3x + 4 ( 3x + 4 )( 3x − 4 )
= ANSWER:
4
4
( 3x + 4 )( 3x − 4 )
( 3x + 4 )( 3x − 4 )
51.
PROBLEM: 2x 1 + 2 2 x −1 x + x SOLUTION: 2x 1 2x 1 + 2 = + 2 x − 1 x + x ( x − 1)( x + 1) x ( x + 1)
LCD = x ( x + 1)( x − 1) To obtain equivalent terms with this common denominator, we will multiply the x x −1 . first term by and the second term by x x −1 2x 1 2x x 1 x −1 + = ⋅ + ⋅ ( x − 1)( x + 1) x ( x + 1) ( x − 1)( x + 1) x x ( x + 1) x − 1 =
x −1 2 x2 + x ( x − 1)( x + 1) x ( x − 1)( x + 1)
2x2 + x −1 = x ( x − 1)( x + 1)
ANSWER:
2x −1 x ( x − 1)
=
( 2 x − 1)( x + 1) x ( x − 1) ( x + 1)
=
2x −1 x ( x − 1)
52.
PROBLEM: x ( 4 x − 1) x − 2 2x + 7 x − 4 4 + x SOLUTION: x ( 4 x − 1) x ( 4 x − 1) x x − = − 2 2 x + 7 x − 4 4 + x ( 2 x − 1)( x + 4 ) x + 4
LCD = ( 2 x − 1)( x + 4 ) To obtain equivalent terms with this common denominator, we will multiply the 1 2x −1 . first term by and the second term by 1 2x −1 x ( 4 x − 1) x ( 4 x − 1) 1 x x 2x −1 − = ⋅ − ⋅ ( 2 x − 1)( x + 4 ) x + 4 ( 2 x − 1)( x + 4 ) 1 x + 4 2 x − 1
ANSWER:
=
4 x2 − x 2 x2 − x − ( 2 x − 1)( x + 4 ) ( 2 x − 1)( x + 4 )
=
4 x2 − x − 2 x2 + x ( 2 x − 1)( x + 4 )
=
2 x2 ( 2 x − 1)( x + 4 )
2 x2 ( 2 x − 1)( x + 4 )
53.
PROBLEM: 3x 2 2x − 2 3x + 5 x − 2 3x − 1 SOLUTION: 3x 2 2x 3x 2 2x − = − 2 3x + 5 x − 2 3x − 1 ( x + 2 )( 3x − 1) 3 x − 1
LCD = ( x + 2 )( 3x − 1) To obtain equivalent terms with this common denominator, we will multiply the 1 x+2 . first term by and the second term by 1 x+2 3x 2 1 2x x + 2 3x 2 2 x2 + 4 x ⋅ − ⋅ = − ( x + 2 )( 3x − 1) 1 3x − 1 x + 2 ( x + 2 )( 3x − 1) ( x + 2 )( 3x − 1) 3x 2 − 2 x 2 − 4 x = ( x + 2 )( 3x − 1) = =
ANSWER:
x ( x − 4)
( x + 2 )( 3x − 1)
x2 − 4 x ( x + 2 )( 3x − 1) x ( x − 4)
( x + 2 )( 3x − 1)
54.
PROBLEM: 2x 11x + 4 − 2 x − 4 x − 2x − 8 SOLUTION: 2x 11x + 4 2x 11x + 4 − 2 = − x − 4 x − 2 x − 8 x − 4 ( x + 2 )( x − 4 )
LCD = ( x + 2 )( x − 4 ) To obtain equivalent terms with this common denominator, we will multiply the x+2 1 and the second term by . first term by x+2 1 2x 11x + 4 2x x + 2 11x + 4 1 − = ⋅ − ⋅ x − 4 ( x + 2 )( x − 4 ) x − 4 x + 2 ( x + 2 )( x − 4 ) 1 =
2x2 + 4x 11x + 4 − ( x + 2 )( x − 4 ) ( x + 2 )( x − 4 )
2 x 2 + 4 x − 11x + 4 = ( x + 2 )( x − 4 )
ANSWER:
2x +1 x+2
=
2 x2 − 7 x + 4 ( x + 2 )( x − 4 )
=
( 2 x + 1)( x − 4 ) ( x + 2 )( x − 4 )
=
2x +1 x+2
55.
PROBLEM: x 6 x − 24 + 2 2x +1 2x − 7x − 4 SOLUTION: x 6 x − 24 x 6 x − 24 + 2 = + 2 x + 1 2 x − 7 x − 4 2 x + 1 ( 2 x + 1)( x − 4 )
LCD = ( 2 x + 1)( x − 4 ) To obtain equivalent terms with this common denominator, we will multiply the x−4 1 and the second term by . first term by x−4 1 6 x − 24 6 x − 24 1 x x x−4 + = ⋅ + ⋅ 2 x + 1 ( 2 x + 1)( x − 4 ) 2 x + 1 x − 4 ( 2 x + 1)( x − 4 ) 1
ANSWER:
x+6 2x +1
=
6 x − 24 x2 − 4 x + ( 2 x + 1)( x − 4 ) ( 2 x + 1)( x − 4 )
=
x 2 − 4 x + 6 x − 24 ( 2 x + 1)( x − 4 )
=
( x − 4 )( x + 6 ) ( 2 x + 1)( x − 4 )
=
x+6 2x +1
56.
PROBLEM: 1 1 + 2 2 x − x − 6 x − 3x − 10 SOLUTION: 1 1 1 1 + 2 = + 2 x − x − 6 x − 3x − 10 ( x − 3)( x + 2 ) ( x − 5)( x + 2 )
LCD = ( x − 5 )( x − 3)( x + 2 ) To obtain equivalent terms with this common denominator, we will multiply the x−5 x−3 . first term by and the second term by x−5 x−3 x−5 x−3 1 1 1 1 + = ⋅ + ⋅ ( x − 3)( x + 2 ) ( x − 5)( x + 2 ) ( x − 3)( x + 2 ) x − 5 ( x − 5)( x + 2 ) x − 3 =
x −5 x −3 + ( x − 5 )( x − 3)( x + 2 ) ( x − 5)( x − 3)( x + 2 )
=
x −5+ x −3 ( x − 5 )( x − 3)( x + 2 )
=
2x − 8 ( x − 5 )( x − 3)( x + 2 )
= ANSWER:
2 ( x − 4)
2 ( x − 4)
( x − 5 )( x − 3)( x + 2 )
( x − 5)( x − 3)( x + 2 )
57.
PROBLEM: x 3 − 2 2 x + 4x + 3 x − 4x − 5 SOLUTION: x 3 x 3 − 2 = − 2 x + 4 x + 3 x − 4 x − 5 ( x + 3)( x + 1) ( x − 5 )( x + 1)
LCD = ( x + 3)( x + 1)( x − 5 ) To obtain equivalent terms with this common denominator, we will multiply the x−5 x+3 . and the second term by first term by x−5 x+3 3 3 x x x −5 x+3 − = ⋅ − ⋅ ( x + 3)( x + 1) ( x − 5)( x + 1) ( x + 3)( x + 1) x − 5 ( x − 5)( x + 1) x + 3
ANSWER:
x −9 ( x + 3)( x − 5)
=
3x + 9 x2 − 5x − ( x + 3)( x + 1)( x − 5) ( x + 3)( x + 1)( x − 5 )
=
x 2 − 5 x − 3x + 9 ( x + 3)( x + 1)( x − 5)
=
x2 − 8x + 9 ( x + 3)( x + 1)( x − 5)
=
( x − 9 )( x + 1) ( x + 3)( x + 1)( x − 5)
=
x−9 ( x + 3)( x − 5)
58.
PROBLEM: y +1 y − 2 2 2 y + 5 y − 3 4 y −1 SOLUTION: y +1 y y +1 y − 2 = − 2 2 y + 5 y − 3 4 y − 1 ( 2 y − 1)( y + 3) ( 2 y − 1)( 2 y + 1)
LCD = ( 2 y − 1)( 2 y + 1)( y + 3) To obtain equivalent terms with this common denominator, we will multiply the 2 y +1 y+3 . first term by and the second term by 2 y +1 y+3 y +1 y y +1 2 y +1 y y +3 − = ⋅ − ⋅ ( 2 y − 1)( y + 3) ( 2 y − 1)( 2 y + 1) ( 2 y − 1)( y + 3) 2 y + 1 ( 2 y − 1)( 2 y + 1) y + 3
ANSWER:
=
2 y2 + 3 y +1 y2 + 3 y − ( 2 y − 1)( 2 y + 1)( y + 3) ( 2 y − 1)( 2 y + 1)( y + 3)
=
2 y2 + 3y +1− y2 − 3y ( 2 y − 1)( 2 y + 1)( y + 3)
=
y2 +1 ( 2 y − 1)( 2 y + 1)( y + 3)
y2 +1 ( 2 y − 1)( 2 y + 1)( y + 3)
59.
PROBLEM: y −1 2 − 2 2 y − 25 y − 10 y + 25 SOLUTION: y −1 y −1 2 2 − 2 = − 2 y − 25 y − 10 y + 25 ( y − 5 )( y + 5) ( y − 5)2
LCD = ( y − 5 ) ( y + 5 ) To obtain equivalent terms with this common denominator, we will multiply the y −5 y+5 . and the second term by first term by y −5 y+5 2 2 y −1 y −1 y −5 y+5 − = ⋅ − ⋅ 2 2 ( y − 5 )( y + 5) ( y − 5) ( y − 5)( y + 5 ) y − 5 ( y − 5) y + 5 2
= = =
ANSWER:
y2 − 6 y + 5
y 2 − 6 y + 5 − 2 y − 10
( y − 5) ( y + 5) 2
y2 − 8 y − 5
( y − 5) ( y + 5)
y2 − 8 y − 5
( y − 5) ( y + 5) 2
−
2 y + 10
( y − 5) ( y + 5) ( y − 5) ( y + 5) 2
2
2
60.
PROBLEM: 3x 2 + 24 12 − 2 x − 2x − 8 x − 4 SOLUTION: 3x 2 + 24 12 3x 2 + 24 12 − = − 2 x − 2 x − 8 x − 4 ( x − 4 )( x + 2 ) x − 4
LCD = ( x − 4 )( x + 2 ) To obtain equivalent terms with this common denominator, we will multiply the 1 x+2 . first term by and the second term by 1 x+2 3x 2 + 24 12 3x 2 + 24 1 12 x + 2 − = ⋅ − ⋅ ( x − 4 )( x + 2 ) x − 4 ( x − 4 )( x + 2 ) 1 x − 4 x + 2 3 x 2 + 24 12 x + 24 = − ( x − 4 )( x + 2 ) ( x − 4 )( x + 2 ) =
3x 2 + 24 − 12 x − 24 ( x − 4 )( x + 2 )
3x 2 − 12 x = ( x − 4 )( x + 2 ) = = ANSWER:
3x x+2
3x ( x − 4 )
( x − 4 )( x + 2 ) 3x x+2
61.
PROBLEM: 4 x 2 + 28 28 − 2 x − 6x − 7 x − 7 SOLUTION: 4 x 2 + 28 28 4 x 2 + 28 28 − = − 2 x − 6 x − 7 x − 7 ( x − 7 )( x + 1) x − 7
LCD = ( x − 7 )( x + 1) To obtain equivalent terms with this common denominator, we will multiply the 1 x +1 . first term by and the second term by x +1 1 4 x 2 + 28 28 4 x 2 + 28 1 28 x + 1 − = ⋅ − ⋅ ( x − 7 )( x + 1) x − 7 ( x − 7 )( x + 1) 1 x − 7 x + 1 4 x 2 + 28 28 x + 28 = − ( x − 7 )( x + 1) ( x − 7 )( x + 1) = = =
ANSWER:
4x x +1
4 x 2 − 28 x ( x − 7 )( x + 1) 4x ( x − 7)
( x − 7 )( x + 1) 4x x +1
62.
PROBLEM: a a 2 − 9a + 18 + 2 4 − a a − 13a + 36 SOLUTION: a a 2 − 9a + 18 a a 2 − 9a + 18 + 2 = + 4 − a a − 13a + 36 4 − a ( a − 4 )( a − 9 )
LCD = ( a − 4 )( a − 9 ) To obtain equivalent terms with this common denominator, we will multiply the a −9 1 first term by and the second term by . a −9 1 2 2 a a − 9a + 18 a a − 9 a − 9a + 18 1 + = ⋅ + ⋅ 4 − a ( a − 4 )( a − 9 ) 4 − a a − 9 ( a − 4 )( a − 9 ) 1
ANSWER:
=
a 2 − 9a + 18 − a 2 + 9a + ( a − 4 )( a − 9 ) ( a − 4 )( a − 9 )
=
− a 2 + 9a + a 2 − 9a + 18 ( a − 4 )( a − 9 )
=
18 ( a − 4 )( a − 9 )
=
18 ( a − 4 )( a − 9 )
18 ( a − 4 )( a − 9 )
63.
PROBLEM: 3a − 12 a+2 − 2 a − 8a + 16 4 − a SOLUTION: 3a − 12 a + 2 3a − 12 a + 2 − = − 2 a − 8a + 16 4 − a ( 4 − a )2 4 − a
LCD = ( 4 − a ) To obtain equivalent terms with this common denominator, we will multiply the 1 4−a . first term by and the second term by 1 4−a 3a − 12 a + 2 3a − 12 1 a + 2 4 − a − = ⋅ − ⋅ 2 2 (4 − a) 4 − a (4 − a) 1 4 − a 4 − a 2
= = =
3a − 12
(4 − a)
2
−
− a 2 + 2a + 8
(4 − a)
3a − 12 + a 2 − 2a − 8
(4 − a)
2
a 2 + a − 20
(4 − a) ( a + 5) ( a − 4 ) = 2 ( a − 4) =
ANSWER:
a+5 a−4
2
2
a+5 a−4
64.
PROBLEM: a 2 − 14 5 − 2 2a − 7 a − 4 1 + 2a SOLUTION: a 2 − 14 5 a 2 − 14 5 − = − 2 2a − 7a − 4 1 + 2a ( 2a + 1)( a − 4 ) 2a + 1
LCD = ( 2a + 1)( a − 4 ) To obtain equivalent terms with this common denominator, we will multiply the 1 a−4 . first term by and the second term by 1 a−4 5 1 5 a 2 − 14 a 2 − 14 a−4 − = ⋅ − ⋅ ( 2a + 1)( a − 4 ) 2a + 1 ( 2a + 1)( a − 4 ) 1 2a + 1 a − 4 =
5a − 20 a 2 − 14 − ( 2a + 1)( a − 4 ) ( 2a + 1)( a − 4 )
=
a 2 − 14 − 5a + 20 ( 2a + 1)( a − 4 )
a 2 − 5a + 6 = ( 2a + 1)( a − 4 ) =
ANSWER:
( a − 2 )( a − 3) ( 2a + 1)( a − 4 )
( a − 2 )( a − 3) ( 2a + 1)( a − 4 )
65.
PROBLEM: 1 x 3 − 2 + 2 x + 3 x − 6x + 9 x − 9 SOLUTION: 1 3 1 3 x x − 2 + 2 = − + 2 x + 3 x − 6 x + 9 x − 9 x + 3 ( x − 3) ( x + 3)( x − 3)
LCD = ( x − 3) ( x + 3) To obtain equivalent terms with this common denominator, we will multiply the 2 x − 3) ( x+3 x−3 , the second term by first term by , and the third term by . 2 x+3 x−3 ( x − 3) 2
1 x 3 1 ( x − 3) x x+3 3 x −3 − + = ⋅ − ⋅ + ⋅ 2 2 2 x + 3 ( x − 3) ( x + 3)( x − 3) x + 3 ( x − 3) ( x − 3) x + 3 ( x + 3)( x − 3) x − 3 2
= =
x2 − 6x + 9
6x
( x − 3) ( x + 3) 2
x 2 + 3x
+
3x − 9
( x − 3) ( x + 3) ( x − 3) ( x + 3) ( x − 3) ( x + 3) 2
2
x 2 − 6 x + 9 − x 2 − 3x + 3x − 9
=−
ANSWER: −
−
( x − 3) ( x + 3) 2
6x
( x − 3) ( x + 3) 2
2
66.
PROBLEM: 3x 2x 23x − 10 − + 2 x + 7 x − 2 x + 5 x − 14 SOLUTION: 3x 2x 23x − 10 3x 2x 23x − 10 − + 2 = − + x + 7 x − 2 x + 5 x − 14 x + 7 x − 2 ( x + 7 )( x − 2 )
LCD = ( x + 7 )( x − 2 ) To obtain equivalent terms with this common denominator, we will multiply the x−2 x+7 1 first term by , the second term by and the third term by . x−2 x+7 1 1 3x 2x 23 x − 10 3x x − 2 2 x x + 7 23 x − 10 − + = ⋅ − ⋅ + ⋅ x + 7 x − 2 ( x + 7 )( x − 2 ) x + 7 x − 2 x − 2 x + 7 ( x + 7 )( x − 2 ) 1
ANSWER:
x+5 x+7
=
3x 2 − 6 x 2 x 2 + 14 x 23 x − 10 − + ( x + 7 )( x − 2 ) ( x + 7 )( x − 2 ) ( x + 7 )( x − 2 )
=
3x 2 − 6 x − 2 x 2 − 14 x + 23 x − 10 ( x + 7 )( x − 2 )
=
x 2 + 3 x − 10 ( x + 7 )( x − 2 )
=
( x + 5)( x − 2 ) ( x + 7 )( x − 2 )
=
x+5 x+7
67.
PROBLEM: x + 3 x − 1 x ( x + 11) + − x − 1 x + 2 x2 + x − 2 SOLUTION: x ( x + 11) x + 3 x − 1 x ( x + 11) x + 3 x − 1 + − 2 = + − x − 1 x + 2 x + x − 2 x − 1 x + 2 ( x − 1)( x + 2 )
LCD = ( x − 1)( x + 2 ) To obtain equivalent terms with this common denominator, we will multiply the x+2 x −1 1 , the second term by first term by , and the third term by . x+2 1 x −1 x ( x + 11) x ( x + 11) 1 x + 3 x −1 x + 3 x + 2 x −1 x −1 + − = ⋅ + ⋅ − ⋅ x − 1 x + 2 ( x − 1)( x + 2 ) x − 1 x + 2 x + 2 x − 1 ( x − 1)( x + 2 ) 1 =
ANSWER:
x−7 x+2
x2 + 5x + 6 x2 − 2x + 1 x 2 + 11x + − ( x − 1)( x + 2 ) ( x − 1)( x + 2 ) ( x − 1)( x + 2 )
=
x 2 + 5 x + 6 + x 2 − 2 x + 1 − x 2 − 11x ( x − 1)( x + 2 )
=
x2 − 8x + 7 ( x − 1)( x + 2 )
=
( x − 7 )( x − 1) ( x − 1)( x + 2 )
=
x−7 x+2
68.
PROBLEM: 4 ( x + 5) −2 x 4 − + 2 3x + 1 x − 2 3x − 5 x − 2 SOLUTION: 4 ( x + 5) 4 ( x + 5) −2 x 4 −2 x 4 − + 2 = − + 3x + 1 x − 2 3x − 5 x − 2 3x + 1 x − 2 ( 3 x + 1)( x − 2 )
LCD = ( 3x + 1)( x − 2 ) To obtain equivalent terms with this common denominator, we will multiply the x−2 3x + 1 1 first term by , the second term by and the third term by . x−2 3x + 1 1 4 ( x + 5) 4 ( x + 5) −2 x −2 x x − 2 1 4 4 3x + 1 − + = ⋅ − ⋅ + ⋅ 3 x + 1 x − 2 ( 3 x + 1)( x − 2 ) 3 x + 1 x − 2 x − 2 3x + 1 ( 3 x + 1)( x − 2 ) 1 =
−2 x 2 + 4 x 12 x + 4 4 x + 20 − + ( 3x + 1)( x − 2 ) ( 3x + 1)( x − 2 ) ( 3x + 1)( x − 2 )
=
−2 x 2 + 4 x − 12 x − 4 + 4 x + 20 ( 3x + 1)( x − 2 )
−2 x 2 − 4 x + 16 = ( 3x + 1)( x − 2 ) =
ANSWER: −
2 ( x + 4)
( 3x + 1)
−2 ( x + 4 )( x − 2 )
( 3x + 1)( x − 2 ) 2 ( x + 4) =− ( 3x + 1)
69.
PROBLEM: 3 ( x + 5) x −1 x + 3 − − 2 4 x − 1 2 x + 3 8 x + 10 x − 3 SOLUTION: 3 ( x + 5) 3 ( x + 5) x −1 x + 3 x −1 x + 3 − − 2 = − − 4 x − 1 2 x + 3 8 x + 10 x − 3 4 x − 1 2 x + 3 ( 4 x − 1)( 2 x + 3)
LCD = ( 4 x − 1)( 2 x + 3) To obtain equivalent terms with this common denominator, we will multiply the 2x + 3 4x −1 1 , the second term by first term by , and the third term by . 2x + 3 4x −1 1 3 ( x + 5) 3 ( x + 5) x −1 x + 3 x −1 2x + 3 x + 3 4x −1 1 − − = ⋅ − ⋅ − ⋅ 4 x − 1 2 x + 3 ( 4 x − 1)( 2 x + 3) 4 x − 1 2 x + 3 2 x + 3 4 x − 1 ( 4 x − 1)( 2 x + 3) 1 =
2x2 + x − 3 4 x 2 + 11x − 3 3 x + 15 − − ( 4 x − 1)( 2 x + 3) ( 4 x − 1)( 2 x + 3) ( 4 x − 1)( 2 x + 3)
=
2 x 2 + x − 3 − 4 x 2 − 11x + 3 − 3 x − 15 ( 4 x − 1)( 2 x + 3)
=
−2 x 2 − 13x − 15 ( 4 x − 1)( 2 x + 3)
=− =−
ANSWER: −
x+5 4x −1
( 2 x + 3)( x + 5 ) ( 4 x − 1) ( 2 x + 3)
x+5 4x −1
70.
PROBLEM: 3x 2 6 x2 − 5x − 9 − − 2x − 3 2x + 3 4x2 − 9 SOLUTION: 3x 2 6 x2 − 5x − 9 3x 2 6 x2 − 5x − 9 − − = − − 2x − 3 2x + 3 4x2 − 9 2 x − 3 2 x + 3 ( 2 x − 3)( 2 x + 3)
LCD = ( 2 x − 3)( 2 x + 3) To obtain equivalent terms with this common denominator, we will multiply the 2x + 3 2x − 3 1 first term by , the second term by and the third term by . 2x + 3 2x − 3 1 2 3x 2 6 x − 5x − 9 3x 2 x + 3 2 2x − 3 6 x2 − 5x − 9 1 − − = ⋅ − ⋅ − ⋅ 2 x − 3 2 x + 3 ( 2 x − 3)( 2 x + 3) 2 x − 3 2 x + 3 2 x + 3 2 x − 3 ( 2 x − 3)( 2 x + 3) 1 6x2 + 9x 4x − 6 6 x2 − 5x − 9 = − − ( 2 x − 3)( 2 x + 3) ( 2 x − 3)( 2 x + 3) ( 2 x − 3)( 2 x + 3) =
6 x2 + 9 x − 4 x + 6 − 6 x2 + 5x + 9 ( 2 x − 3)( 2 x + 3)
=
10 x + 15 ( 2 x − 3)( 2 x + 3)
= =
ANSWER:
5 2x − 3
5 ( 2 x + 3)
( 2 x − 3)( 2 x + 3) 5 2x − 3
71.
PROBLEM: 1 1 2 + + 2 y +1 y y −1 SOLUTION: 1 1 2 1 1 2 + + 2 = + + y + 1 y y − 1 y + 1 y ( y + 1)( y − 1)
LCD = y ( y + 1)( y − 1) To obtain equivalent terms with this common denominator, we will multiply the y ( y − 1) ( y − 1)( y + 1) , and the third term by y . , the second term by first term by y y ( y − 1) ( y − 1)( y + 1) 1 1 2 1 y ( y − 1) 1 ( y − 1)( y + 1) 2 y + + = ⋅ + ⋅ + ⋅ y + 1 y ( y + 1)( y − 1) y + 1 y ( y − 1) y ( y − 1)( y + 1) ( y + 1)( y − 1) y
ANSWER:
2 y −1 y ( y − 1)
=
y2 − y + y2 −1+ 2 y y ( y + 1)( y − 1)
=
2 y2 + y −1 y ( y + 1)( y − 1)
=
( 2 y − 1)( y + 1) y ( y + 1)( y − 1)
=
2 y −1 y ( y − 1)
72.
PROBLEM: 1 1 1 − + y y +1 y −1 SOLUTION: LCD = y ( y − 1)( y + 1)
To obtain equivalent terms with this common denominator, we will multiply the ( y − 1)( y + 1) , the second term by y ( y − 1) and the third term by first term by y ( y − 1) ( y − 1)( y + 1) y ( y + 1) y ( y + 1)
.
1 1 1 1 ( y − 1)( y + 1) 1 y ( y − 1) 1 y ( y + 1) − + = ⋅ − ⋅ + ⋅ y y + 1 y − 1 y ( y − 1)( y + 1) y + 1 y ( y − 1) y − 1 y ( y + 1) =
y2 −1 y2 − y y2 + y − + y ( y − 1)( y + 1) y ( y − 1)( y + 1) y ( y − 1)( y + 1)
=
y2 −1 − y2 + y + y2 + y y ( y − 1)( y + 1)
=
y2 + 2 y −1 y ( y − 1)( y + 1)
=
y2 + 2 y −1 y ( y − 1)( y + 1)
ANSWER:
y2 + 2 y −1 y ( y − 1)( y + 1)
73.
PROBLEM: 5−2 + 2−1 SOLUTION: 1 1 1 1 5−2 + 2−1 = 2 + = + 5 2 25 2 LCD = 50 To obtain equivalent terms with this common denominator, we will multiply the 2 25 first term by and the second term by . 2 25 1 1 1 2 1 25 2 25 2 + 25 27 + = ⋅ + ⋅ = + = = 25 2 25 2 2 25 50 50 50 50 ANSWER:
74.
27 50
PROBLEM: 6−1 + 4−2 SOLUTION: 1 1 1 1 6−1 + 4−2 = + 2 = + 6 4 6 16 LCD = 48 To obtain equivalent terms with this common denominator, we will multiply the 8 3 first term by and the second term by . 8 3 1 1 1 8 1 3 8 3 8 + 3 11 + = ⋅ + ⋅ = + = = 6 16 6 8 16 3 48 48 48 48 11 ANSWER: 48
75.
PROBLEM: x −1 + y −1 SOLUTION: 1 1 x −1 + y −1 = + x y LCD = xy To obtain equivalent terms with this common denominator, we will multiply the y x first term by and the second term by . y x 1 1 1 y 1 x y x x+ y + = ⋅ + ⋅ = + = x y x y y x xy xy xy ANSWER:
76.
x+ y xy
PROBLEM: x −2 − y −1 SOLUTION: 1 1 x −2 − y −1 = 2 − x y 2 LCD = x y To obtain equivalent terms with this common denominator, we will multiply the y x2 first term by and the second term by 2 . y x 1 1 1 y 1 x2 y x2 y − x2 − = ⋅ − ⋅ = − = x2 y x2 y y x2 x2 y x2 y x2 y y − x2 ANSWER: 2 x y
77.
PROBLEM: −1 ( 2 x − 1) − x −2 SOLUTION:
( 2 x − 1)
−1
− x −2 =
1 1 − 2 2x −1 x
LCD = x 2 ( 2 x − 1) To obtain equivalent terms with this common denominator, we will multiply the x2 2x −1 . first term by 2 and the second term by 2x −1 x x2 1 2 x −1 2 1 1 − 2 = ⋅ 2− 2⋅ x 2x −1 2x −1 x 2x −1 x 2 x 2x −1 = 2 − 2 x ( 2 x − 1) x ( 2 x − 1) =
x2 − 2 x + 1 x 2 ( 2 x − 1)
( x − 1) = 2 x ( 2 x − 1) 2 x − 1) ( ANSWER: 2 x ( 2 x − 1) 2
78.
PROBLEM: −1 ( x − 4 ) − ( x + 1)−1 SOLUTION:
( x − 4)
−1
1 1 − x − 4 x +1
− ( x + 1) −1 =
LCD = ( x − 4 )( x + 1) To obtain equivalent terms with this common denominator, we will multiply the x +1 x−4 and the second term by first term by . x−4 x +1 1 1 x +1 x−4 − = − x − 4 x + 1 ( x − 4 )( x + 1) ( x − 4 )( x + 1) = = ANSWER:
79.
x +1− x + 4 ( x − 4 )( x + 1) 5
( x − 4 )( x + 1) 5
( x − 4 )( x + 1)
PROBLEM: 3x 2 ( x − 1) −1 − 2 x SOLUTION:
3x 2 3x ( x − 1) − 2 x = − 2x x −1 LCD = x − 1 To obtain equivalent terms with this common denominator, we will multiply the 1 x −1 first term by and the second term by . 1 x −1 x − 1 3x 2 2 x 2 − 2 x 3x 2 3x 2 1 − 2x = ⋅ − 2x ⋅ = − x −1 x −1 x −1 x −1 1 x −1 2 2 2 3x − 2 x + 2 x x + 2 x = = x −1 x −1 x ( x + 2) = x −1 2
−1
ANSWER:
x ( x + 2) x −1
80.
PROBLEM: −2 −1 2 ( y − 1) − ( y − 1) SOLUTION: 2 ( y − 1) − ( y − 1) = −2
2
−1
( y − 1)
2
−
1 y −1
LCD = ( y − 1) To obtain equivalent terms with this common denominator, we will multiply the 1 y −1 first term by and the second term by . 1 y −1 1 2 1 2 1 y −1 − = ⋅ − ⋅ 2 2 ( y − 1) y − 1 ( y − 1) 1 y − 1 y − 1 2
= = =
ANSWER:
2
( y − 1)
2
−
2 − y +1
( y − 1)
2
3− y
( y − 1)
3− y
( y − 1)
2
2
y −1
( y − 1)
2
Part C: Adding and Subtracting Rational Functions Calculate ( f + g ) ( x) and ( f − g ) ( x) and state the restrictions to the domain.
81.
PROBLEM: 1 1 f ( x) = and g ( x ) = 3x x−2 SOLUTION:
1 1 + 3x x − 2 LCD = 3x ( x − 2 ) To obtain equivalent terms with this common denominator, we will multiply the x−2 3x and the second term by . first term by x−2 3x 1 1 1 x−2 1 3x 3x x−2 + = ⋅ + ⋅ = + 3x x − 2 3x x − 2 x − 2 3x 3x ( x − 2 ) 3x ( x − 2 )
( f + g ) ( x) =
=
x − 2 + 3x 3x ( x − 2 )
=
4x − 2 3x ( x − 2 )
=
2 ( 2 x − 1)
3x ( x − 2 )
2 ( 2 x − 1)
( f + g ) ( x) =
3x ( x − 2 )
1 1 − 3x x − 2 1 1 1 x−2 1 3x 3x x−2 − = ⋅ − ⋅ = − 3x x − 2 3x x − 2 x − 2 3x 3x ( x − 2 ) 3x ( x − 2 )
( f − g ) ( x) =
=
x − 2 − 3x 3x ( x − 2 )
=
−4 x − 2 3x ( x − 2 )
=−
( f − g ) ( x)
2 ( 2 x + 1)
3x ( x − 2 )
=−
2 ( 2 x + 1)
3x ( x − 2 )
In this case, the domain of f ( x) consists of all real numbers except 0, and the domain of g ( x) consists all real numbers except 2. ANSWER: ( f + g ) ( x) =
2 ( 2 x − 1)
, ( f − g ) ( x) = −
2 ( 2 x + 1)
3x ( x − 2 ) 3x ( x − 2 ) The domain consists of any real number x where x ≠ 0 and x ≠ 2 . 82.
PROBLEM: 1 1 f ( x) = and g ( x ) = x −1 x+5 SOLUTION:
1 1 + x −1 x + 5 LCD = ( x − 1)( x + 5) To obtain equivalent terms with this common denominator, we will multiply the x+5 x −1 and the second term by first term by . x+5 x −1 1 1 1 x+5 1 x −1 x+5 x −1 + = ⋅ + ⋅ = + x − 1 x + 5 x − 1 x + 5 x + 5 x − 1 ( x − 1)( x + 5 ) ( x − 1)( x + 5 )
( f + g ) ( x) =
=
x + 5 + x −1 ( x − 1)( x + 5)
=
2x + 4 ( x − 1)( x + 5)
=
2 ( x + 2)
( x − 1)( x + 5) 2 ( x + 2) ( f + g ) ( x) = ( x − 1)( x + 5)
( f − g ) ( x) =
1 1 − x −1 x + 5
x+5 x −1 1 1 1 x+5 1 x −1 − = ⋅ − ⋅ = − x − 1 x + 5 x − 1 x + 5 x + 5 x − 1 ( x − 1)( x + 5 ) ( x − 1)( x + 5 ) = =
( f − g ) ( x) =
x + 5 − x +1 ( x − 1)( x + 5 ) 6
( x − 1)( x + 5 ) 6
( x − 1)( x + 5)
In this case, the domain of f ( x) consists of all real numbers except –1, and the domain of g ( x) consists all real numbers except –5. ANSWER: ( f + g ) ( x) =
( f − g ) ( x) =
2 ( x + 2)
( x − 1)( x + 5)
6
( x − 1)( x + 5)
The domain consists of any real number x where x ≠ −1 and x ≠ −5 .
83.
PROBLEM: x 1 f ( x) = and g ( x ) = 4− x x−4 SOLUTION:
x 1 + x−4 4− x x 1 x 1 x −1 + = − = x−4 4− x x−4 x−4 x−4 x −1 ( f + g ) ( x) = x−4
( f + g ) ( x) =
x 1 − x−4 4− x x 1 x 1 x +1 − = + = x−4 4− x x−4 x−4 x−4 x +1 ( f − g ) ( x) = x−4 In this case, the domain of f ( x) consists of all real numbers except 4, and the domain of g ( x) consists all real numbers except 4.
( f − g ) ( x) =
x −1 x +1 , ( f − g ) ( x) = x−4 x−4 The domain consists of any real number x where x ≠ 4 . ANSWER: ( f + g ) ( x) =
84.
PROBLEM: x 1 f ( x) = and g ( x ) = 2x − 3 x −5 SOLUTION:
x 1 + x − 5 2x − 3 LCD = ( x + 5)( 2 x − 3) To obtain equivalent terms with this common denominator, we will multiply the 2x − 3 x−5 and the second term by first term by . 2x − 3 x−5 1 1 x x 2x − 3 x − 5 2 x 2 − 3x + x − 5 + = ⋅ + ⋅ = x − 5 2 x − 3 x − 5 2 x − 3 2 x − 3 x − 5 ( x + 5 )( 2 x − 3)
( f + g ) ( x) =
=
( f + g ) ( x) =
2 x2 − 2x − 5 ( x + 5)( 2 x − 3)
2 x2 − 2 x − 5 ( x + 5)( 2 x − 3)
x 1 − x − 5 2x − 3 1 1 x x 2x − 3 x − 5 2 x 2 − 3x − x + 5 − = ⋅ − ⋅ = x − 5 2 x − 3 x − 5 2 x − 3 2 x − 3 x − 5 ( x + 5 )( 2 x − 3)
( f − g ) ( x) =
=
2x2 − 4x + 5 ( x + 5)( 2 x − 3)
2 x2 − 4 x + 5 ( x + 5 )( 2 x − 3) In this case, the domain of f ( x) consists of all real numbers except 5, and the 3 domain of g ( x) consists all real numbers except . 2 2 2x − 2x − 5 ANSWER: ( f + g ) ( x) = ( x + 5)( 2 x − 3)
( f − g ) ( x) =
( f − g ) ( x) =
2 x2 − 4 x + 5 ( x + 5 )( 2 x − 3)
The domain consists of any real number x where x ≠ 5 and x ≠
3 . 2
85.
PROBLEM: x −1 4 f ( x) = 2 and g ( x ) = 2 x −4 x − 6 x − 16 SOLUTION:
x −1 4 + 2 2 x − 4 x − 6 x − 16 x −1 4 x −1 4 + 2 = + 2 x − 4 x − 6 x − 16 ( x − 2 )( x + 2 ) ( x − 8 )( x + 2 )
( f + g ) ( x) =
LCD = ( x − 2 )( x + 2 )( x − 8) To obtain equivalent terms with this common denominator, we will multiply the x −8 x−2 and the second term by first term by . x −8 x−2 x −1 4 x −1 x −8 4 x−2 + = ⋅ + ⋅ ( x − 2 )( x + 2 ) ( x − 8)( x + 2 ) ( x − 2 )( x + 2 ) x − 8 ( x − 8)( x + 2 ) x − 2 =
4x − 8 x2 − 9 x + 8 + ( x − 2 )( x + 2 )( x − 8) ( x − 2 )( x + 2 )( x − 8)
x2 − 9x + 8 + 4x − 8 = ( x − 2 )( x + 2 )( x − 8) = =
x2 − 5x ( x − 2 )( x + 2 )( x − 8) x ( x − 5)
( x − 2 )( x + 2 )( x − 8)
( f + g ) ( x) =
x( x − 5) ( x − 2 )( x + 2 )( x − 8)
( f − g ) ( x) =
x −1 4 − 2 2 x − 4 x − 6 x − 16
x −1 4 x −1 x −8 4 x−2 − = ⋅ − ⋅ ( x − 2 )( x + 2 ) ( x − 8 )( x + 2 ) ( x − 2 )( x + 2 ) x − 8 ( x − 8 )( x + 2 ) x − 2 =
x2 − 9 x + 8 4x − 8 − ( x − 2 )( x + 2 )( x − 8) ( x − 2 )( x + 2 )( x − 8)
=
x2 − 9x + 8 − 4x + 8 ( x − 2 )( x + 2 )( x − 8)
x 2 − 13 x + 16 = ( x − 2 )( x + 2 )( x − 8)
( f − g ) ( x) =
x 2 − 13 x + 16 ( x − 2 )( x + 2 )( x − 8 )
In this case, the domain of f ( x) consists of all real numbers except 2 and –2, and the domain of g ( x) consists all real numbers except –2 and 8. ANSWER: ( f + g ) ( x) =
x( x − 5) , ( x − 2 )( x + 2 )( x − 8 )
x 2 − 13 x + 16 ( x − 2 )( x + 2 )( x − 8 ) The domain consists of any real number x where x ≠ ±2 and x ≠ 8 .
( f − g ) ( x) =
86.
PROBLEM: 5 f ( x) = and g ( x ) = 3x + 4 x+2 SOLUTION:
( f + g ) ( x) =
5 + 3x + 4 x+2
LCD = x + 2 To obtain equivalent terms with this common denominator, we will multiply the 1 x+2 first term by and the second term by . 1 x+2 5 5 1 x+2 + 3x + 4 = ⋅ + 3x + 4 ⋅ x+2 x+2 1 x+2 2 5 3 x + 10 x + 8 = + x+2 x+2 2 5 + 3 x + 10 x + 8 = x+2 2 3 x + 10 x + 13 = x+2 2 3 x + 10 x + 13 ( f + g ) ( x) = x+2
5 − ( 3x + 4 ) x+2 5 5 1 x+2 − ( 3x + 4 ) = ⋅ − ( 3x + 4 ) ⋅ x+2 x+2 1 x+2 2 5 3 x + 10 x + 8 = − x+2 x+2 2 5 − 3 x − 10 x − 8 = x+2 −3 x 2 − 10 x − 3 = x+2 2 3 x + 10 x + 3 ( f − g ) ( x) = − x+2 In this case, the domain of f ( x) consists of all real numbers except –2, and the domain of g ( x) consists all real numbers.
( f − g ) ( x) =
ANSWER: ( f + g ) ( x) =
3 x 2 + 10 x + 13 x+2
3 x 2 + 10 x + 3 x+2 The domain consists of any real number x where x ≠ −2 .
( f − g ) ( x) = −
Calculate ( f + f ) ( x) and state the restrictions to the domain.
87.
PROBLEM: 1 f ( x) = x SOLUTION:
( f + f )( x ) =
1 1 + x x
1 1 2 + = x x x 2 x In this case, the domain of f ( x) consists of all real numbers except 0.
( f + f )( x ) =
2 x The domain consists of any real number x where x ≠ 0 . ANSWER: ( f + f )( x ) =
88.
PROBLEM: 1 f ( x) = 2x SOLUTION:
1 1 + 2x 2x 1 1 2 1 + = = 2x 2x 2x x 1 ( f + f )( x ) = x In this case, the domain of f ( x) consists of all real numbers except 0.
( f + f )( x ) =
1 x The domain consists of any real number x where x ≠ 0 . ANSWER: ( f + f )( x ) =
89.
PROBLEM: x f ( x) = 2x −1 SOLUTION:
x x + 2x −1 2x −1 x x 2x + = 2x −1 2x −1 2x −1 2x ( f + f )( x ) = 2x −1
( f + f )( x ) =
In this case, the domain of f ( x) consists of all real numbers except ANSWER: ( f + f )( x ) =
2x 2x −1
The domain consists of any real number x where x ≠ 90.
1 . 2
1 . 2
PROBLEM: 1 f ( x) = x+2 SOLUTION:
1 1 + x+2 x+2 1 1 2 + = x+2 x+2 x+2 2 ( f + f )( x ) = x+2 In this case, the domain of f ( x) consists of all real numbers except –2.
( f + f )( x ) =
2 x+2 The domain consists of any real number x where x ≠ −2 . ANSWER: ( f + f )( x ) =
Part D: Discussion Board
91.
PROBLEM:
Explain to a classmate why this is incorrect:
1 2 3 + 2 = 2. 2 2x x x
ANSWER: 1 2 3 + 2 = 2 2 x x x The sum of the two rational expressions on the right hand side of the equation is 3 3 not . 2 2x 2 x
92.
PROBLEM: Explain to a classmate how to find the common denominator when adding algebraic expressions. Give an example. ANSWER: Adding and subtracting rational expressions is similar to adding and subtracting fractions. Recall that if the denominators are the same, we can add or subtract the numerators and write the result over the common denominator. 3 7 3+ 7 + = 13 13 13 10 = 13 x−5 1 x − 5 −1 − = Simplify the numerator. 2x −1 2x −1 2x −1 x−6 = 2x −1
To add rational expressions with unlike denominators, we need to first find equivalent expressions with common denominators. We do this just as we have with fractions. If the denominators of fractions are relatively prime, then the LCD is their product. For example, 1 1 + ⇒ LCD = 3*5 = 15 3 5 1 1 1 y 1 x + = ⋅ + ⋅ x y x y y x y x = + Equivalent terms with xy xy a common denominator. =
y+x xy
Add the numerators and place the result over the common denominator xy.
7.4 Complex Rational Expressions
Part A: Complex Rational Expressions Simplify. (Assume all denominators are nonzero.)
1.
PROBLEM: 1 2 5 4 SOLUTION: 1 2 = 1⋅4 = 2 5 2 5 5 4 ANSWER:
2.
2 5
PROBLEM: 7 8 5 4 SOLUTION: 7 8 = 7⋅4 = 7 5 8 5 10 4 7 ANSWER: 10
3.
PROBLEM: 10 3 20 9 SOLUTION: 10 3 = 10 ⋅ 9 = 3 20 3 20 2 9 ANSWER:
4.
3 2
PROBLEM: 4 − 21 8 7 SOLUTION: 4 − 21 = − 4 ⋅ 7 = − 1 8 21 8 6 7 1 ANSWER: − 6
5.
PROBLEM: 2 3 5 6 SOLUTION: 2 3 = 2⋅6 = 4 5 3 5 5 6 ANSWER:
4 5
6.
PROBLEM: 7 4 14 3 SOLUTION: 7 4 = 7⋅ 3 = 3 14 4 14 8 3 3 ANSWER: 8
7.
PROBLEM: 3 1− 2 5 1 − 4 3 SOLUTION: 3 1 1− − 2 = 2 = − 1 ⋅ 12 = − 6 5 1 11 2 11 11 − 4 3 12 ANSWER: −
8.
6 11
PROBLEM: 1 −5 2 1 1 + 2 3 SOLUTION: 1 9 −5 − 9 6 27 2 = 2 =− ⋅ =− 1 1 5 2 5 5 + 2 3 6 ANSWER: −
27 5
9.
PROBLEM: 3 1+ 2 1 1− 4 SOLUTION: 3 5 1+ 2 = 2 = 5 ⋅ 4 = 10 1 3 2 3 3 1− 4 4 ANSWER:
10.
10 3
PROBLEM: 1 2− 2 3 1+ 4 SOLUTION: 1 3 2− 2 = 2 = 3⋅4 = 6 3 7 2 7 7 1+ 4 4 ANSWER:
11.
6 7
PROBLEM: 5x2 x +1 25 x x +1 SOLUTION: 5x2 2 x + 1 = 5x ⋅ x + 1 = x 25 x x + 1 25 x 5 x +1 ANSWER:
x 5
12.
PROBLEM: 7+ x 7x x+7 14 x 2 SOLUTION: 7+ x 2 7 x = x + 7 ⋅ 14 x = 2 x x+7 7x x + 7 2 14 x ANSWER: 2x
13.
PROBLEM: 3y x y2 x −1 SOLUTION: 3y x = 3 y ⋅ x − 1 = 3 ( x − 1) y2 x y2 xy x −1 ANSWER:
14.
3 ( x − 1) xy
PROBLEM: 5a 2 b −1 15a 3
( b − 1)
2
SOLUTION: 5a 2 2 2 b − 1 = 5a ⋅ ( b − 1) = b − 1 15a 3 b − 1 15a 3 3a 2 ( b − 1) b −1 ANSWER: 3a
15.
PROBLEM: 1 1+ x 1 2− x SOLUTION: 1 x 1 x 1 x +1 1+ 1⋅ + + x = x x = x x = x 1 x 1 2x 1 2x −1 2− 2⋅ − − x x x x x x x +1 x = ⋅ x 2x −1 x +1 = 2x −1 ANSWER:
16.
x +1 2x −1
PROBLEM: 2 +1 x 1 3− x SOLUTION: 2 2 x 2 x 2+ x +1 + 1⋅ + x x = x x = x = x 1 x 1 3x 1 3x − 1 3− 3⋅ − − x x x x x x x 2+ x = ⋅ x 3x − 1 x+2 = 3x − 1 x+2 ANSWER: 3x − 1
17.
PROBLEM: 2 −4 3y 1 6− y SOLUTION: 2 2 3y 2 12 y 2 − 12 y −4 − 4⋅ − 3y 3y 3y 3y 3y 3y = = = y 1 1 6y 1 6 y −1 6− 6⋅ − − y y y y y y −2 ( 6 y − 1) −2 ( 6 y − 1) y 3y = = ⋅ 6 y −1 3y 6 y −1 y 2 =− 3
ANSWER: −
18.
2 3
PROBLEM: 5 1 − y 2 10 − y y2 SOLUTION: y 5 1 5 2 1 y 10 10 − y − ⋅ − ⋅ − y 2 y 2 2 y 2y 2y 2y = = = 10 − y 10 − y 10 − y 10 − y 2 2 2 y y y y2 10 − y y 2 ⋅ 2 y 10 − y y = 2 y ANSWER: 2 =
19.
PROBLEM: 1 1 − 5 x 1 1 − 25 x 2 SOLUTION: 1 1 1 x 1 5 x 5 x−5 − ⋅ − ⋅ − 5 x = 5 x x 5 = 5x 5x = 5x 1 1 x2 x 2 − 25 1 x 2 1 25 25 − 2 ⋅ 2− 2⋅ − 25 x 25 x x 25 25 x 2 25 x 2 25 x 2 x −5 5x = ( x − 5)( x + 5) 25 x 2 x−5 25 x 2 = ⋅ 5 x ( x − 5 )( x + 5 ) =
5x x+5
ANSWER:
5x x+5
20.
PROBLEM: 1 1 + x 5 1 1 − 25 x 2 SOLUTION: 1 1 1 x 1 5 x 5 x+5 + ⋅ + ⋅ + 5 x = 5 x x 5 = 5x 5x = 5x 1 1 x2 x 2 − 25 1 x 2 1 25 25 − 2 ⋅ 2− 2⋅ − 25 x 25 x x 25 25 x 2 25 x 2 25 x 2 x+5 5x = ( x − 5)( x + 5) 25 x 2 x+5 25 x 2 = ⋅ 5 x ( x − 5 )( x + 5 ) 5x x −5 5x ANSWER: x−5 =
21.
PROBLEM: 1 1 − x 3 1 1 − 9 x2 SOLUTION: 1 1 1 3 1 x 3 x 3− x − ⋅ − ⋅ − x 3 = x 3 3 x = 3x 3x = 3x 2 1 1 x2 9 x2 − 9 − 2 1 ⋅ x 2 − 12 ⋅ 9 − 9 x 9 x x 9 9 x2 9 x2 9 x2 3− x x−3 − x −3 9x2 3 x 3 x = 2 = =− ⋅ x − 9 ( x − 3)( x + 3) 3x ( x − 3)( x + 3) 9x2 9x2 3x =− x+3 ANSWER: −
3x x+3
22.
PROBLEM: 1 1 + 4 x 1 1 − x 2 16 SOLUTION: 1 1 1 x 1 4 x 4 + ⋅ + ⋅ + 4 x = 4 x x 4 = 4x 4x 1 1 x2 1 16 1 x 2 16 − ⋅ − ⋅ − x 2 16 x 2 16 16 x 2 16 x 2 16 x 2 x+4 = 4x 2 16 − x 16 x 2 x+4 4x = ( 4 − x )( 4 + x ) 16 x 2 x+4 16 x 2 = ⋅ 4 x ( 4 − x )( 4 + x ) =
4x 4− x
ANSWER:
4x 4− x
23.
PROBLEM: 1 16 − 2 x 1 −4 x SOLUTION: 1 x 2 1 16 x 2 1 16 x 2 − 1 − 2 16 − 2 16 ⋅ 2 − 2 x = x x = x2 x = x2 1 1 1 4x 1 − 4x x −4 −4 ⋅ − x x x x x x ( 4 x − 1)( 4 x + 1) ( 4 x − 1)( 4 x + 1) x2 x2 = = 1 − 4x 4x −1 − x x ( 4 x − 1)( 4 x + 1) ⋅ ⎛ − x ⎞ = ⎜ ⎟ x2 ⎝ 4x −1 ⎠ 4x +1 =− x ANSWER: −
4x +1 x
24.
PROBLEM: 1 2− y 1 1− 2 4y SOLUTION: 1 2y 1 2 y −1 y 1 − 2− 2⋅ − y y y y y y = = = 2 2 2 1 1 1 − 2 1⋅ 4 y 2 − 1 2 4 y 2 − 1 2 4 y − 2 4y 4y 4y 4y 4y 4y 2 y −1 y = ( 2 y − 1)( 2 y + 1) 4 y2 =
2 y −1 4 y2 ⋅ y ( 2 y − 1)( 2 y + 1)
4y 2 y +1 4y ANSWER: 2 y +1 =
25.
PROBLEM: 1 1 + x y 1 1 − y2 x2 SOLUTION: y 1 1 1 y 1 x + ⋅ + ⋅ + xy x y x y y x = = 2 2 2 1 1 1 1 x y x − ⋅ 2− 2⋅ 2 − 2 y 2 x2 y x x y x2 y 2 x+ y xy = ( x + y )( x − y ) x2 y 2
( x + y) ⋅ = xy
=
xy x− y
ANSWER:
xy x− y
x2 y2 ( x + y )( x − y )
x+ y x xy xy = 2 2 y x − y2 x2 y2 x2 y2
26.
PROBLEM: 1 4 − 2x 3 1 16 − 4x2 9 SOLUTION: 1 4 1 3 4 2x 3 8x − ⋅ − ⋅ − 2x 3 = 2x 3 3 2x = 6x 6x 1 16 1 9 16 4 x 2 9 64 x 2 − ⋅ − ⋅ − 4x2 9 4 x 2 9 9 4 x 2 36 x 2 36 x 2 3 − 8x 6x = 9 − 64 x 2 36 x 2 3 − 8x 6x = ( 3 − 8 x )( 3 + 8 x ) 36 x 2 3 − 8x 36 x 2 = ⋅ 6 x ( 3 − 8 x )( 3 + 8 x ) =
6x 3 + 8x
ANSWER:
6x 3 + 8x
27.
PROBLEM: 2 1 − 2 25 2 x 1 1 − 5 2x SOLUTION: 2 1 2 2x2 1 25 4x2 25 − 2 ⋅ 2− 2⋅ − 2 25 2 x = 25 2 x 2 x 25 = 50 x 50 x 2 1 1 1 2x 1 5 2x 5 − ⋅ − ⋅ − 5 2x 5 2x 2x 5 10 x 2 x 2 4 x − 25 2 = 50 x 2x − 5 10 x ( 2 x − 5)( 2 x + 5) 50 x 2 = 2x − 5 10 x ( 2 x − 5)( 2 x + 5) 10 x = ⋅ 50 x 2 ( 2 x − 5)
=
2x + 5 5x
ANSWER:
2x + 5 5x
28.
PROBLEM: 4 1 − 2 25 4 x 1 1 + 5 4x SOLUTION: 4 1 4 4 x2 1 25 16 x 2 25 − 2 ⋅ 2− 2⋅ − 2 25 4 x = 25 4 x 4 x 25 = 100 x 100 x 2 1 1 1 4x 1 5 4x 5 + ⋅ + ⋅ + 5 4x 5 4x 4x 5 20 x 20 x 2 16 x − 25 2 = 100 x 4x + 5 20 x ( 4 x + 5)( 4 x − 5) 100 x 2 = 4x + 5 20 x ( 4 x + 5)( 4 x − 5) ⋅ 20 x = 100 x 2 4x + 5 4x − 5 = 5x ANSWER:
4x − 5 5x
29.
PROBLEM: 1 1 − y x 2 4− xy SOLUTION: 1 1 1 x 1 y x y x− y − ⋅ − ⋅ − y x y x x y xy xy xy = = = 2 xy 2 4 xy 2 4 xy − 2 4− 4⋅ − − xy xy xy xy xy xy x− y xy = ⋅ xy 4 xy − 2 x− y = 4 xy − 2 ANSWER:
30.
x− y 4 xy − 2
PROBLEM: 1 +2 ab 1 1 + a b SOLUTION: 1 1 ab 1 2ab 1 + 2ab +2 + 2⋅ + ab ab = ab ab = ab = ab 1 1 1 b 1 a b a a+b + ⋅ + ⋅ + a b a b b a ab ab ab 1 + 2ab 1 + 2ab ab ab = = ⋅ a+b ab a+b ab 1 + 2ab = a+b ANSWER:
1 + 2ab a+b
31.
PROBLEM: 1 1 + y x xy SOLUTION: 1 1 1 x 1 y x y x+ y + ⋅ + ⋅ + y x y x x y xy xy xy = = = xy xy xy xy x+ y 1 = ⋅ xy xy x+ y = 2 2 x y ANSWER:
32.
x+ y x2 y2
PROBLEM: 3x 1 1 − 3 x SOLUTION: 3x 3x 3x 3x = = = 1 1 1 x 1 3 x 3 x−3 − ⋅ − ⋅ − 3 x 3 x x 3 3x 3x 3x 3x = 3x ⋅ x−3 2 9x = x −3 ANSWER:
9x2 x−3
33.
PROBLEM: 4 21 1− − 2 x x 2 15 1− − 2 x x SOLUTION: 4 21 ⎛ 4 21 ⎞ 2 1 − − 2 ⎜1 − − 2 ⎟ x 2 x x = ⎝ x x ⎠ = x − 4 x − 21 2 2 15 2 15 1 − − 2 ⎜⎛1 − − 2 ⎟⎞ x 2 x − 2 x − 15 x x ⎝ x x ⎠
34.
=
( x − 7 )( x + 3) ( x − 5)( x + 3)
=
x−7 x −5
ANSWER:
x−7 x−5
PROBLEM: 3 4 1− − 2 x x 16 1− 2 x SOLUTION: 3 4 ⎛ 3 4⎞ 2 1 − − 2 ⎜1 − − 2 ⎟ x 2 x x = ⎝ x x ⎠ = x − 3x − 4 16 x 2 − 16 ⎛ 16 ⎞ 2 1− 2 1 x − ⎜ 2 ⎟ x ⎝ x ⎠ ( x − 4 )( x + 1) = ( x − 4 )( x + 4 ) =
x +1 x+4
ANSWER:
x +1 x+4
35.
PROBLEM: 1 1 3− − 2 2x 2x 2 1 2− + 2 x 2x SOLUTION: 1 1 ⎞ ⎛ 1 1 − 2 ⎟ 2 x2 − 2 ⎜3− 3− 6x2 − x −1 2x 2x ⎠ 2x 2x = ⎝ = 2 2 1 2 1 ⎞ 4x − 4x +1 ⎛ 2− + 2 2 − + 2 ⎟ 2x2 ⎜ x 2x x 2x ⎠ ⎝
ANSWER:
36.
=
( 3x + 1)( 2 x − 1) 2 ( 2 x − 1)
=
3x + 1 2x −1
3x + 1 2x −1
PROBLEM: 1 5 12 − + 2 x x2 1 6 18 − + 2 x x2 SOLUTION: 1 5 12 ⎛ 1 − 5 + 12 ⎞ 2x 2 − + ⎜ 2 ⎟ x 2 − 10 x + 24 2 x x2 = ⎝ 2 x x ⎠ = 1 6 18 ⎛ 1 6 18 ⎞ 2 x 2 − 12 x + 36 − + 2 ⎜ − + 2 ⎟ 2x 2 x x ⎝2 x x ⎠ ( x − 6 )( x − 4 ) = 2 ( x − 6)
=
ANSWER:
x−4 x−6
x−4 x−6
37.
PROBLEM: 1 4 − x 3x 2 8 16 3− + 2 x 3x SOLUTION: ⎛1 4 ⎞ 2 1 4 − 2 ⎜ − 2 ⎟ 3x 3x − 4 1 x 3x = ⎝ x 3x ⎠ = = 2 8 16 8 16 3x − 4 3 − + 2 ⎛⎜ 3 − + 2 ⎞⎟ 3x 2 ( 3x − 4 ) x 3x x 3x ⎠ ⎝ ANSWER:
38.
1 3x − 4
PROBLEM: 3 1 1+ − 10 x 10 x 2 3 1 1 − − 2 5 10 x 5 x SOLUTION: 3 1 ⎞ ⎛ 3 1 − 10 x 2 − 1+ ⎜1 + 2 ⎟ 2 10 x 2 + 3x − 1 10 x 10 x = ⎝ 10 x 10 x ⎠ = 3 1 1 3 1 1 ⎞ 6 x2 − x − 2 − − 2 ⎛⎜ − − 2 ⎟10 x 2 5 10 x 5 x ⎝ 5 10 x 5 x ⎠ ( 2 x + 1)( 5 x − 1) = ( 2 x + 1)( 3x − 2 ) =
ANSWER:
5x − 1 3x − 2
5x − 1 3x − 2
39.
PROBLEM: x −1 4 5 1+ − 2 x x SOLUTION: x − 1) x 2 x 2 ( x − 1) ( x −1 = = 2 4 5 4 5 1 + − 2 ⎛⎜1 + − 2 ⎞⎟ x 2 x + 4 x − 5 x x ⎝ x x ⎠ x 2 ( x − 1) = ( x + 5)( x − 1)
40.
=
x2 x+5
ANSWER:
x2 x+5
PROBLEM: 5 3 2− − 2 2x x 4x + 3 SOLUTION: 5 3 ⎞ 5 3 ⎛ − 2 ⎟ 2x2 − 2 ⎜2− 2− 4 x2 − 5x − 6 2x x ⎠ 2x x = ⎝ = 2 4x + 3 2 x ( 4 x + 3) ( 4 x + 3) 2 x 2
ANSWER:
=
( 4 x + 3)( x − 2 ) 2 x 2 ( 4 x + 3)
=
x−2 2x2
x−2 2 x2
41.
PROBLEM: 1 2 + x −3 x 1 3 − x x −3 SOLUTION:
2x − 6 x 1 2 1 x 2 x −3 + + ⋅ + ⋅ x − 3 x = x − 3 x x x − 3 = x ( x − 3) x ( x − 3) 1 3 1 x −3 3 x 3x x −3 − ⋅ − ⋅ − x x − 3 x x − 3 x − 3 x x ( x − 3) x ( x − 3) 3( x − 2) 3x − 6 x + 2x − 6 x ( x − 3) x ( x − 3) x ( x − 3) = = = 2x + 3 x − 3 − 3x −2 x − 3 − x ( x − 3) x ( x − 3) x ( x − 3) 3 ( x − 2 ) ⎛ x ( x − 3) ⎞ ⋅⎜ − ⎟ x ( x − 3) ⎝ 2 x + 3 ⎠ 3 ( x − 2) =− 2x + 3
=
ANSWER: −
3( x − 2) 2x + 3
42.
PROBLEM: 1 1 + 2 4x − 5 x 1 1 + 2 x 3 x − 10 SOLUTION: 1 1 1 x2 1 4 x − 5 + 2 ⋅ 2+ 2⋅ 4x − 5 x = 4x − 5 x x 4x − 5 1 1 1 3 x − 10 1 x2 + ⋅ + ⋅ x 2 3 x − 10 x 2 3 x − 10 3 x − 10 x 2 x2 4x − 5 + 2 2 x ( 4 x − 5) x ( 4 x − 5) = x2 3x − 10 + x 2 ( 3 x − 10 ) x 2 ( 3 x − 10 )
( x + 5)( x − 1) x2 + 4 x − 5 2 x ( 4 x − 5) x 2 ( 4 x − 5) = 2 = x + 3 x − 10 ( x + 5 ) ( x − 2 ) x 2 ( 3 x − 10 ) x 2 ( 3 x − 10 ) x + 5 )( x − 1) x 2 ( 3 x − 10 ) ( = 2 ⋅ x ( 4 x − 5 ) ( x + 5 )( x − 2 ) ( x − 1)( 3x − 10 ) = ( x − 2 )( 4 x − 5 ) ANSWER:
( x − 1)( 3x − 10 ) ( x − 2 )( 4 x − 5 )
43.
PROBLEM: 1 4 + x+5 x−2 2 1 − x−2 x+5 SOLUTION: 1 4 1 x−2 4 x+5 + ⋅ + ⋅ x+5 x−2 = x+5 x−2 x−2 x+5 2 1 2 x+5 1 x−2 − − x−2 x+5 x−2 x+5 x+5 x−2 4 x + 20 x−2 + ( x + 5)( x − 2 ) ( x − 2 )( x + 5) = 2 x + 10 x−2 − ( x + 5)( x − 2 ) ( x + 5)( x − 2 ) x − 2 + 4 x + 20 ( x + 5)( x − 2 ) = 2 x + 10 − x + 2 ( x + 5)( x − 2 ) 5 x + 18 ( x + 5)( x − 2 ) = x + 12 ( x + 5)( x − 2 ) =
( x + 5)( x − 2 ) 5 x + 18 ⋅ x + 12 ( x + 5)( x − 2 )
=
5 x + 18 x + 12
ANSWER:
5 x + 18 x + 12
44.
PROBLEM: 3 2 − x −1 x + 3 2 1 + x +3 x −3 SOLUTION:
3x + 9 2x − 2 3x + 9 − 2 x + 2 3 2 3 x+3 2 x −1 − − ⋅ − ⋅ x − 1 x + 3 = x − 1 x + 3 x + 3 x − 1 = ( x − 1)( x + 3) ( x − 1)( x + 3) = ( x − 1)( x + 3) 2 1 2 x −3 1 x+3 2x − 6 2x − 6 + x + 3 x+3 + + + x +3 x −3 x +3 x −3 x −3 x +3 ( x + 3)( x − 3) ( x + 3)( x − 3) ( x − 3)( x + 3) x + 11 ( x − 1)( x + 3) = 3x − 3 ( x − 4 )( x + 3) x + 11 ( x − 1)( x + 3) = 3 ( x − 1) ( x − 3)( x + 3) =
( x − 3)( x + 3) x + 11 ⋅ ( x − 1)( x + 3) 3 ( x − 1)
=
( x + 11)( x − 3) 2 3 ( x − 1)
ANSWER:
( x + 11)( x − 3) 2 3 ( x − 1)
45.
PROBLEM: 2 x − x +1 x + 3 x 1 + 3x + 4 x + 1 SOLUTION:
2x + 2 x 2 + 3x 2 2 x +1 x x x+3 − − ⋅ − ⋅ x + 1 x + 3 = x + 1 x + 3 x + 3 x + 1 = ( x + 1)( x + 3) ( x + 1)( x + 3) 1 x x x + 1 1 3x + 4 3x + 4 x2 + x + ⋅ + ⋅ + 3 x + 4 x + 1 3 x + 4 x + 1 x + 1 3 x + 4 ( 3 x + 4 )( x + 1) ( 3x + 4 )( x + 1) x 2 + 3x − 2 x − 2 ( x + 1)( x + 3) = 2 x + x + 3x + 4 ( 3x + 4 )( x + 1) x2 + x − 2 ( x + 1)( x + 3) = x2 + 4x + 4 ( 3x + 4 )( x + 1)
( x + 2 )( x − 1) ( x + 1)( x + 3) = 2 ( x + 2) ( 3x + 4 )( x + 1) ( x + 2 )( x − 1) ⋅ ( 3x + 4 )( x + 1) = ( x + 1)( x + 3) ( x + 2 )2 ( x − 1)( 3x + 4 ) = ( x + 3)( x + 2 ) ANSWER:
( x − 1)( 3x + 4 ) ( x + 3)( x + 2 )
46.
PROBLEM: x 2 + x − 9 x +1 x 1 − 7x − 9 x +1 SOLUTION:
2 x − 18 x2 + x 2 2 x −9 x x x +1 + + ⋅ + ⋅ x − 9 x + 1 = x − 9 x + 1 x + 1 x − 9 = ( x − 9 )( x + 1) ( x − 9 )( x + 1) 1 1 7x − 9 x x x +1 7x − 9 x2 + x − ⋅ − ⋅ − 7 x − 9 x + 1 7 x − 9 x + 1 x + 1 7 x − 9 ( 7 x − 9 )( x + 1) ( 7 x − 9 )( x + 1) x 2 + x + 2 x − 18 ( x − 9 )( x + 1) = 2 x + x − 7x + 9 ( 7 x − 9 )( x + 1) x 2 + 3x − 18 ( x − 9 ) ( x + 1) = x2 − 6x + 9 ( 7 x − 9 )( x + 1)
( x + 6 )( x − 3) ( x − 9 )( x + 1) = 2 ( x − 3) ( 7 x − 9 )( x + 1) ( x + 6 )( x − 3) ⋅ ( 7 x − 9 )( x + 1) = 2 ( x − 9 )( x + 1) ( x − 3) ( x + 6 )( 7 x − 9 ) = ( x − 9 )( x − 3) ANSWER:
( x + 6 )( 7 x − 9 ) ( x − 9 )( x − 3)
47.
PROBLEM: 1 x − 3x + 2 x + 2 2 x − x+2 x+2 SOLUTION:
3x + 2 x2 + 2x 1 1 3x + 2 x x x+2 − − ⋅ − ⋅ 3 x + 2 x + 2 = 3 x + 2 x + 2 x + 2 3 x + 2 = ( 3 x + 2 )( x + 2 ) ( 3x + 2 )( x + 2 ) 2 2 2 x x x − − − x+2 x+2 x+2 x+2 x+2 x+2 2 x + 2 x − 3x − 2 ( 3x + 2 )( x + 2 ) = x−2 x+2 2 x −x−2 ( 3x + 2 )( x + 2 ) = x−2 x+2 ( x − 2 )( x + 1) ( 3x + 2 )( x + 2 ) = x−2 x+2 ( x − 2 )( x + 1) ⋅ x + 2 = ( 3x + 2 )( x + 2 ) x − 2 =
ANSWER:
x +1 3x + 2
x +1 3x + 2
48.
PROBLEM: 1 x + x−4 x+2 1 x + 3x + 4 x + 2 SOLUTION:
x2 + 2x x−4 1 1 x−4 x x x+2 + + ⋅ + ⋅ x − 4 x + 2 = x − 4 x + 2 x + 2 x − 4 = ( x − 4 )( x + 2 ) ( x − 4 )( x + 2 ) 1 1 3x + 4 x x x+2 3x + 4 x2 + 2x + + ⋅ + 3 x + 4 x + 2 3 x + 4 x + 2 x + 2 3 x + 4 ( 3x + 4 )( x + 2 ) ( 3x + 4 )( x + 2 ) x2 + 2x + x − 4 ( x − 4 )( x + 2 ) = 2 x + 2 x + 3x + 4 ( 3x + 4 )( x + 2 ) x 2 + 3x − 4 ( x − 4 )( x + 2 ) = x2 + 5x + 4 ( 3x + 4 )( x + 2 )
( x + 4 )( x − 1) ( x − 4 )( x + 2 ) = ( x + 4 )( x + 1) ( 3x + 4 )( x + 2 ) ( x + 4 )( x − 1) ⋅ ( 3x + 4 )( x + 2 ) = ( x − 4 )( x + 2 ) ( x + 4 )( x + 1) ( 3x + 4 )( x − 1) = ( x − 4 )( x + 1) ANSWER:
( 3x + 4 )( x − 1) ( x − 4 )( x + 1)
49.
PROBLEM: a 3 − 8b3 27 a − 2b SOLUTION: 2 2 a 3 − 8b3 ( a − 2b ) ( a + 2ab + 4b ) 27 = a − 2b
27 a − 2b a 2 + 2ab + 4b 2 = 27
=
( a − 2b ) ( a 2 + 2ab + 4b 2 ) 27
⋅
1 a − 2b
a 2 + 2ab + 4b 2 ANSWER: 27 50.
PROBLEM: 27a 3 + b3 ab 3a + b SOLUTION:
(
)
2 2 27a 3 + b3 ( 3a + b ) 9a − 3ab + b ( 3a + b ) 9a 2 − 3ab + b2 1 ab ab = = ⋅ 3a + b 3a + b ab 3a + b 2 2 9a − 3ab + b = ab
ANSWER:
9a 2 − 3ab + b 2 ab
(
)
51.
PROBLEM: 1 1 + b3 a 3 1 1 + b a SOLUTION: a3 b3 a 3 + b3 1 1 1 a 3 1 b3 + ⋅ + ⋅ + b3 a 3 = b3 a 3 a 3 b3 = a 3b3 a 3b3 = a 3b3 1 1 1 a 1 b a b a+b + ⋅ + ⋅ + b a b a a b ab ab ab
( a + b ) ( a 2 − ab + b2 ) a 3b3 a+b ab
=
=
( a + b ) ( a 2 − ab + b2 ) 3 3
ab a − ab + b 2 = a 2b 2 2
a 2 − ab + b 2 ANSWER: a 2b 2
⋅
ab a+b
52.
PROBLEM: 1 1 − b3 a 3 1 1 − a b SOLUTION: 1 1 1 a 3 1 b3 a3 b3 a 3 − b3 − ⋅ − ⋅ − b3 a 3 = b3 a 3 a 3 b3 = a 3b3 a 3b3 = a3b3 b a b−a 1 1 1 b 1 a − ⋅ − ⋅ − a b a b b a ab ab ab 2 2 ( a − b ) a + ab + b
(
ab b−a ab
=
=
)
3 3
( a − b ) ( a 2 + ab + b2 )
=−
3 3
ab
a + ab + b 2 a 2b 2 2
ANSWER: −
a 2 + ab + b 2 a 2b 2
⎛ ab ⎞ ⋅⎜ − ⎟ ⎝ a −b ⎠
53.
PROBLEM: x2 + y 2 +2 xy x2 − y 2 2 xy SOLUTION: x2 + y 2 x2 + y 2 xy x 2 + y 2 2 xy x 2 + y 2 + 2 xy +2 + 2⋅ + xy xy xy xy xy xy = = = 2 2 2 2 2 2 2 x −y x −y x −y x − y2 2 xy 2 xy 2 xy 2 xy
( x + y) =
2
xy
( x + y )( x − y ) 2 xy
( x + y) =
2
xy
=
ANSWER:
2( x + y) x− y 2( x + y) x− y
⋅
2 xy ( x + y )( x − y )
54.
PROBLEM: x 4y +4+ y x x 2y +3+ y x SOLUTION: 4y x x x xy 4 y y x 2 4 xy 4 y 2 +4+ ⋅ + 4⋅ + ⋅ + + y x y x xy x y xy xy xy = = 2 2y x x x xy 2 y y x 3xy 2 y 2 +3+ ⋅ + 3⋅ + ⋅ + + y x y x xy x y xy xy xy
x 2 + 4 xy + 4 y 2 xy = 2 x + 3xy + 2 y 2 xy
( x + 2y) =
2
xy ( x + 2 y )( x + y ) xy
( x + 2y) = xy
=
x + 2y x+ y
ANSWER:
x + 2y x+ y
2
⋅
xy ( x + 2 y )( x + y )
55.
PROBLEM: 1 1+ 1 1+ 2 SOLUTION: 1 1 1 2 2 1+ = 1+ = 1 + = 1 + 1⋅ = 1 + 1 2 1 3 3 3 1+ 1⋅ + 2 2 2 2 2 3 2 5 = 1 + = 1⋅ + = 3 3 3 3 ANSWER:
56.
5 3
PROBLEM: 1 2− 1 1+ 3 SOLUTION: 1 1 1 3 4 3 = 2− = 2 − = 2 − 1⋅ = 2 ⋅ − 2− 1 3 1 4 4 4 4 1+ 1⋅ + 3 3 3 3 8−3 5 = = 4 4 ANSWER:
5 4
57.
PROBLEM: 1 1 1+ 1+ x SOLUTION: 1 1 1 = = 1 1+ x 1 1+ x 1 + + 1+ 1⋅ 1+ x 1+ x 1+ x 1+ x 1+ x 1 = 2+ x 1+ x x +1 = 1⋅ x+2 x +1 = x+2 ANSWER:
58.
x +1 x+2
PROBLEM: x +1 x 1 1− x +1 SOLUTION: x +1 x +1 x +1 x +1 x x x = = = x x +1 x +1−1 x 1 1 − 1− 1⋅ x +1 x +1 x +1 x +1 x +1 x +1 x +1 ⎛ x +1⎞ = ⋅ =⎜ ⎟ x x ⎝ x ⎠
⎛ x +1 ⎞ ANSWER: ⎜ ⎟ ⎝ x ⎠
2
2
59.
PROBLEM: 1 1− x 1 x− x SOLUTION: 1 x 1 x −1 x −1 1− 1⋅ − x = x x = x = x 1 x 1 x 2 − 1 ( x − 1)( x + 1) x− x⋅ − x x x x x x −1 x = ⋅ x ( x − 1)( x + 1)
= 60.
1 x +1
ANSWER:
1 x +1
PROBLEM: 1 −x x x −1 x2 SOLUTION: 1 1 x 1 x 2 1 − x 2 − ( x − 1)( x + 1) −x − x⋅ − x x = x x = x = x = x x −1 x −1 x −1 x −1 x −1 x2 x2 x2 x2 x2 − ( x − 1)( x + 1) x 2 = ⋅ x x −1 = − x ( x + 1) ANSWER: − x ( x + 1)
Part B: Discussion Board Topics
61.
PROBLEM: Choose a problem from this exercise set and clearly work it out on paper, explaining each step in words. Scan your page and post it on the discussion board. ANSWER: (Students answers may vary.) 1 1 + x y 1 1 − y2 x2 1 1 1 y 1 x y + ⋅ + ⋅ + x y x y y x xy = = 2 2 2 1 1 x y x 1 1 − ⋅ 2− 2⋅ 2 − 2 y 2 x2 y x x y x2 y 2 x+ y xy = ( x + y )( x − y ) x2 y 2
62.
=
( x + y) ⋅
=
xy x− y
xy
x x+ y xy xy = 2 2 y x − y2 x2 y 2 x2 y2
x2 y2 ( x + y )( x − y )
PROBLEM: Explain why we need to simplify the numerator and denominator to a single algebraic fraction before multiplying by the reciprocal of the divisor. ANSWER: The goal of the simplification is to obtain a single algebraic fraction divided by another single algebraic fraction.
63.
PROBLEM: Two methods for simplifying complex rational expressions have been presented in this section. Which of the two methods do you feel is more efficient, and why? ANSWER: (Students answers may vary.) Simplifying complex rational expressions using division is more efficient, as it involves less number of steps compared to the simplifying method using LCD.
7.5 Solving Rational Equations Part A: Rational Equations Solve.
1.
PROBLEM: 1 1 1 + = 2 x 8 SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, 2x. 1 1 1 + = 2 x 8 1 ⎛1 1 ⎞ ⎜ + ⎟ 2x = ⋅ 2x 8 ⎝2 x ⎠ x x+2= 4 4x + 8 = x
4x − x + 8 = x − x 3x + 8 = 0 3x + 8 − 8 = 0 − 8 3x = −8 8 x=− 3 8 We can check our answer by substituting − in for x to see if we obtain a true 3 statement. 1 1 1 + = 2 x 8 1 1 1 + = 2 −8 8 3 1 3 1 − = 2 8 8 4−3 1 = 8 8 1 1 = 8 8 8 ANSWER: The solution is − . 3
2.
PROBLEM: 1 1 2 − = 3 x 9 SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, 3x. 1 1 2 − = 3 x 9 2 ⎛1 1⎞ ⎜ − ⎟ 3x = ⋅ 3x 9 ⎝3 x⎠ 2x x −3 = 3 3x − 9 = 2 x
3x − 2 x − 9 = 2 x − x x −9 = 0 x −9+9 = 0+9 x=9 We can check our answer by substituting 9 in for x to see if we obtain a true statement. 1 1 2 − = 3 x 9 1 1 2 − = 3 9 9 3 −1 2 = 9 9 2 2 = 9 9 ANSWER: The solution is 9 .
3.
PROBLEM: 1 2 1 − = 3x 3 x SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, 3x. ⎛ 1 2⎞ ⎛1⎞ ⎜ − ⎟ 3x = ⎜ ⎟ 3x ⎝ 3x 3 ⎠ ⎝ x⎠ 1 − 2x = 3
1 −1 − 2x = 3 −1 −2 x = 2 x = −1 We can check our answer by substituting −1 in for x to see if we obtain a true statement. 1 2 1 − = 3x 3 x 1 2 1 − = 3 ( −1) 3 ( −1)
−3 = −1 3 −1 = −1 ANSWER: The solution is −1 .
4.
PROBLEM: 2 1 3 − = 5 x x 10 SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, 3x. 3 ⎛ 2 1⎞ ⎜ − ⎟ 5x = ⋅ 5x 10 ⎝ 5x x ⎠ 15 x 2−5 = 10 15 x −3 = 10 3 ⋅10 x=− 15 x = −2 We can check our answer by substituting −2 in for x to see if we obtain a true statement. 2 1 3 − = 5 x x 10 2 1 3 − = 5 ( −2 ) ( −2 ) 10
1 1 3 − + = 5 2 10 5−2 3 = 10 10 3 3 = 10 10 ANSWER: The solution is −2 .
5.
PROBLEM: 1 =5 2x +1
1 SOLUTION: We first note that x ≠ − . 2 Multiply both sides by the LCD, 2 x + 1 . 1 ⋅ 2 x + 1 = 5 ( 2 x + 1) 2x +1 1 = 10 x + 5
1 − 5 = 10 x + 5 − 5 10 x = −4 4 x=− 10 2 x=− 5 We can check our answer by substituting − statement. 1 =5 2x +1 1 =5 ⎛ 2⎞ 2 ⎜ − ⎟ +1 ⎝ 5⎠ 1 =5 4 − +1 5 1 =5 −4 + 5 5 5=5 2 ANSWER: The solution is − . 5
2 in for x to see if we obtain a true 5
6.
PROBLEM: 3 +4=5 3x − 1 SOLUTION: 3 +4=5 3x − 1 3 + 4−4 = 5−4 3x − 1 3 =1 3x − 1 Multiply both sides by the LCD, 3 x − 1 . 3 ⋅ 3x − 1 = 1( 3x − 1) 3x − 1 3 = 3x − 1
3 + 1 = 3x − 1 + 1 4 = 3x x=
4 3
We can check our answer by substituting statement. 3 +4=5 3x − 1 3 +4=5 4 3 ⋅ −1 3 3 +4=5 3 1+ 4 = 5
5=5 4 ANSWER: The solution is . 3
4 in for x to see if we obtain a true 3
7.
PROBLEM: 2x − 3 2 = x+5 x+5 SOLUTION: We first note that x ≠ −5 . Multiply both sides by the LCD, x + 5 . 2x − 3 2 ⋅x+5 = ⋅x+5 x+5 x+5 2x − 3 = 2
2x − 3 + 3 = 2 + 3 2x = 5 5 x= 2 We can check our answer by substituting statement. 2x − 3 2 = x+5 x+5 ⎛5⎞ 2⎜ ⎟ − 3 2 ⎝2⎠ = 5 5 +5 +5 2 2 5−3 2 = 15 15 2 2 2 2 = 15 15 2 2 4 4 = 15 15 5 ANSWER: The solution is . 2
5 in for x to see if we obtain a true 2
8.
PROBLEM: 5x x −1 = 2x −1 2x −1
1 . 2 Multiply both sides by the LCD, 2 x − 1 . x −1 5x = 2x −1 2x −1 x −1 5x ⋅ 2x −1 = ⋅ 2x −1 2x −1 2x −1 5x = x − 1 SOLUTION: We first note that x ≠
5x − x = x − x −1 4 x = −1 1 x=− 4 We can check our answer by substituting − statement. 5x x −1 = 2x −1 2x −1 ⎛ 1⎞ ⎛ 1⎞ 5⎜ − ⎟ ⎜ − ⎟ −1 ⎝ 4⎠ = ⎝ 4⎠ ⎛ 1⎞ ⎛ 1⎞ 2 ⎜ − ⎟ −1 2 ⎜ − ⎟ −1 ⎝ 4⎠ ⎝ 4⎠ 5 5 − − 4= 4 6 6 − − 4 4 5 5 = 6 6 1 ANSWER: The solution is − . 4
1 in for x to see if we obtain a true 4
9.
PROBLEM: 5 6 = x −7 x−9 SOLUTION: We first note that x ≠ 7 and x ≠ 9 . Multiply both sides by the LCD, ( x − 7 )( x − 9 ) .
5 6 ⋅ ( x − 7 )( x − 9 ) = ⋅ ( x − 7 )( x − 9 ) x−7 x −9 5 x − 45 = 6 x − 42 5 x − 6 x − 45 = 6 x − 6 x − 42 − x − 45 = −42 − x − 45 + 45 = −42 + 45 −x = 3 x = −3 We can check our answer by substituting −3 in for x to see if we obtain a true statement. 5 6 = x −7 x −9 5 6 = −3 − 7 −3 − 9 5 6 = −10 −14 5 ⋅ ( −14 ) = 6 ⋅ ( −10 )
−60 = −60 ANSWER: The solution is −3 .
10.
PROBLEM: 5 3 = x + 5 x +1 SOLUTION: We first note that x ≠ −5 and x ≠ −1 . Multiply both sides by the LCD, ( x + 5 )( x + 1) .
5 3 ⋅ ( x + 5 )( x + 1) = ⋅ ( x + 5)( x + 1) x+5 x +1 5 x + 5 = 3x + 15 5 x − 3x + 5 = 3x − 3x + 15 2 x + 5 = 15 2 x + 5 − 5 = 15 − 5 2 x = 10 x=5 We can check our answer by substituting 5 in for x to see if we obtain a true statement. 5 3 = x + 5 x +1 5 3 = 5 + 5 5 +1 5 3 = 10 6 1 1 = 2 2 ANSWER: The solution is 5 .
11.
PROBLEM: x 6 − =0 6 x SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, 6x . x 6 ⋅ 6x − ⋅ 6x = 0 x 6 2 x − 36 = 0
( x − 6 )( x + 6 ) = 0 x = 6 or x = − 6 Check x = 6 6 6 − =0 6 6 1 −1 = 0 0=0 ANSWER: The solution is −6 or 6 .
Check x = −6 −6 6 − =0 6 −6 −1 + 1 = 0 0=0
12.
PROBLEM: 5 x + = −2 x 5 SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, 5x . ⎛5 x⎞ ⎜ + ⎟ 5 x = −2 ⋅ 5 x ⎝ x 5⎠ 25 + x 2 = −10 x
x 2 + 10 x + 25 = −10 x + 10 x x 2 + 10 x + 25 = 0
( x + 5)
2
=0
x+5 = 0 x = −5 We can check our answer by substituting −5 in for x to see if we obtain a true statement. 5 x + = −2 x 5 5 −5 + = −2 −5 5 −1 − 1 = −2
−2 = −2 ANSWER: The solution is −5 .
13.
PROBLEM: x 2 = x + 12 x SOLUTION: We first note that x ≠ 0 and x ≠ 12 . Multiply both sides by the LCD, ( x + 12 ) x .
2 ⎛ x ⎞ ⎜ ⎟ ( x + 12 ) x = ⋅ ( x + 12 ) x x ⎝ x + 12 ⎠ x 2 = 2 x + 24 x 2 − 2 x − 24 = 2 x − 2 x + 24 − 24 x 2 − 2 x − 24 = 0
( x − 6 )( x + 4 ) = 0 x = 6 or x = −4 Check x = 6 6 2 = 6 + 12 6 6 2 = 18 6 1 1 = 3 3 ANSWER: The solution is −4 or 6 .
Check x = −4 −4 2 = −4 + 12 −4 4 2 − =− 8 4 1 1 − =− 2 2
14.
PROBLEM: 2x 1 = x+5 6− x SOLUTION: We first note that x ≠ −5 and x ≠ −6 . Multiply both sides by the LCD, ( x + 5 )( 6 − x ) .
⎛ 2x ⎞ ⎛ 1 ⎞ ⎜ ⎟ ( x + 5 )( 6 − x ) = ⎜ ⎟ ( x + 5)( 6 − x ) ⎝ x+5⎠ ⎝ 6− x ⎠ 12 x − 2 x 2 = x + 5 −2 x 2 + 12 x − x − 5 = x + 5 − x − 5 −2 x 2 + 11x − 5 = 0 − ( 2 x − 1)( x − 5) = 0
( 2 x − 1)( x − 5) = 0 x=
1 and x = 5 2 Check x =
1 2
⎛1⎞ 2⎜ ⎟ ⎝2⎠ = 1 1 1 +5 6− 2 2 1 1 = 11 11 2 2 2 2 = 11 11 1 ANSWER: The solution is or 5 . 2
Check x = 5 2 ( 5) 1 = 5+5 6−5 10 1 = 10 1 1 1 = 1 1
15.
PROBLEM: 1 x + =0 x 2x + 1 SOLUTION: We first note that x ≠ 0 and x ≠ −
Multiply both sides by the LCD, x ( 2 x + 1) .
1 . 2
x 1 ⋅ x ( 2 x + 1) + ⋅ x ( 2 x + 1) = 0 x 2x +1 2 x + 1 + x2 = 0 x2 + 2 x + 1 = 0
( x + 1)
2
=0
x +1 = 0 x = −1 We can check our answer by substituting −1 in for x to see if we obtain a true statement. 1 −1 + =0 −1 2 ( −1) + 1
−1 + 1 = 0 0=0 ANSWER: The solution is −1 .
16.
PROBLEM: 9x 4 − =0 3x − 1 x
1 SOLUTION: We first note that x ≠ and x ≠ 0 . 3 Multiply both sides by the LCD, ( 3 x − 1) x .
9x 4 ⋅ ( 3x − 1) x − ⋅ ( 3 x − 1) x = 0 3x − 1 x 2 9 x − 12 x + 4 = 0
( 3x − 2 )
2
=0
3x − 2 = 0 x=
2 3
We can check our answer by substituting statement. ⎛2⎞ 9⎜ ⎟ ⎝3⎠ − 4 =0 2 ⎛2⎞ 3⎜ ⎟ −1 3 ⎝3⎠ 6−6 = 0
0=0 2 ANSWER: The solution is . 3
2 in for x to see if we obtain a true 3
17.
PROBLEM: 2 48 1− = 2 x x SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, x 2 . ⎛ 2 ⎞ 2 48 2 ⎜1 − ⎟ ⋅ x = 2 ⋅ x x ⎝ x⎠
x 2 − 2 x = 48 x 2 − 2 x − 48 = 0
( x − 8)( x + 6 ) = 0 x = 8 or x = −6 Check x = 8 2 48 1− = 2 8 8 6 48 = 8 64 3 3 = 4 4 ANSWER: The solution is 8 or − 6 .
Check x = −6 −4 2 = −4 + 12 −4 4 2 − =− 8 4 1 1 − =− 2 2
18.
PROBLEM: 9 5 2− = 2 x x SOLUTION: We first note that x ≠ 0 . Multiply both sides by the LCD, x 2 . 9⎞ 2 5 2 ⎛ ⎜2 − ⎟⋅ x = 2 ⋅ x x⎠ x ⎝
2 x2 − 9 x = 5 2 x2 − 9 x − 5 = 0
( 2 x + 1)( x − 5) = 0 x=−
1 or x = 5 2 Check x = −
Check x = 5 9 5 2− = 2 5 5 1 1 = 5 5
1 2 5
9 = 1 ⎛ 1 ⎞2 − 2 ⎜⎝ − 2 ⎟⎠ 2 + 18 = 5 ⋅ 4 20 = 20 2−
ANSWER: The solution is −
1 or 5 . 2
19.
PROBLEM: 12 12 1+ = x x−2 SOLUTION: We first note that x ≠ 0 and x ≠ 2 . Multiply both sides by the LCD, x ( x − 2 ) .
⎛ 12 ⎞ ⎛ 12 ⎞ ⎜1 + ⎟ ⋅ x ( x − 2 ) = ⎜ ⎟ ⋅ x ( x − 2) x⎠ ⎝ ⎝ x−2⎠ x 2 − 2 x + 12 x − 24 = 12 x x 2 − 2 x − 24 = 0
( x − 6 )( x + 4 ) = 0 x = 6 or x = −4 Check x = 6 12 12 1+ = 6 6−2 3=3
ANSWER: The solution is 6 or − 4 .
Check x = −4 12 12 1+ = −4 − 4 − 2 2 − = −2
20.
PROBLEM: 3x − 5 1 1− =− x(3x − 4) x SOLUTION: We first note that x ≠ 0 and x ≠
Multiply both sides by the LCD, x ( 3 x − 4 ) .
1⋅ x(3x − 4) −
4 . 3
3x − 5 1 ⋅ x(3x − 4) = − ⋅ x(3x − 4) x(3x − 4) x
3x 2 − 4 x − 3 x + 5 = −3x + 4 3x 2 − 4 x + 1 = 0
( 3x − 1)( x − 1) = 0 x=
1 or x = 1 3 Check x =
1 3
⎛1⎞ 3⎜ ⎟ − 5 1 ⎝3⎠ 1− =− ⎛1⎞ ⎛ 1 ⎞⎛ ⎛ 1 ⎞ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 3⎜ ⎟ − 4 ⎟ ⎝3⎠ ⎝ 3 ⎠⎝ ⎝ 3 ⎠ ⎠ −4 1− = −3 −1 −3 = −3 ANSWER: The solution is
1 or 1 . 3
Check x = 1 3 (1) − 5 1 1− =− (1) (3 (1) − 4) 1 −2 = −1 −1 − 1 = −1
1−
21.
PROBLEM: x 14 = 2 x+3 SOLUTION: We first note that x ≠ −3 . Multiply both sides by the LCD, 2 ( x + 3) .
14 x ⋅ 2 ( x + 3) 2 ( x + 3) ) = ( x+3 2 x 2 + 3x = 28 x 2 + 3x − 28 = 0
( x + 7 )( x − 4 ) = 0 x = −7 or x = 4 Check x = −7 7 14 − = 2 −7 + 3 7 14 − =− 2 4 7 7 − =− 2 2 ANSWER: The solution is −7 or 4 .
Check x = 4 4 14 = 2 4+3 14 2= 7 2=2
22.
PROBLEM: 3x x + 1 = 2 3− x SOLUTION: We first note that x ≠ 3 . Multiply both sides by the LCD, 2 ( 3 − x ) .
x +1 3x ⋅ 2 (3 − x ) = ⋅ 2 (3 − x ) 2 3− x 9 x − 3x 2 = 2 x + 2 3x 2 − 7 x + 2 = 0
( 3x − 1)( x − 2 ) = 0 x=
1 or x = 2 3 Check x =
Check x = 2 3⋅ 2 2 +1 = 2 3− 2 3=3
1 3
⎛1⎞ 3⎜ ⎟ 1 +1 ⎝3⎠ = 3 1 2 3− 3 4 1 3 = 2 8 3 1 1 = 2 2
ANSWER: The solution is
1 or 2 . 3
23.
PROBLEM: −3x + 3 6= x −1 SOLUTION: We first note that x ≠ 1 . Multiply both sides by the LCD, x − 1 . −3x + 3 6 ( x − 1) = ⋅ x −1 x −1 6 x − 6 = −3x + 3
9x = 9 x =1 x ≠ 1. ANSWER: Hence, the solution is ∅.
24.
PROBLEM: 12 6(4 − x) = 2+ x−2 x−2 SOLUTION: We first note that x ≠ 2 . Multiply both sides by the LCD, x − 2 . 12 6(4 − x) ⎞ ⎛ ⋅x −2 = ⎜2+ ⎟ ( x − 2) x−2 x−2 ⎠ ⎝ 12 = 2 x − 4 + 24 − 6 x
4x = 8 x=2 x ≠ 2. ANSWER: Hence, the solution is ∅.
25.
PROBLEM: 2x 3( x − 1) = 2+ x −3 x −3 SOLUTION: We first note that x ≠ 3 . Multiply both sides by the LCD, x − 3 . 2x ⎞ 3( x − 1) ⎛ ⋅x −3 ⎜2+ ⎟ ( x − 3) = x −3⎠ x −3 ⎝ 2 x − 6 + 2 x = 3x − 3
x=3 x ≠ 3. ANSWER: Hence, the solution is ∅.
26.
PROBLEM: x 1 x + = x − 1 6 x − 1 ( x − 1)( 6 x − 1)
1 . 6 Multiply both sides by the LCD, ( x − 1)( 6 x − 1) . SOLUTION: We first note that x ≠ 1 and x ≠
1 ⎞ x ⎛ x + ⋅ ( x − 1)( 6 x − 1) ⎜ ⎟ ( x − 1)( 6 x − 1) = ( x − 1)( 6 x − 1) ⎝ x −1 6x −1 ⎠ 6 x2 − x + x − 1 = x 6 x2 − x − 1 = 0
( 3x + 1)( 2 x − 2 ) = 0 x=−
1 1 or x = 3 2 Check x = −
1 3
Check x =
1 2
1 1 1 ⎛ 1⎞ ⎛ 1⎞ + = − ⎟ ⎜− ⎟ ⎜ 1 1 ⎛1⎞ ⎞ 1 ⎛ 1 ⎝ 3⎠ + ⎝ 3⎠ 1− 6 ⎜ ⎟ − 1 ⎛⎜ 1 − ⎞⎟ ⎜ 6 ⎛⎜ ⎞⎟ − 1⎟ = 2 ⎝2⎠ ⎛ 1⎞ ⎛ 1⎞ ⎛⎛ 1 ⎞ ⎞⎛ ⎛ 1 ⎞ ⎞ ⎝ 2 ⎠⎝ ⎝ 2 ⎠ ⎠ ⎜ − ⎟ − 1 6 ⋅ ⎜ − ⎟ − 1 ⎜ ⎜ − ⎟ − 1⎟ ⎜ 6 ⋅ ⎜ − ⎟ − 1⎟ ⎝ 3⎠ ⎝ 3⎠ 1 1 ⎝⎝ 3 ⎠ ⎠⎝ ⎝ 3 ⎠ ⎠ 2+ = 2 1 1 1 − − 2 3 −1 = 3 4 3 4 2=2 − − ⋅ ( −3) 3 3 1 1 1 − =− 4 3 3⋅ 4 1 1 − =− 12 12
ANSWER: The solution is −
1 1 or . 3 2
27.
PROBLEM: 12 1 2 = − 2 x − 81 x + 9 x − 9 SOLUTION: 12 1 2 = − 2 x − 81 x + 9 x − 9 12 1 2 = − ( x + 9 )( x − 9 ) x + 9 x − 9
We first note that x ≠ ±9 . Multiply both sides by the LCD, ( x + 9 )( x − 9 ) .
12 2 ⎞ ⎛ 1 ⋅ ( x + 9 )( x − 9 ) = ⎜ − ⎟ ( x + 9 )( x − 9 ) ( x + 9 )( x − 9 ) ⎝ x +9 x −9 ⎠ 12 = x − 9 − 2 x − 18 x = −39 We can check our answer by substituting −39 in for x to see if we obtain a true statement. 12 1 2 = − 2 ( −39 ) − 81 −39 + 9 −39 − 9 12 1 2 =− + 1440 30 48 12 −8 + 10 = 1440 240 12 2 = 1440 240 1 1 = 120 120 ANSWER: The solution is −39 .
28.
PROBLEM: 14 2 3 = − 2 x − 49 x − 7 x + 7 SOLUTION: 14 2 3 = − 2 x − 49 x − 7 x + 7 14 2 3 = − ( x − 7 )( x + 7 ) x − 7 x + 7
1 . 6 Multiply both sides by the LCD, ( x − 7 )( x + 7 ) .
We first note that x ≠ ±7 and x ≠
14 3 ⎞ ⎛ 2 ⋅ ( x − 7 )( x + 7 ) = ⎜ − ⎟ ( x − 7 )( x + 7 ) ( x − 7 )( x + 7 ) ⎝ x−7 x+7⎠ 14 = 2 x + 14 − 3x + 21 x = 21 We can check our answer by substituting 21 in for x to see if we obtain a true statement. 14 2 3 = − 2 21 − 49 21 − 7 21 + 7 14 2 3 = − 392 14 28 1 1 = 28 28 ANSWER: The solution is 21.
29.
PROBLEM: 6x 4 3x + = 2 x +3 x −3 x −9 SOLUTION: 6x 4 3x + = 2 x +3 x −3 x −9 6x 4 3x + = x + 3 x − 3 ( x − 3)( x + 3)
We first note that x ≠ ±3 . Multiply both sides by the LCD, ( x + 3)( x − 3) .
4 ⎞ 3x ⎛ 6x + ⋅ ( x − 3)( x + 3) ⎜ ⎟ ( x − 3)( x + 3) = ( x − 3)( x + 3) ⎝ x +3 x −3⎠ 6 x 2 − 18 x + 4 x + 12 = 3 x 6 x 2 − 17 x + 12 = 0
( 3x − 4 )( 2 x − 3) = 0 x=
4 3 or x = 3 2 Check x =
4 3
Check x =
3 2
6⋅
4 4 3⋅ 3 + 4 = 3 2 4 4 +3 −3 ⎛ 4 ⎞ −9 ⎜ ⎟ 3 3 ⎝3⎠
3 3 3⋅ 2 + 4 = 2 2 3 3 +3 −3 ⎛ 3 ⎞ −9 ⎜ ⎟ 2 2 ⎝2⎠
24 12 4⋅9 − =− 13 5 65 120 − 12 ⋅13 4⋅9 =− 65 65 36 36 − =− 65 65
9 8 2− = 2 3 − 27 4 2 2 − =− 3 3
ANSWER: The solution is
4 3 or . 3 2
6⋅
30.
PROBLEM: 3x 17 48 − =− 2 x+2 x−2 x −4 SOLUTION: 3x 17 48 − =− 2 x+2 x−2 x −4 3x 17 48 − =− x+2 x−2 ( x + 2 )( x − 2 )
We first note that x ≠ ±2 . Multiply both sides by the LCD, ( x + 2 )( x − 2 ) .
17 ⎞ 48 ⎛ 3x − ⋅ ( x + 2 )( x − 2 ) ⎜ ⎟ ( x + 2 )( x − 2 ) = − ( x + 2 )( x − 2 ) ⎝ x+2 x−2⎠ 3x 2 − 6 x − 17 x − 34 = −48 3x 2 − 23x + 14 = 0
Check x = 7 21 17 −48 − = 9 5 45 48 48 =− =− 45 45
Check x =
3 2 − 17 = −48 2 3 3 +2 −3 ⎛ 3 ⎞ −4 ⎜ ⎟ 2 2 ⎝2⎠ 3
=
ANSWER: The solution is 7 or
3 2
3 . 2
192 192 = 7 7
31.
PROBLEM: x −1 + 3 = 0 SOLUTION: x −1 + 3 = 0
1 +3= 0 x We first note that x ≠ 0 . Multiply both sides by the LCD, x . ⎛1 ⎞ ⎜ + 3⎟ x = 0 ⋅ x ⎝x ⎠ 1 + 3x = 0
x=−
1 3
We can check our answer by substituting − statement. 1 +3= 0 1 − 3 −3 + 3 = 0
0=0 1 ANSWER: The solution is − . 3
1 in for x to see if we obtain a true 3
32.
PROBLEM: 4 − y −1 = 0 SOLUTION: 4 − y −1 = 0
1 =0 y We first note that y ≠ 0 . Multiply both sides by the LCD, y . 4−
⎛ 1⎞ ⎜4 − ⎟ y = 0⋅ y y⎠ ⎝ 4 y −1 = 0 1 y= 4 We can check our answer by substituting statement. 1 4− = 0 1 4 4−4 = 0
0=0 1 ANSWER: The solution is . 4
1 in for x to see if we obtain a true 4
33.
PROBLEM: y −2 − 4 = 0 SOLUTION: y −2 − 4 = 0
1 −4 =0 y2 We first note that y ≠ 0 . Multiply both sides by the LCD, y 2 .
⎛ 1 ⎞ 2 2 ⎜ 2 − 4⎟ y = 0⋅ y y ⎝ ⎠ 2 1− 4 y = 0
(1 − 2 y )(1 + 2 y ) = 0 y=
1 1 or y = − 2 2 Check y = 1 2
1 2
−4 = 0
⎛1⎞ ⎜ ⎟ ⎝2⎠ 4−4 = 0 0=0
1 ANSWER: The solution is ± . 2
Check y = − 1
1 2
−4=0 2 ⎛ 1⎞ ⎜− ⎟ ⎝ 2⎠ 4−4 = 0 0=0
34.
PROBLEM: 9 x −2 − 1 = 0 SOLUTION: 9 x −2 − 1 = 0
9 −1 = 0 x2 We first note that x ≠ 0 . Multiply both sides by the LCD, x 2 . ⎛ 9 ⎞ 2 2 ⎜ 2 − 1⎟ x = 0 ⋅ x ⎝x ⎠ 2 9− x = 0
( 3 − x )( 3 + x ) = 0 x = 3 or x = −3 Check x = 3 9 −1 = 0 32 1 −1 = 0 0=0 ANSWER: The solution is ±3 .
Check x = −3 9 −1 = 0 2 ( −3 ) 1−1 = 0 0=0
35.
PROBLEM: −1 3 ( x − 1) + 5 = 0 SOLUTION: −1 3 ( x − 1) + 5 = 0 3 +5 = 0 x −1 We first note that x ≠ 1 . Multiply both sides by the LCD, x − 1 . ⎛ 3 ⎞ + 5 ⎟ ( x − 1) = 0 ⋅ ( x − 1) ⎜ ⎝ x −1 ⎠ 3 + 5x − 5 = 0
5x − 2 = 0 5x = 2 2 x= 5 We can check our answer by substituting statement. 3 +5 = 0 2 −1 5 3 +5 = 0 3 − 5 −5 + 5 = 0
0=0 2 ANSWER: The solution is . 5
2 in for x to see if we obtain a true 5
36.
PROBLEM: −1 5 − 2 ( 3x + 1) = 0 SOLUTION: −1 5 − 2 ( 3x + 1) = 0 5−
2 =0 3x + 1
1 We first note that x ≠ − . 3 Multiply both sides by the LCD, 3 x + 1 . 2 ⎞ ⎛ ⎜5 − ⎟ ( 3x + 1) = 0 ⋅ ( 3x + 1) ⎝ 3x + 1 ⎠ 15 x + 5 − 2 = 0
x=−
1 5
We can check our answer by substituting − statement. 2 5− =0 ⎛ 1⎞ 3⎜ − ⎟ +1 ⎝ 5⎠ 2 5− = 0 2 5 5−5 = 0
0=0 1 ANSWER: The solution is − . 5
1 in for x to see if we obtain a true 5
37.
PROBLEM: 2 2 3+ = x −3 x −3 SOLUTION: We first note that x ≠ 3 . Multiply both sides by the LCD, x − 3 . 2 ⎞ 2 ⎛ ⋅ ( x − 3) ⎜3+ ⎟ ( x − 3) = x −3⎠ x −3 ⎝ 3x − 9 + 2 = 2
3x = 9 x=3 x ≠ 3. ANSWER: Hence, the solution is ∅.
38.
PROBLEM: 1 1 = x x +1 SOLUTION: We first note that x ≠ 0 and x ≠ 1 . Multiply both sides by the LCD, x ( x + 1) .
1 1 ⋅ x ( x + 1) = ⋅ x ( x + 1) x x +1 x +1 = x 1≠ 0 ANSWER: The solution is ∅.
39.
PROBLEM: x x +1 = x +1 x SOLUTION: We first note that x ≠ 0 and x ≠ −1 . Multiply both sides by the LCD, x ( x + 1) .
⎛ x ⎞ ⎛ x +1 ⎞ ⎜ ⎟ ⋅ x ( x + 1) = ⎜ ⎟ ⋅ x ( x + 1) ⎝ x +1 ⎠ ⎝ x ⎠ x2 = x2 + 2 x + 1 2 x = −1 x=−
1 2
We can check our answer by substituting − statement. 1 1 − − +1 2 = 2 1 1 − +1 − 2 2 1 1 − 2= 2 1 1 − 2 2 −1 = −1 1 ANSWER: The solution is- . 2
1 in for x to see if we obtain a true 2
40.
PROBLEM: 3x − 1 x = 3x x+3 SOLUTION: We first note that x ≠ 0 and x ≠ −3 . Multiply both sides by the LCD, x ( x + 1) .
x ⎛ 3x − 1 ⎞ ⋅ 3 x ( x + 3) ⎜ ⎟ ⋅ 3 x ( x + 3) = x+3 ⎝ 3x ⎠ 3x 2 + 8 x − 3 = 3x 2 8x − 3 = 0 x=
3 8
We can check our answer by substituting statement. ⎛3⎞ 3 3⎜ ⎟ − 1 ⎝8⎠ = 8 3 ⎛ 3⎞ +3 3⎜ ⎟ 8 ⎝8⎠ 1 3 8= 8 9 27 8 8 1 1 = 9 9 3 ANSWER: The solution is . 8
3 in for x to see if we obtain a true 8
41.
PROBLEM: 4 x − 7 3x − 2 = x−5 x −5 SOLUTION: We first note that x ≠ 5 . Multiply both sides by the LCD, x − 5 . 4x − 7 3x − 2 ⋅x−5 = ⋅x−5 x−5 x −5 4 x − 7 = 3x − 2
x=5 x ≠ 5. ANSWER: Hence, the solution is ∅.
42.
PROBLEM: x 1 = 2 x −9 x −3 SOLUTION: x 1 = 2 x −9 x−3 x 1 = ( x − 3)( x + 3) x − 3
We first note that x ≠ ±3 . Multiply both sides by the LCD, ( x − 3)( x + 3) .
x
( x − 3)( x + 3)
⋅ ( x − 3)( x + 3) =
x = x+3 0≠3 ANSWER: The solution is ∅.
1 ⋅ ( x − 3)( x + 3) x−3
43.
PROBLEM: 3x + 4 2 − =1 x −8 8− x SOLUTION: 3x + 4 2 − =1 x −8 8− x 3x + 4 2 + =1 x −8 x −8 We first note that x ≠ 8 . Multiply both sides by the LCD, x − 8 . ⎛ 3x + 4 ⎞ ⎛ 2 ⎞ ⎜ ⎟ ( x − 8) + ⎜ ⎟ ( x − 8) = 1 ⋅ ( x − 8) ⎝ x −8 ⎠ ⎝ x −8⎠ 3x + 4 + 2 = x − 8
2 x = −14 x = −7 We can check our answer by substituting −7 in for x to see if we obtain a true statement. 3 ( −7 ) + 4 2 + =1 −7 − 8 −7 − 8 17 2 − =1 15 15 1=1 ANSWER: The solution is −7 .
44.
PROBLEM: 1 6 = x x ( x + 3) SOLUTION: We first note that x ≠ 0 and x ≠ 3 . Multiply both sides by the LCD, x ( x − 3) .
1 6 ⋅ x ( x + 3) = ⋅ x ( x + 3) x x ( x + 3)
x+3= 6 x=3 We can check our answer by substituting 3 in for x to see if we obtain a true statement. 1 6 = 3 3 ( 3 + 3) 1 6 = 3 18 1 1 = 3 3 ANSWER: The solution is 3 .
45.
PROBLEM: 3 1 13 = + x x + 1 x ( x + 1) SOLUTION: We first note that x ≠ 0 and x ≠ −1 . Multiply both sides by the LCD, x ( x + 1) .
⎛ 1 3 13 ⎞ x x + 1) ⋅ x ( x + 1) = ⎜ + ⎜ x + 1 x ( x + 1) ⎟⎟ ( x ⎝ ⎠ 3x + 3 = x + 13 2 x = 10 x=5 We can check our answer by substituting 5 in for x to see if we obtain a true statement. 3 1 13 = + 5 5 + 1 5 ( 5 + 1) 3 1 13 = + 5 6 30 3 3 = 5 5 ANSWER: The solution is 5 .
46.
PROBLEM: x 3 9x − = x − 1 4 x − 1 ( 4 x − 1)( x − 1)
1 . 4 Multiply both sides by the LCD, ( 4 x − 1)( x − 1) .
SOLUTION: We first note that x ≠ 1 and x ≠
⎛ ⎞ 3 ⎞ 9x ⎛ x x x 4 1 1 4 x − 1)( x − 1) − − − = ( )( ) ⎜ ⎜ ⎟ ⎜ ( 4 x − 1)( x − 1) ⎟⎟ ( ⎝ x − 1 4x − 1 ⎠ ⎝ ⎠ 4 x 2 − x − 3x + 3 = 9 x 4 x 2 − 13x + 3 = 0
( 4 x − 1)( x − 3) = 0 1 or x = 3 4 1 x≠ . 4
x=
ANSWER: Hence, the only solution is 3. We can check our answer by substituting 3 in for x to see if we obtain a true statement. 9 ( 3) 3 3 − = 3 − 1 4 ( 3) − 1 ( 4 ( 3) − 1) ( 3 − 1)
3 3 27 − = 2 11 11 ⋅ 2 27 27 = 22 22
47.
PROBLEM: 1 x 2 + = 2 x − 4 x − 2 x − 6x + 8 SOLUTION: 1 x 2 + = 2 x − 4 x − 2 x − 6x + 8 1 x 2 + = x − 4 x − 2 ( x − 4 )( x − 2 )
We first note that x ≠ 4 and x ≠ 2 . Multiply both sides by the LCD, ( x − 4 )( x − 2 ) .
⎛ ⎞ x ⎞ 2 ⎛ 1 + ⎟⎟ ( x − 4 )( x − 2 ) ⎜ ⎟ ( x − 4 )( x − 2 ) = ⎜⎜ ⎝ x−4 x−2⎠ ⎝ ( x − 4 )( x − 2 ) ⎠ x − 2 + x2 − 4 x = 2 x 2 − 3x − 4 = 0
( x − 4 )( x + 1) = 0 x = 4 or x = −1 x ≠ 4. ANSWER: Hence, the only solution is −1 . We can check our answer by substituting −1 in for x to see if we obtain a true statement. 1 1 2 + = −1 − 4 2 + 1 ( −1 − 4 )( −1 − 2 )
1 1 2 − = 3 5 15 2 2 = 15 15
48.
PROBLEM: x x −1 5 + 2 = x − 5 x − 11x + 30 x − 6 SOLUTION: x x −1 5 + 2 = x − 5 x − 11x + 30 x − 6 x x −1 5 + = x − 5 ( x − 6 )( x − 5) x − 6 We first note that x ≠ 5 and x ≠ 6 . Multiply both sides by the LCD, ( x − 6 )( x − 5 ) .
⎛ x ⎞ x −1 ⎛ 5 ⎞ + ⎜⎜ ⎟⎟ ( x − 6 )( x − 5) = ⎜ ⎟ ( x − 6 )( x − 5) x x x x 5 6 5 6 − − − − ( )( ) ⎝ ⎠ ⎝ ⎠ x 2 − 6 x + x − 1 = 5 x − 25 x 2 − 10 x + 24 = 0
( x − 6 )( x − 4 ) = 0 x = 6 or x = 4 x ≠ 6. ANSWER: Hence, the only solution is 4. We can check our answer by substituting 3 in for x to see if we obtain a true statement. 4 4 −1 5 + = 4 − 5 ( 4 − 6 )( 4 − 5) 4 − 6
3 5 =− 2 2 5 5 − =− 2 2 −4 +
49.
PROBLEM: x 6 5 − 2 =− 5x − 1 x + 1 5x + 4 x − 1 SOLUTION: x 6 5 − 2 =− x + 1 5x + 4x − 1 5x − 1 x 6 5 − =− 5x − 1 x + 1 ( x + 1)( 5 x − 1)
1 and x ≠ −1 . 5 Multiply both sides by the LCD, ( x + 1)( 5 x − 1) .
We first note that x ≠
⎛ x ⎞ 6 5 ⎞ ⎛ − ⎜⎜ ⎟⎟ ( x + 1)( 5 x − 1) = ⎜ − ⎟ ( x + 1)( 5 x − 1) 1 5 1 x x x x + + 1 5 − 1 − ( )( ) ⎝ ⎠ ⎝ ⎠ 2 5 x − x − 6 = −5 x − 5 5x2 + 4 x − 1 = 0
( 5x − 1)( x + 1) = 0 1 or x = −1 5 1 x ≠ and x ≠ −1 . 5
x=
ANSWER: Hence, the solution is ∅.
50.
PROBLEM: −8 2( x + 2) 1 + 2 = 2 x − 4 x − 12 x + 4 x − 60 x + 2 SOLUTION: −8 2( x + 2) 1 + 2 = 2 x − 4 x − 12 x + 4 x − 60 x + 2 −8 2( x + 2) 1 + = ( x − 6 )( x + 2 ) ( x + 10 )( x − 6 ) x + 2
We first note that x ≠ 6, x ≠ −2, and x ≠ −10 . Multiply both sides by the LCD, ( x + 10 )( x − 6 )( x + 2 ) .
⎛ ⎞ −8 2( x + 2) 1 + ⋅ ( x + 10 )( x − 6 )( x + 2 ) ⎜⎜ ⎟⎟ ( x + 10 )( x − 6 )( x + 2 ) = x+2 ⎝ ( x − 6 )( x + 2 ) ( x + 10 )( x − 6 ) ⎠ −8 x − 80 + 2 x 2 + 8 x + 8 = x 2 + 4 x − 60 x 2 − 4 x − 12 = 0
( x − 6 )( x + 2 ) = 0 x = 6 or x = −2 x ≠ 6 and x ≠ −2 .
ANSWER: Hence, the solution is∅.
51.
PROBLEM: x 20 4 − 2 =− x+2 x − x−6 x−3 SOLUTION: x 20 4 − 2 =− x+2 x − x−6 x−3 x 20 4 − =− x + 2 ( x + 2 )( x − 3) x −3 We first note that x ≠ −2 and x ≠ 3 . Multiply both sides by the LCD, ( x + 2 )( x − 3) .
⎛ x ⎞ 20 4 ⎞ ⎛ − ⎜⎜ ⎟⎟ ( x + 2 )( x − 3) = ⎜ − ⎟ ( x + 2 )( x − 3) ⎝ x −3⎠ ⎝ x + 2 ( x + 2 )( x − 3) ⎠ x 2 − 3x − 20 = −4 x − 8 x 2 + x − 12 = 0
( x + 4 )( x − 3) = 0 x = −4 or x = 3 x ≠ 3. ANSWER: Hence, the only solution is –4. We can check our answer by substituting −4 in for x to see if we obtain a true statement. −4 20 4 − =− −4 + 2 ( −4 + 2 )( −4 − 3) −4 − 3
10 4 = 7 7 4 4 = 7 7
2−
52.
PROBLEM: x + 7 x −1 4 + = 2 x −1 x + 1 x −1 SOLUTION: x + 7 x −1 4 + = 2 x −1 x +1 x −1 x + 7 x −1 4 + = x − 1 x + 1 ( x − 1)( x + 1) We first note that x ≠ ±1 . Multiply both sides by the LCD, ( x + 1)( x − 1) .
⎛ ⎞ 4 ⎛ x + 7 x −1⎞ + ⎟⎟ ( x + 1)( x − 1) ⎜ ⎟ ( x + 1)( x − 1) = ⎜⎜ − + 1 1 x x ( )( ) ⎝ x −1 x +1⎠ ⎝ ⎠ x2 + 8x + 7 + x2 − 2 x + 1 = 4 2 x2 + 6 x + 4 = 0 2 ( x + 2 )( x + 1) = 0 x = −2 or x = −1 x ≠ ±1 . ANSWER: Hence, the only solution is –2.
We can check our answer by substituting −2 in for x to see if we obtain a true statement. −2 + 7 −2 − 1 4 + = −2 − 1 −2 + 1 ( −2 − 1)( −2 + 1)
5 4 − +3= 3 3 4 4 = 3 3
53.
PROBLEM: x − 1 x − 3 −x + 5 + = x − 3 x −1 x−3 SOLUTION: We first note that x ≠ 3 and x ≠ 1 . Multiply both sides by the LCD, ( x − 3)( x − 1) .
⎛ x −1 x − 3 ⎞ ⎛ −x + 5 ⎞ + ⎜ ⎟ ( x − 3)( x − 1) = ⎜ ⎟ ( x − 3)( x − 1) ⎝ x − 3 x −1 ⎠ ⎝ x−3 ⎠ x2 − 2 x + 1 + x2 − 6 x + 9 = − x2 + 6 x − 5 3x 2 − 14 x + 15 = 0
( x − 3)( 3x − 5) = 0 x = 3 or x = x ≠ 3.
5 3
5 ANSWER: Hence, the only solution is . 3 We can check our answer by substituting statement. 5 5 5 −1 −3 − +5 3 +3 = 3 5 5 5 −3 −1 −3 3 3 3 2 4 10 − 3 + 3= 3 4 2 4 − − 3 3 3 1 5 − −2= − 2 2 5 5 − =− 2 2
5 in for x to see if we obtain a true 3
54.
PROBLEM: x−2 x−5 8− x − = x−5 x−2 x−5 SOLUTION: We first note that x ≠ 5 and x ≠ 2 . Multiply both sides by the LCD, ( x − 5 )( x − 2 ) .
⎛ x−2 x −5⎞ ⎛8− x⎞ − ⎜ ⎟ ( x − 5)( x − 2 ) = ⎜ ⎟ ( x − 5)( x − 2 ) ⎝ x−5 x−2⎠ ⎝ x−5⎠ x 2 − 4 x + 4 − x 2 + 10 x − 25 = − x 2 + 10 x − 16
x2 − 4 x − 5 = 0
( x − 5)( x + 1) = 0 x = 5 or x = −1 x ≠ 5. ANSWER: Hence, the only solution is −1 .
We can check our answer by substituting −1 in for x to see if we obtain a true statement. −1 − 2 −1 − 5 8 − ( −1) − = −1 − 5 −1 − 2 ( −1) − 5
1 3 −2= − 2 2 3 3 − =− 2 2
55.
PROBLEM: x+7 81 9 − 2 = x − 2 x + 5 x − 14 x + 7 SOLUTION: x+7 81 9 − 2 = x − 2 x + 5 x − 14 x + 7 x+7 81 9 − = x − 2 ( x − 2 )( x + 7 ) x + 7 We first note that x ≠ 2 and x ≠ −7 . Multiply both sides by the LCD, ( x − 2 )( x + 7 ) .
⎛ x+7 ⎞ 81 ⎛ 9 ⎞ − ⎜⎜ ⎟⎟ ( x − 2 )( x + 7 ) = ⎜ ⎟ ( x − 2 )( x + 7 ) x x x x 2 2 7 7 − − + + ( )( ) ⎝ ⎠ ⎝ ⎠ x 2 + 14 x + 49 − 81 = 9 x − 18 x 2 + 5 x − 14 = 0
( x + 7 )( x − 2 ) = 0 x = −7 or x = 2 x ≠ 2 and x ≠ −7 . ANSWER: Hence, the solution is ∅.
56.
PROBLEM: x 5 x + 30 +1 = x−6 36 − x 2 SOLUTION: x 5 x + 30 +1 = x−6 36 − x 2 x 5 x + 30 +1 = − x−6 ( x + 6 )( x − 6 ) We first note that x ≠ ±6 . Multiply both sides by the LCD, ( x − 6 )( x + 6 ) .
⎛ 5 x + 30 ⎞ ⎛ x ⎞ + 1⎟ ( x + 6 )( x − 6 ) = ⎜ − x + 6 )( x − 6 ) ⎜ ⎜ ( x + 6 )( x − 6 ) ⎟⎟ ( ⎝ x−6 ⎠ ⎝ ⎠
x 2 + 6 x + x 2 − 36 = −5 x − 30 2 x 2 + 11x − 6 = 0
( 2 x − 1)( x + 6 ) = 0 1 or x = −6 2 x ≠ ±6 .
x=
1 ANSWER: Hence, the only solution is . 2 We can check our answer by substituting −1 in for x to see if we obtain a true statement. ⎛1⎞ 1 5 ⎜ ⎟ + 30 ⎝2⎠ 2 +1 = − 1 ⎛1 ⎞⎛ 1 ⎞ −6 + 6 ⎟⎜ − 6 ⎟ ⎜ 2 ⎝2 ⎠⎝ 2 ⎠ 1 130 − +1 = 11 143 10 10 = 11 11
57.
PROBLEM: 2x 4 −7 − = 2 x +1 4x − 3 4x + x − 3 SOLUTION: 2x 4 −7 − = 2 x + 1 4x − 3 4x + x − 3 2x 4 −7 − = x + 1 4 x − 3 ( 4 x − 3)( x + 1)
3 . 4 Multiply both sides by the LCD, ( 4 x − 3)( x + 1) .
We first note that x ≠ −1 and x ≠
⎛ ⎞ −7 4 ⎞ ⎛ 2x x x − − + = 4 3 1 ( )( ) ⎜ ⎟⎟ ( 4 x − 3)( x + 1) ⎜ ⎟ ⎜ ⎝ x +1 4x − 3 ⎠ ⎝ ( 4 x − 3)( x + 1) ⎠ 8 x 2 − 6 x − 4 x − 4 = −7 8 x 2 − 10 x + 3 = 0
( 4 x − 3)( 2 x − 1) = 0 3 1 or x = 4 2 3 x≠ . 4
x=
1 . 2 1 We can check our answer by substituting in for x to see if we obtain a true 2 statement. ⎛1⎞ 2⎜ ⎟ 4 −7 ⎝2⎠ − = 1 1 + 1 4 ⎜⎛ ⎟⎞ − 3 ⎜⎛ 4 ⎜⎛ 1 ⎟⎞ − 3 ⎟⎞ ⎜⎛ 1 + 1⎟⎞ 2 ⎝2⎠ ⎝ ⎝ 2 ⎠ ⎠⎝ 2 ⎠
ANSWER: Hence, the only solution is
2 14 +4= 3 3 14 14 = 3 3
58.
PROBLEM: x −5 5 5x + =− 2 x − 10 x − 5 x − 15 x + 50 SOLUTION: x−5 5 5x + =− 2 x − 10 x − 5 x − 15 x + 50 x−5 5 5x + =− x − 10 x − 5 ( x − 10 )( x − 5) We first note that x ≠ 10 and x ≠ 5 . Multiply both sides by the LCD, ( x − 10 )( x − 5 ) .
⎛ ⎞ 5 ⎞ 5x ⎛ x−5 + ⎟⎟ ( x − 10 )( x − 5) ⎜ ⎟ ( x − 10 )( x − 5 ) = ⎜⎜ − ⎝ x − 10 x − 5 ⎠ ⎝ ( x − 10 )( x − 5) ⎠ x 2 − 10 x + 25 + 5 x − 50 = −5 x x 2 − 25 = 0
( x − 5)( x + 5) = 0 x = ±5 x ≠ 5. ANSWER: Hence, the only solution is −5 . We can check our answer by substituting −5 in for x to see if we obtain a true statement. 5 ( −5) −5 − 5 5 + =− −5 − 10 −5 − 5 ( −5 − 10)( −5 − 5)
2 1 1 − = 3 2 6 1 1 = 6 6
59.
PROBLEM: 5 x +1 5 + 2 = 2 2 x + 5 x + 4 x + 3x − 4 x − 1 SOLUTION: 5 x +1 5 + 2 = 2 2 x + 5 x + 4 x + 3x − 4 x − 1 5 x +1 5 + = ( x + 1)( x + 4 ) ( x + 4 )( x − 1) ( x − 1)( x + 1) We first note that x ≠ ±1 and x ≠ −4 . Multiply both sides by the LCD, ( x − 1)( x + 1)( x + 4 ) .
⎛ ⎞ ⎛ ⎞ x +1 5 5 + ⎜⎜ ⎟⎟ ( x − 1)( x + 1)( x + 4 ) = ⎜⎜ ⎟⎟ ( x − 1)( x + 1)( x + 4 ) x x x x x x 1 4 4 1 1 1 + + + − − + ( )( ) ( )( ) ( )( ) ⎝ ⎠ ⎝ ⎠ 2 5 x − 5 + x + 2 x + 1 = 5 x + 20 x 2 + 2 x − 24 = 0
( x + 6 )( x − 4 ) = 0 x = −6 or x = 4 Check x = −6 5
( −6 + 1)( −6 + 4 )
+
Check x = 4
5 4 +1 5 5 −6 + 1 + = = ( −6 + 4 )( −6 − 1) ( −6 − 1)( − ( 4 + 1)( 4 + 4 ) ( 4 + 4 )( 4 − 1) ( 4 − 1)( 4
1 5 1 − = 2 14 7 2 1 = 14 7 1 1 = 7 7
ANSWER: The solution is −6 or 4 .
1 5 1 + = 8 24 3 8 1 = 24 3 1 1 = 3 3
60.
PROBLEM: 1 x −9 1 + 2 = 2 2 x − 2 x − 63 x + 10 x + 21 x − 6 x − 27 SOLUTION: x −9 1 1 + 2 = 2 2 x − 2 x − 63 x + 10 x + 21 x − 6 x − 27 x −9 1 1 + = ( x + 7 )( x − 9 ) ( x + 3)( x + 7 ) ( x − 9 )( x + 3)
We first note that x ≠ −7, x ≠ 9, and x ≠ −3 . Multiply both sides by the LCD, ( x − 9 )( x + 3)( x + 7 ) .
⎛ ⎞ ⎛ ⎞ x −9 1 1 + ⎜⎜ ⎟⎟ ⋅ ( x − 9 )( x + 3)( x + 7 ) = ⎜⎜ ⎟⎟ ( x − 9 )( x + 3)( x + 7 ) ⎝ ( x + 7 )( x − 9 ) ( x + 3)( x + 7 ) ⎠ ⎝ ( x − 9 )( x + 3) ⎠ x + 3 + x 2 − 18 x + 81 = x + 7 x 2 − 18 x + 77 = 0
( x − 11)( x − 7 ) = 0 x = 11 or x = 7 Check x = 11 1
(11 + 7 )(11 − 9 )
+
11 − 9 1 = (11 + 3)(11 + 7 ) (11 − 9 )(11
1 1 1 + = 36 126 28 1 1 = 28 28
ANSWER: The solution is 11 or 7 .
Check x = 7 1
( 7 + 7 )( 7 − 9 )
+
7−9 1 = ( 7 + 3)( 7 + 7 ) ( 7 − 9 )( 7 +
1 1 1 − =− 28 70 20 1 1 − =− 20 20 −
61.
PROBLEM: 2 ( x − 2) x+2 4 + 2 = 2 2 x − 4 x − 4 x − 12 x − 8 x + 12 SOLUTION: 2 ( x − 2) x+2 4 + 2 = 2 2 x − 4 x − 4 x − 12 x − 8 x + 12 2 ( x − 2) x+2 4 + = ( x + 2 )( x − 2 ) ( x − 6 )( x + 2 ) ( x − 6 )( x − 2 ) We first note that x ≠ ±2 and x ≠ 6 . Multiply both sides by the LCD, ( x − 6 )( x − 2 )( x + 2 ) .
⎛ ⎛ ⎞ 2 ( x − 2) ⎞ 4 x+2 + ⎜⎜ ⎟⎟ ( x − 6 )( x − 2 )( x + 2 ) = ⎜⎜ ⎟⎟ ( x − 6 )( x − 2 )( x + 2 ) ⎝ ( x + 2 )( x − 2 ) ( x − 6 )( x + 2 ) ⎠ ⎝ ( x − 6 )( x − 2 ) ⎠ 4 x − 24 + 2 x 2 − 8 x + 8 = x 2 + 4 x + 4 x 2 − 8 x − 20 = 0
( x − 10 )( x + 2 ) = 0 x = 10 or x = −2 x ≠ ±2 .
ANSWER: Hence, the only solution is 10. We can check our answer by substituting 10 in for x to see if we obtain a true statement. 2 (10 − 2 ) 4 10 + 2 + = (10 + 2 )(10 − 2 ) (10 − 6 )(10 + 2 ) (10 − 6 )(10 − 2 )
1 1 3 + = 24 3 8 3 3 = 8 8
62.
PROBLEM: x+2 x+2 x −1 + 2 = 2 2 x − 5x + 4 x + x − 2 x − 2x − 8 SOLUTION: x+2 x+2 x −1 + 2 = 2 2 x − 5x + 4 x + x − 2 x − 2 x − 8 x+2 x+2 x −1 + = ( x − 4 )( x − 1) ( x + 2 )( x − 1) ( x + 2 )( x − 4 )
We first note that x ≠ 4, x ≠ 1, and x ≠ −2 . Multiply both sides by the LCD, ( x + 2 )( x − 4 )( x − 1) .
⎛ ⎞ ⎛ ⎞ x+2 x+2 x −1 + ⎜⎜ ⎟⎟ ( x + 2 )( x − 4 )( x − 1) = ⎜⎜ ⎟⎟ ( x + 2 )( x − 4 )( x − 1) x x x x x x 4 1 2 1 2 4 − − + − + − ( )( ) ( )( ) ( )( ) ⎝ ⎠ ⎝ ⎠ 2 2 2 x + 4x + 4 + x − 2x − 8 = x − 2x +1 x2 + 4 x − 5 = 0
( x + 5)( x − 1) = 0 x = −5 or x = 1 x ≠ 1. ANSWER: Hence, the only solution is –5. We can check our answer by substituting −5 in for x to see if we obtain a true statement. −5 + 2 −5 + 2 −5 − 1 + = ( −5 − 4 )( −5 − 1) ( −5 + 2 )( −5 − 1) ( −5 + 2 )( −5 − 4 )
1 1 2 − =− 18 6 9 2 2 − =− 9 9 −
63.
PROBLEM: 6x 11x + 1 6x − 2 = x −1 2x − x −1 2x +1 SOLUTION: 6x 11x + 1 6x − 2 = x −1 2x − x −1 2x +1 6x 11x + 1 6x − = x − 1 ( 2 x + 1)( x − 1) 2 x + 1
1 . 2 Multiply both sides by the LCD, ( 2 x + 1)( x − 1) .
We first note that x ≠ 1 and x ≠ −
⎛ 6x ⎞ 11x + 1 ⎛ 6x ⎞ − ⎜⎜ ⎟⎟ ( 2 x + 1)( x − 1) = ⎜ ⎟ ( 2 x + 1)( x − 1) − + − + 1 2 1 1 2 1 x x x x ( )( ) ⎝ ⎠ ⎝ ⎠ 2 12 x + 6 x − 11x − 1 = 6 x 2 − 6 x 6 x2 + x − 1 = 0
( 3x − 1)( 2 x + 1) = 0 1 1 or x = − 3 2 1 x≠− . 2
x=
1 ANSWER: Hence, the only solution is . 3 We can check our answer by substituting statement. 1 1 1 6⋅ 11⋅ + 1 6⋅ 3− 3 3 = 1 1 1 1 − 1 ⎛⎜ 2 ⋅ + 1⎞⎟ ⎛⎜ − 1⎞⎟ 2 ⋅ + 1 3 3 ⎝ 3 ⎠⎝ 3 ⎠ 14 6 3 −3 − = 5 ⎛ 2⎞ 5 ⋅⎜ − ⎟ 3 ⎝ 3⎠ 21 6 −3 + = 5 5 6 6 = 5 5
1 in for x to see if we obtain a true 3
64.
PROBLEM: 8x 4x 1 + 2 = 2x − 3 2x − 7 x + 6 x − 2 SOLUTION: 8x 4x 1 + 2 = 2x − 3 2x − 7x + 6 x − 2 8x 4x 1 + = 2 x − 3 ( 2 x − 3)( x − 2 ) x − 2
3 and x ≠ 2 . 2 Multiply both sides by the LCD, ( 2 x − 3)( x − 2 ) .
We first note that x ≠
⎛ 8x ⎞ 4x ⎛ 1 ⎞ + ⎜⎜ ⎟⎟ ( 2 x − 3)( x − 2 ) = ⎜ ⎟ ( 2 x − 3)( x − 2 ) ⎝ x−2⎠ ⎝ 2 x − 3 ( 2 x − 3)( x − 2 ) ⎠ 8 x 2 − 16 x + 4 x = 2 x − 3 8 x 2 − 14 x + 3 = 0
( 4 x − 1)( 2 x − 3) = 0 1 3 or x = 4 2 3 x≠ . 2
x=
1 ANSWER: Hence, the only solution is . 4 1 We can check our answer by substituting in for x to see if we obtain a true 4 statement. 1 1 8⋅ 4⋅ 1 4 + 4 = 1 1 1 1 2 ⋅ − 3 ⎛⎜ 2 ⋅ − 3 ⎞⎟ ⎛⎜ − 2 ⎞⎟ −2 4 ⎝ 4 ⎠⎝ 4 ⎠ 4 4 28 4 − + =− 5 35 7 4 4 − =− 7 7
Part B: Literal Equations Solve for the indicated variable.
65.
PROBLEM:
Solve for r: t =
D r
SOLUTION: Multiply both sides by the LCD, r . D t ⋅r = ⋅r r tr = D
tr D = t t D r= t ANSWER: r =
66.
D t
PROBLEM:
Solve for b: h =
2A b
SOLUTION: Multiply both sides by the LCD, b . 2A h ⋅b = ⋅b b hb = 2 A
hb 2 A = h h 2A b= h ANSWER: b =
2A h
67.
PROBLEM:
Solve for P: t =
I Pr
SOLUTION: Multiply both sides by the LCD, Pr . I t ⋅ Pr = ⋅ Pr Pr Ptr = I
Ptr I = tr tr I P= tr ANSWER: P =
68.
I tr
PROBLEM:
Solve for π : r =
C 2π
SOLUTION: Multiply both sides by the LCD, 2π . C r= 2π C r ⋅ 2π = ⋅ 2c 2π 2π r = C
2π r C = 2r 2r C π= 2r ANSWER: π =
C 2r
69.
PROBLEM: 1 1 1 Solve for c: = + a b c SOLUTION: Multiply both sides by the LCD, abc. 1 ⎛1 1⎞ ⋅ abc = ⎜ + ⎟ ⋅ abc a ⎝b c⎠ bc = ac + ab
bc − ac = ac − ac + ab bc − ac = ab c ( b − a ) = ab c (b − a )
(b − a )
c=
=
ab b−a
ab b−a
ANSWER: c =
70.
PROBLEM:
Solve for y: m =
ab b−a y − y1 x − x1
SOLUTION: Multiply both sides by the LCD, x − x1 . ⎛ y − y1 ⎞ m ( x − x1 ) = ⎜ ⎟ ( x − x1 ) ⎝ x − x1 ⎠ m ( x − x1 ) = y − y1 y − y1 + y1 = m ( x − x1 ) + y1 y = m ( x − x1 ) + y1
ANSWER: y = m ( x − x1 ) + y1
71.
PROBLEM: Solve for w: P = 2(l + w) SOLUTION: P = 2(l + w)
P = 2l + 2w P − 2l = 2l − 2l + 2w 2w = P − 2l 2w P − 2l = 2 2 P − 2l w= 2 ANSWER: w =
72.
P − 2l 2
PROBLEM: Solve for t: A = P(1 + rt ) SOLUTION: A = P(1 + rt )
A P(1 + rt ) = P P A = 1 + rt P A − 1 = 1 − 1 + rt P A− P rt = p rt A − P = r r⋅ p A− P t= pr ANSWER: t =
A− P pr
73.
PROBLEM:
1 n+m
Solve for m: s =
SOLUTION: Multiply both sides by the LCD, n + m . 1 s ( n + m) = ⋅ ( n + m) n+m sn + sm = 1
ms + ns − ns = 1 − ns ms = 1 − ns ms 1 − ns = s s 1 − ns m= s ANSWER: m =
74.
1 − ns s
PROBLEM:
Solve for S: h =
S −r 2π r
SOLUTION: Multiply both sides by the LCD, 2π r . ⎛ S ⎞ h ⋅ 2π r = ⎜ − r ⎟ 2π r ⎝ 2π r ⎠
2π rh = S − 2π r 2
S − 2π r 2 + 2π r 2 = 2π rh + 2π r 2 S = 2π rh + 2π r 2 ANSWER: S = 2π rh + 2π r 2
75.
PROBLEM:
Solve for x: y =
x x+2
SOLUTION: Multiply both sides by the LCD, x + 2 . x ⋅x+2 y ( x + 2) = x+2 xy + 2 y = x
xy − xy + 2 y = x − xy 2 y = x (1 − y ) x (1 − y ) 2y = 1 − y (1 − y ) x=
2y 1− y
ANSWER: x =
76.
PROBLEM:
Solve for x: y =
2y 1− y
2x +1 5x
SOLUTION: Multiply both sides by, 5x . 2x +1 y ⋅ 5x = ⋅ 5x 5x 5 xy = 2 x + 1 5 xy − 2 x = 2 x − 2 x + 1
x (5 y − 2) = 1 x (5 y − 2)
(5 y − 2)
x=
=
1 5y − 2
1 5y − 2
ANSWER: x =
1 5y − 2
77.
PROBLEM:
Solve for R :
1 1 1 = + R R1 R2
SOLUTION: Multiply both sides by LCD, RR1 R2 . ⎛ 1 1 1 ⎞ ⋅ RR1 R2 = ⎜ + ⎟ RR1 R2 R ⎝ R1 R2 ⎠ R1 R2 = RR2 + RR1 R1 R2 = R ( R2 + R1 )
R ( R2 + R1 ) R1 R2 = R2 + R1 ( R2 + R1 ) R=
R1 R2 R2 + R1
ANSWER: R =
78.
R1 R2 R2 + R1
PROBLEM: 1 1 1 = + f S1 S 2 SOLUTION: Multiply both sides by LCD, fS1S2 . ⎛1 1 ⎞ 1 ⋅ fS1S 2 = ⎜ + ⎟ fS1S 2 f ⎝ S1 S 2 ⎠ S1S 2 = fS 2 + fS1 S1S 2 − fS1 = fS 2 + fS1 − fS1 S1 ( S 2 − f ) = fS 2 S1 ( S 2 − f )
( S2 − f )
S1 =
=
fS 2 S2 − f
fS 2 S2 − f
ANSWER: S1 =
fS2 S2 − f
Part C: Discussion Board Solve for the indicated variable.
79.
PROBLEM: Explain why multiplying both sides of an equation by the LCD sometimes produces extraneous solutions. ANSWER: When both the sides are multiplied by the LCD, the solutions that go missing while arriving at the rational expression, emerges again.
80.
PROBLEM: Explain the connection between the technique of cross multiplication and multiplying both sides by the LCD. ANSWER: Cross multiplication is a simplified version of the LCD method. 5 3 = For e.g.: x + 5 x +1 Cross multiplication method. 5 3 = x + 5 x +1 5 ( x + 1) = 3 ( x + 5)
5 x + 5 = 3x + 15 2 x = 10 x=5 The LCD method. We first note that x ≠ −5 and x ≠ −1 . Multiply both sides by the LCD, ( x + 5 )( x + 1) .
5 3 ⋅ ( x + 5 )( x + 1) = ⋅ ( x + 5 )( x + 1) x+5 x +1 5 x + 5 = 3x + 15 5 x − 3x + 5 = 3x − 3x + 15 2 x + 5 = 15 2 x + 5 − 5 = 15 − 5 2 x = 10 x=5
81.
PROBLEM: Explain how we can tell the difference between a rational expression and a rational equation. How do we treat them differently? ANSWER: A rational expression is in the form X/Y, where Y is not equal to zero A rational equation equates two rational expressions. A/B=C/D, where B and D are not equal to zero.
7.6 Applications of Rational Equations
Part A: Number Problems Use algebra to solve the following applications.
1.
PROBLEM: A positive integer is twice that of another. The sum of the reciprocals of the two positive integers is 3/10. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: 2n 1 1 3 + = n 2n 10 We first note that n ≠ 0. Multiply both sides by the LCD, 20n. 3 ⎛1 1 ⎞ ⎜ + ⎟ ⋅ 20n = ⋅ 20n 10 ⎝ n 2n ⎠ 20 + 10 = 6n 6n = 30 n=5
Larger integer: 2n = 2 ⋅ 5 = 10 ANSWER: The two integers are 5 and 10.
2.
PROBLEM: A positive integer is twice that of another. The sum of the reciprocals of the two positive integers is 3/12. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: 2n 1 1 3 + = n 2n 12 We first note that n ≠ 0. Multiply both sides by the LCD, 12n. 3 ⎛1 1 ⎞ ⎜ + ⎟ ⋅12n = ⋅12n 12 ⎝ n 2n ⎠ 12 + 6 = 3n 3n = 18 n=6 Larger integer: 2n = 2 ⋅ 6 = 12 ANSWER: The two integers are 6 and 12.
3.
PROBLEM: A positive integer is twice that of another. The difference of the reciprocals of the two positive integers is 1/8. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: 2n 1 1 1 − = n 2n 8 We first note that n ≠ 0. Multiply both sides by the LCD, 16n. 1 ⎛1 1 ⎞ ⎜ − ⎟ ⋅16n = ⋅16n 8 ⎝ n 2n ⎠ 16 − 8 = 2n 2n = 8 n=4 Larger integer: 2n = 8 ANSWER: The two integers are 4 and 8.
4.
PROBLEM: A positive integer is twice that of another. The difference of the reciprocals of the two positive integers is 1/18. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: 2n 1 1 1 − = n 2n 18 We first note that n ≠ 0. Multiply both sides by the LCD, 36n. 1 ⎛1 1 ⎞ ⎜ − ⎟ ⋅ 36n = ⋅ 36n 18 ⎝ n 2n ⎠ 36 − 18 = 2n 2n = 18 n=9 Larger integer: 2n = 2 ⋅ 9 = 18 ANSWER: The two integers are 9 and 18.
5.
PROBLEM: A positive integer is 2 less than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is 5/12, then find the two integers. SOLUTION: Larger positive integer is represented by n. Smaller integer: n − 2 1 2 5 + = n − 2 n 12 We first note that n ≠ 0 and n ≠ 2 . Multiply both sides by the LCD, 12n ( n − 2 ) .
2⎞ 5 ⎛ 1 + ⎟12n ( n − 2 ) = ⋅12n ( n − 2 ) ⎜ 12 ⎝ n−2 n⎠ 12n + 24n − 48 = 5n 2 − 10n 5n 2 − 46n + 48 = 0
( 5n − 6 )( n − 8) = 0 6 or n = 8 5 The problem speaks about positive integer. Hence, the larger integer is 8. Smaller integer: n − 2 = 8 − 2 = 6 n=
ANSWER: The integers are 6 and 8.
6.
PROBLEM: A positive integer is 2 more than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is 17/35, then find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer is n + 2. 1 2 17 + = n n + 2 35 We first note that n ≠ 0 and n ≠ −2. Multiply both sides by the LCD, 35n ( n + 2 ) .
2 ⎞ 17 ⎛1 ⎜ + ⎟ 35n ( n + 2 ) = ⋅ 35n ( n + 2 ) 35 ⎝n n+2⎠ 2 35n + 70 + 70n = 17 n + 34n 17 n 2 − 71n + 70 = 0
(17n + 14 )( n − 5) 14 or n = 5 17 The problem speaks about positive integers. Hence, the smaller integer is 5. Larger integer: n + 2 = 5 + 2 = 7 n=−
ANSWER: The two integers are 5 and 7.
7.
PROBLEM: The sum of the reciprocals of two consecutive positive even integers is 11/60. Find the two even integers. SOLUTION: The first even integer is represented by n. The second even integer is n + 2. 1 1 11 + = n n + 2 60 We first note that n ≠ 0 or n ≠ −2. Multiply both the sides by the LCD, 60n ( n + 2 ) .
1 ⎞ 11 ⎛1 ⎜ + ⎟ 60n ( n + 2 ) = ⋅ 60n ( n + 2 ) 60 ⎝n n+2⎠ 2 60n + 120 + 60n = 11n + 22n 11n 2 − 98n − 120 = 0
( n − 10 )(11n + 12 ) = 0 12 11 The problem speaks about positive integers. Hence, the first even integer is 10. The next even integer: n + 2 = 10 +2 = 12 n = 10 or n = −
ANSWER: The two consecutive even integers are 10 and 12.
8.
PROBLEM: The sum of the reciprocals of two consecutive positive odd integers is 16/63. Find the integers. SOLUTION: The first odd integer is represented by n. The consecutive odd integer: n + 2 1 1 16 + = n n + 2 63 We first note that n ≠ 0 and n ≠ −2 . Multiply both the sides by the LCD, 63n ( n + 2 ) . 1 ⎞ 16 ⎛1 ⎜ + ⎟ 63n ( n + 2 ) = ⋅ 63n ( n + 2 ) 63 ⎝n n+2⎠ 2 63n + 126 + 63n = 16n + 32n 16n 2 − 94n − 126 = 0
(
)
2 8n 2 − 47 n − 63 = 0 8n 2 − 47 n − 63 = 0
( n − 7 )(8n + 9 ) = 0 9 8 The problem speaks about positive integers. Hence, the first odd integer is 7. Consecutive odd integer: n + 2 = 7 + 2 = 9 n = 7 or n = −
ANSWER: The two consecutive odd integers are 7 and 9.
9.
PROBLEM: The difference of the reciprocals of two consecutive positive even integers is 1/24. Find the two even integers. SOLUTION: The first even integer is represented by n. Consecutive even integer: n + 2 1 1 1 − = n n + 2 24 We first note that n ≠ 0 or n ≠ −2. Multiply both the sides by the LCD, 24n ( n + 2 ) .
1 ⎞ 1 ⎛1 ⋅ 24n ( n + 2 ) ⎜ − ⎟ 24n ( n + 2 ) = 24 ⎝n n+2⎠ 24n + 48 − 24n = n 2 + 2n n 2 + 2n − 48 = 0
( n + 8)( n − 6 ) = 0 n = −8 or n = 6 The problem speaks about positive integers. Hence, the first even integer is 8. The next even integer: n + 2 = 6 +2 = 8 ANSWER: The two consecutive even integers are 6 and 8.
10.
PROBLEM: The difference between the reciprocals of two consecutive positive odd integers is 2/99. Find the integers. SOLUTION: The first odd integer is represented by n. The consecutive odd integer: n + 2 1 1 2 − = n n + 2 99 We first note that n ≠ 0 and n ≠ −2 . Multiply both the sides by the LCD, 99n ( n + 2 ) .
1 ⎞ 2 ⎛1 ⎜ − ⎟ 99n ( n + 2 ) = ⋅ 99n ( n + 2 ) 99 ⎝n n+2⎠ 2 99n + 198 − 99n = 2n + 4n 2n 2 + 4n − 198 = 0
(
)
2 n 2 + 2n − 99 = 0 n 2 + 2n − 99 = 0
( n + 11)( n − 9 ) = 0 n = −11 or n = 9 The problem speaks about positive integers. Hence, the first odd integer is 9. Consecutive odd integer: 9 + 2 = 9 + 2 = 11 ANSWER: The two consecutive odd integers are 9 and 11.
11.
PROBLEM: If 3 times the reciprocal of the larger of two consecutive integers is subtracted from 2 times the reciprocal of the smaller, then the result is 1/2. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: n + 1 2 3 1 − = n n +1 2 We first note that n ≠ 0 and n = −1 . Multiply both sides by the LCD, 2n ( n + 1) . 3 ⎞ 1 ⎛2 ⎜ − ⎟ 2n ( n + 1) = ⋅ 2n ( n + 1) 2 ⎝ n n +1 ⎠ 4n + 4 − 6 n = n 2 + n n 2 + 3n − 4 = 0
( n + 4 )( n − 1) = 0 n = −4 or n = 1 n + 1 = −4 + 1 or n + 1 = 1 + 1 n + 1 = −3 or n + 1 = 2
ANSWER: The two consecutive integers are –4 and –3 or 1 and 2.
12.
PROBLEM: If 3 times the reciprocal of the smaller of two consecutive integers is subtracted from 7 times the reciprocal of the larger, then the result is 1/2. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: n + 1 7 3 1 − = n +1 n 2 We first note that n ≠ 0 and n = −1 . Multiply both sides by the LCD, 2n ( n + 1) . 3⎞ 1 ⎛ 7 − ⎟ 2n ( n + 1) = ⋅ 2n ( n + 1) ⎜ 2 ⎝ n +1 n ⎠ 14n − 6n − 6 = n 2 + n n 2 − 7n + 6 = 0
( n − 1)( n − 6 ) = 0 n = 1 or n = 6 n + 1 = 1 + 1 or n + 1 = 6 + 1 n +1 = 2
or n + 1 = 7
ANSWER: The two consecutive integers are 1 and 2 or 6 and 7.
13.
PROBLEM: A positive integer is 5 less than another. If the reciprocal of the smaller integer is subtracted from 3 times the reciprocal of the larger, then the result is 1/12. Find the two integers. SOLUTION: Larger integer is represented by n. Smaller integer: n – 5 3 1 1 − = n n − 5 12 We first note that n ≠ 0 and n = 5 . Multiply both sides by the LCD, 12n ( n − 5 ) . 1 ⎞ 1 ⎛3 ⎜ − ⎟12n ( n − 5 ) = ⋅12n ( n − 5 ) 12 ⎝ n n−5⎠ 36n − 180 − 12n = n 2 − 5n n 2 − 29n + 180
( n − 20 )( n − 9 ) = 0 n = 20 or n = 9 n − 5 = 20 − 5 or n − 5 = 9 − 5 n − 5 = 15
or n − 5 = 4
ANSWER: The two consecutive integers are 20 and 15 or 9 and 4.
14.
PROBLEM: A positive integer is 6 less than another. If the reciprocal of the smaller integer is subtracted from 10 times the reciprocal of the larger, then the result is 3/7. Find the two integers. SOLUTION: Larger integer is represented by n. Smaller integer: n – 6 10 1 3 − = n n−6 7 We first note that n ≠ 0 and n ≠ 6. Multiply both sides by the LCD, 7 n ( n − 6 ) . 1 ⎞ 3 ⎛ 10 ⎜ − ⎟ 7n ( n − 6 ) = ⋅ 7n ( n − 6 ) 7 ⎝ n n−6⎠ 70n − 420 − 7 n = 3n 2 − 18n 3n 2 − 81n + 420 = 0
(
)
3 n 2 − 27 n + 140 = 0 n 2 − 27 n + 140 = 0
( n − 20 )( n − 7 ) = 0 n = 20 or n = 7 n − 6 = 20 − 6 = 14 or n − 6 = 7 − 6 = 1
ANSWER: The two integers are 20 and 14 or 7 and 1.
Part B: Uniform Motion Problems Use algebra to solve the following applications.
15.
PROBLEM: James can jog twice as fast as he can walk. He was able to jog the first 9 miles to his grandmother’s house, but then he tired and walked the remaining 1.5 miles. If the total trip took 2 hours, then what was his average jogging speed? SOLUTION: Let x represent James’ average walking speed. James average jogging speed: 2x D 9 = Time taken by James to jog 9 miles: r 2x D 1.5 Time taken by James to walk 1.5 miles: = r x Total time taken = 2 hours Time taken by jogging + Time taken by walking = Total time taken 9 1.5 + =2 2x x 9 15 + =2 2 x 10 x We first note that x ≠ 0 . Multiply both sides with the LCD, 10x. ⎛ 9 15 ⎞ ⎜ + ⎟10 x = 2 ⋅10 x ⎝ 2 x 10 x ⎠ 45 + 15 = 20 x 60 = 20 x x = 3 miles/hour
Jogging speed = 2x = 2 ⋅ 3 = 6 miles/hour ANSWER: Jogging speed = 2x = 2 ⋅ 3 = 6 miles/hour
16.
PROBLEM: On a business trip, an executive traveled 720 miles by jet aircraft and then another 80 miles by helicopter. If the jet averaged 3 times the speed of the helicopter and the total trip took 4 hours, then what was the average speed of the jet? SOLUTION: Speed of the helicopter is represented by x. Speed of the jet: 3x D 80 Time taken by the helicopter to cover 80 miles: t = = r x D 720 240 Time taken by the jet to cover 720 miles: t = = = r 3x x Total time taken = 4 hours We first note that x ≠ 0. Time taken by the helicopter + Time taken by the jet = Total time taken 80 240 + =4 x x 320 =4 x 320 x= 4 x = 80 miles/hour Jet speed: 3x = 3 ⋅ 80 = 240 miles/hour ANSWER: Jet speed: 3x = 3 ⋅ 80 = 240 miles/hour
17.
PROBLEM: Sally was able to drive an average of 20 mph faster in her car when the traffic cleared. She drove 23 miles in traffic before it cleared and then drove another 99 miles. If the total trip took 2 hours, then what was her average speed in traffic? SOLUTION: Speed of Sally’s car in traffic is represented by x. Speed of Sally’s car when the traffic clears: x + 20
Time taken by Sally to drive 99 miles when the traffic clears: t = Time taken by Sally to drive 23 miles in traffic: t =
D 99 = r x + 20
D 23 = r x
Total time taken = 2 hours Time taken in traffic + Time taken when the traffic clears = Total time taken 23 99 + =2 x x + 20 We first note that x ≠ 0 and x ≠ −20 . Multiplying both sides with the LCD, x ( x + 20 ) . 99 ⎞ ⎛ 23 ⎜ + ⎟ x ( x + 20 ) = 2 ⋅ x ( x + 20 ) ⎝ x x + 20 ⎠ 23 x + 460 + 99 x = 2 x 2 + 40 x 2 x 2 − 82 x − 460 = 0
(
2 x 2 − 41x − 230
)
x 2 − 41x − 230 = 0
( x + 5 )( x − 46 ) = 0 x = −5 or x = 46 The problem speaks about time taken. Hence, only postive values are considered. ANSWER: Average speed of Sally’s car in traffic is 46 miles/hour.
18.
PROBLEM: Harry traveled 15 miles on the bus and then another 72 miles on a train. If the train was 18 mph faster than the bus and the total trip took 2 hours, then what was the average speed of the train? SOLUTION: Speed of the bus is represented by x. Speed of the train is represented by x + 18. D 72 Time taken by the train to cover 72 miles: t = = r x + 18 D 15 Time taken by the bus to cover 15 miles: t = = r x Total time taken = 2 hours
Time taken in the rain + Time taken in the bus = Total time taken 72 15 + =2 x + 18 x We first note that x ≠ 0 and x = −18 . Multiply both sides with the LCD, x ( x + 18 ) . 15 ⎞ ⎛ 72 + ⎟ x ( x + 18 ) = 2 ⋅ x ( x + 18 ) ⎜ ⎝ x + 18 x ⎠ 72 x + 15 x + 270 = 2 x 2 + 36 x 2 x 2 − 51x − 270 = 0
( x − 30 )( 2 x + 9 ) = 0 9 2 The problem speaks about time. Hence, the positive integer is the solution. x = 30 or x = −
ANSWER: Speed of the train: x + 18 = 30 + 18 = 48 miles/hour.
19.
PROBLEM: A bus is on average 6 mph faster than a trolley. If the bus can travel 90 miles in the same time it takes the trolley to travel 75 miles, then what was the speed of each? SOLUTION: Speed of the trolley is represented by x. Speed of the bus: x + 6 D 75 Time taken by the trolley to cover 75 miles: t = = r x D 90 Time taken by the bus to cover 90 miles: t = = r x+6 Time taken by the trolley to cover 75 miles = Time taken by the bus to cover 90 miles 75 90 = x x+6 We first note that x ≠ 0 and x ≠ −6 . 75 90 = x x+6 75 ( x + 6 ) = 90 ⋅ x
75 x + 450 = 90 x 15 x = 450 x = 30 Speed of the trolley: x = 30 miles/hour Speed of the bus: x + 6 = 30 + 6 = 36 miles/hour ANSWER: Speed of the trolley: x = 30 miles/hour
Speed of the bus: x + 6 = 30 + 6 = 36 miles/hour
20.
PROBLEM: A passenger car is on average 16 mph faster than the bus. If the bus can travel 56 miles in the same time it takes the passenger car to travel 84 miles, then what was the speed of each? SOLUTION: Speed of the bus is represented by x. Speed of the passenger car: x + 16 D 56 Time taken by the bus to cover 56 miles: t = = r x
D 84 = r x + 16 Time taken by the passenger car to = Time taken by the bus to cover 56 miles cover 84 miles 84 56 = x + 16 x 84 x = 56 ( x + 16 )
Time taken by the passenger car to cover 84 miles: t =
84 x = 56 x + 896 28 x = 896 x = 32 Speed of the bus: x = 32 miles/hour Speed of the passenger car: x + 16 = 32 + 16 = 48 miles/hour ANSWER: Speed of the bus: x = 32 miles/hour Speed of the passenger car: x + 16 = 32 + 16 = 48 miles/hour
21.
PROBLEM: A light aircraft travels 2 mph less than twice as fast as a passenger car. If the passenger car can travel 231 miles in the same time it takes the aircraft to travel 455 miles, then what was the average speed of each? SOLUTION: Speed of the passenger car is represented by x. Speed of the light aircraft: 2x – 2 D 231 Time taken by passenger car to cover 231 miles: t = = r x D 455 Time taken by light aircraft to cover 455 miles: t = = r 2x − 2 Time taken by passenger car to cover 231 miles = Time taken by light aircraft to cover 455 miles 231 455 = x 2x − 2 231( 2 x − 2 ) = 455 x
462 x − 462 = 455 x 7 x = 462 x = 66 Speed of the passenger car: x = 66 miles/hour Speed of the light aircraft: 2x − 2 = 2 ⋅ 66 − 2 = 130 miles/hour ANSWER: Speed of the passenger car: x = 66 miles/hour Speed of the light aircraft: 2x − 2 = 2 ⋅ 66 − 2 = 130 miles/hour
22.
PROBLEM: Mary can run 1 mph more than twice as fast as Bill can walk. If Bill can walk 3 miles in the same time it takes Mary to run 7.2 miles, then what is Bill’s average walking speed? SOLUTION: Bill’s average walking speed is represented by x. Mary’s running speed: 2x + 1 D 3 Time taken by Bill to walk 3 miles: t = = r x D 7.2 Time taken by Mary to run 7.2 miles: t = = r 2x +1 Time taken by Bill to walk 3 miles = Time taken by Mary to run 7.2 miles 3 7.2 = x 2x +1 3 ( 2 x + 1) = 7.2 x
6 x + 3 = 7.2 x 60 x + 30 = 72 x 72 x − 60 x = 30 12 x = 30 30 = 2.5 x= 12 Bill's walking speed: x = 2.5 miles/hour ANSWER: Bill's walking speed: x = 2.5 miles/hour
23.
PROBLEM: An airplane, traveling with a 20 mph tailwind, was able to cover 270 miles. On the return trip, against the wind, it was only able to cover 190 miles in the same amount of time. What was the speed of the airplane in still air? SOLUTION: Speed of the airplane in still air is represented by x. Speed of the airplane along the direction of the tailwind: x + 20 Speed of the airplane against the direction of the tailwind: x – 20
D 270 = r x + 20 D 190 Time taken to cover 190 miles against the direction of the wind: t = = r x − 20 Time taken to cover 270 miles = Time taken to cover 190 miles against the along the direction of the wind direction of the wind 270 190 = x + 20 x − 20 270 ( x − 20 ) = 190 ( x + 20 ) Time taken to cover 270 miles along the direction of the wind: t =
270 x − 5400 = 190 x + 3800 80 x = 9200 x = 115 Speed of the airplane in still air: x = 115 miles/hour ANSWER: Speed of the airplane in still air: x = 115 miles/hour
24.
PROBLEM: A jet airliner, traveling with a 30 mph tailwind, was able to cover 525 miles in the same amount of time it was able to travel 495 miles after the tailwind eased to 10 mph. What was the speed of the airliner in still air? SOLUTION: Speed of the jet airliner in still air is represented by x. Speed of the jet airliner along the direction of the tailwind 30 mph: x + 30 Speed of the jet airliner after the tail wind eases to 10 mph: x + 10 D 525 Time taken to cover 525 miles along the direction of the: t = = r x + 30 tailwind 30 mph D 495 Time taken to cover 495 miles along the direction of the: t = = r x + 10 tailwind 10 mph Time taken to cover 525 miles along the = Time taken to cover 495 miles along along the direction of the tailwind 30 mph the direction of the tailwind 10 mph 525 495 = x + 30 x + 10 525 ( x + 10 ) = ( x + 30 ) 495
525 x + 5250 = 495 x + 14850 525 x − 495 x = 14850 − 5250 30 x = 9600 x = 320 Speed of the jet airliner in still air : x = 320 miles/hour ANSWER: Speed of the jet airliner in still air : x = 320 miles/hour
25.
PROBLEM: A boat can average 16 mph in still water. With the current, the boat can travel 95 miles in the same time it can travel 65 miles against it. What is the speed of the current? SOLUTION: Speed of the current is represented by x. Speed of the boat in the direction of the current: 16 + x Speed of the boat against the direction of the current: 16 – x D 95 along the direction of Time taken by the boat to cover 95 miles: t = = r 16 + x the current
Time taken by the boat to cover 65 miles: t =
D 65 against the direction of = r 16 − x
the current Time taken by the boat to cover 95 miles = Time taken by the boat to cover 65 miles along the direction of the current against the direction of the current 95 65 = 16 + x 16 − x 95 (16 − x ) = 65 (16 + x ) 1520 − 95 x = 65 x + 1040 160 x = 480 x=3 ANSWER: Speed of the current is 3 miles/hour.
26.
PROBLEM: A river tour boat can average 7 mph in still water. If the total 24 mile tour downriver and 24 miles back took 7 hours, then how fast was the river current? SOLUTION: Speed of the river current is represented by x. Speed of the tour boat along the direction of the river current: 7 + x Speed of the tour boat against the direction of the river current: 7 – x 24 Time taken by the tour boat to cover 24 miles downstream: 7+ x
Time taken by the tour boat to cover 24 miles upstream:
24 7−x
Total time taken = 7 hours Time taken to travel upstream + Time taken to travel downstream = Total time taken 24 24 + =7 7+ x 7−x We first note that x = ±7 . Multiply both sides with the LCD, ( 7 + x )( 7 − x ) . 24 ⎞ ⎛ 24 + ⎜ ⎟ ( 7 + x )( 7 − x ) = 7 ⋅ ( 7 + x )( 7 − x ) ⎝7+ x 7−x⎠ 168 − 24 x + 168 + 24 x = 343 − 7 x 2 7 x2 − 7 = 0 7 ( x + 1)( x − 1) = 0 x = ±1 The solution is the positive integer. ANSWER: Speed of the river current: x = 1 mile/hour
27.
PROBLEM: If the river current flows at an average 3 mph, then a tour boat can make the 9 mile tour downstream with the current and back the 9 miles against the current in 4 hours. What is the average speed of the boat in still water? SOLUTION: Speed of the boat in still water is represented by x. Speed of the current = 3 mph Speed of the boat along the direction of the current: x + 3 Speed of the boat against the direction of the current: x – 3 D 9 Time taken by the boat to cover 9 miles downstream: t = = r x+3 D 9 Time taken by the boat to cover 9 miles upstream: t = = r x −3 Total time taken = 4 hours Time taken by the boat to + Time taken by the boat to = Total time taken cover 9 miles downstream cover 9 miles upstream 9 9 + =4 x+3 x−3 We first note that x ≠ ±3 . Multiply both the sides with the LCD, ( x + 3)( x − 3) . 9 ⎞ ⎛ 9 + ⎜ ⎟ ( x + 3)( x − 3) = 4 ⋅ ( x + 3)( x − 3) ⎝ x+3 x−3⎠ 9 x − 27 + 9 x + 27 = 4 x 2 − 36 4 x 2 − 18 x − 36 = 0
(
)
2 2 x 2 − 9 x − 18 = 0 2 x 2 − 9 x − 18 = 0
( 2 x + 3)( x − 6 ) = 0 3 or x = 6 2 The positive integer is the solution. x=−
ANSWER: Speed of the boat in still water is 6 miles/hour.
28.
PROBLEM: Jane rowed her canoe, against a 1 mph current, upstream 12 miles and then returned the 12 miles back downstream. If the total trip took 5 hours, then at what speed can Jane row in still water? SOLUTION: Speed at which Jane can row in still water is represented by x. Speed of the current = 1 mph Speed at which Jane can row upstream: x – 1 Speed at which Jane can row downstream: x + 1 D 12 Time taken by Jane to cover 12 miles against the current: t = = r x −1 D 12 Time taken by Jane to cover 12 miles along the current: t = = r x +1 Total time taken = 5 hours Time taken by Jane to cover + Time taken by Jane to cover = Total time taken 12 miles against the current 12 miles along the current 12 12 + =5 x −1 x +1 We first note that x = ±1 . Multiply both sides with the LCD, ( x − 1)( x + 1) .
12 ⎞ ⎛ 12 + ⎜ ⎟ ( x − 1)( x + 1) = 5 ⋅ ( x − 1)( x + 1) ⎝ x −1 x +1 ⎠ 12 x + 12 + 12 x − 12 = 5 x 2 − 5 5 x 2 − 24 x − 5 = 0
( 5 x + 1)( x − 5) = 0 1 or x = 5 5 The solution is the positive integer. x=−
ANSWER: Speed at which Jane can row in still water is 5 miles/hour.
29.
PROBLEM: Jose drove 15 miles to pick up his sister and then returned home. On the return trip, he was able to average 15 mph faster than he did on the trip to pick her up. If the total trip took 1 hour, then at what average speed was Jose able to average on the return trip? SOLUTION: Speed at which Jose drives to pick his sister is represented by x
The speed Jose was able to average on the return trip: x + 15. D 15 Time taken by Jose to reach the destination: t = = r x D 15 Time taken by Jose on his return trip: t = = r x + 15 Total time taken = 1 hour Time taken by Jose + Time taken by Jose on his return trip = Total time taken to reach the destination 15 15 + =1 x x + 15 We first note that x ≠ 0 and x ≠ −15 . Multiply both the sides with the LCD, x ( x + 15 ) . 15 ⎞ ⎛ 15 ⎜ + ⎟ x ( x + 15 ) = 1 ⋅ x ( x + 15 ) ⎝ x x + 15 ⎠ 15 x + 225 + 15 x = x 2 + 15 x x 2 − 15 x − 225 = 0 a = 1; b = −15; c = −225 x= x=
−b ± b 2 − 4ac 2a − ( −15 ) ±
( −15)
2
− 4 (1)( −225 )
2 ⋅1 x = 24.25 or x = −9.125 The positive number is the solution.
=
15 ± 33.5 2
ANSWER: The speed Jose was able to average on the return trip: x + 15 = 24.25 + 15 = 39.25 ~ 40 miles/hour.
30.
PROBLEM: Barry drove the 24 miles to town and then back in 1 hour. On the return trip, he was able to average 14 mph faster than he was on the trip to town. What was his average speed on the trip to town? SOLUTION: Speed at which Barry drove to town is represented by x. Speed of the return trip: x + 14 D 24 Time taken by Barry to reach the town: t = = r x D 24 Time taken by Barry on his return trip: t = = r x + 14 Total time taken = 1 hour Time taken by Barry + Time taken by Barry = Total time taken to reach the town on his return trip 24 24 + =1 x x + 14 We first note that x ≠ 0 and x ≠ −14 . Multiply both the sides with the LCD, x ( x + 14 ) .
24 ⎞ ⎛ 24 ⎜ + ⎟ x ( x + 14 ) = 1 ⋅ x ( x + 14 ) ⎝ x x + 14 ⎠ 24 x + 336 + 24 x = x 2 + 14 x x 2 − 34 x − 336 = 0
( x − 42 )( x + 8 ) = 0 x = 42 or x = −8 The positive integer is the solution. Speed of the return trip: x + 14 = 42 + 14 = 56 mile/hour ANSWER: Speed of the trip to the town : x = 42
31.
PROBLEM: Jerry paddled his kayak, upstream against a 1 mph current, for 12 miles. The return trip, downstream with the 1 mph current, took one hour less time. How fast can Jerry paddle the kayak in still water? SOLUTION: The speed at which Jerry can paddle the kayak in still water is represented by x. Speed of the current = 1 mph Speed of the kayak along the direction of the current: x +1 Speed of the kayak against the direction of the current: x – 1 12 Time taken by Jerry to cover 12 miles upstream: x −1 12 Time taken by Jerry to cover 12 miles downstream: x +1 Time taken by Time taken by Jerry to cover = Jerry to cover – 1 12 miles downstream 12 miles upstream 12 12 = −1 x −1 x +1 We first note that x ≠ ±1 . Multiply both the side with the LCD, ( x + 1)( x − 1) .
⎛ 12 ⎞ ⎛ 12 ⎞ − 1⎟ ( x + 1)( x − 1) ⎜ ⎟ ( x + 1)( x − 1) = ⎜ ⎝ x −1 ⎠ ⎝ x +1 ⎠ 12 x + 12 = 12 x − 12 − x 2 + 1 x 2 − 25 = 0
( x + 5)( x − 5) = 0 x = ±5 The positive integer is the solution. ANSWER: The speed at which Jerry can paddle the kayak in still water is 5 miles/hour.
32.
PROBLEM: It will take a light aircraft 1 hour more time to fly 360 miles against a 30 mph headwind than it will to fly the same distance with it. What is the speed of the aircraft in calm air? SOLUTION: The speed of the aircraft in calm air is represented by x. Speed of the headwind = 30 mph Speed of the aircraft along the direction of the headwind: x + 30 Speed of the aircraft against the direction of the headwind: x – 30 D 360 Time taken by the aircraft to cover 360 miles: t = = r x + 30 along the direction of the wind D 360 Time taken by the aircraft to cover 360 miles: t = = r x − 30 against the direction of the wind
Time taken by the aircraft to = Time taken by the aircraft cover 360 miles against the to cover 360 miles along direction of the wind the direction of the wind 360 360 = +1 x − 30 x + 30 We first note that x ≠ ±30 . Multiply both sides with LCD, ( x − 30 )( x + 30 ) .
+
⎛ 360 ⎞ ⎛ 360 ⎞ + 1⎟ ( x − 30 )( x + 30 ) ⎜ ⎟ ( x − 30 )( x + 30 ) = ⎜ ⎝ x − 30 ⎠ ⎝ x + 30 ⎠ 360 x + 10800 = 360 x − 10800 + x 2 − 900 x 2 − 22500 = 0
( x + 150 )( x − 150 ) = 0 x = ±150 The positive integer is the solution. ANSWER: Speed of the aircraft in calm air is 150 miles/hour.
1
Part C: Work Rate Problems Use algebra to solve the following applications.
33.
PROBLEM: James can paint the office, by himself, in 7 hours and Manny can do it in 10 hours. How long will it take them to paint the office working together? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 7 hours; t2 = 10 hours 1 1 ⋅t + ⋅t = 1 7 10 17t =1 70 17t = 70 70 t= 17 2 t=4 17
ANSWER: James and Manny can paint the office in 4
2 hours. 17
34.
PROBLEM: Barry can lay a brick driveway, by himself, in 12 hours and Robert can do it in 10 hours. How long will it take them to lay the brick driveway working together? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 12 hours; t2 = 10 hours 1 1 ⋅t + ⋅t =1 12 10 22t =1 120 22t = 120 120 t= 22 60 t= 11 5 t =5 11
ANSWER: Barry and Robert can lay the brick driveway in 5
together.
5 hours working 11
35.
PROBLEM: Jerry can detail a car, by himself, in 50 minutes and Sally can do it in one hour. How long will it take them to detail a car working together? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 50 minutes; t2 = 1 hour = 60 minutes 1 1 ⋅t + ⋅t = 1 50 60 110t =1 3000 110t = 3000 300 t= 11 3 t = 27 11
ANSWER: Jerry and Sally can detail the car together in 27
36.
3 minutes. 11
PROBLEM: Jose can build a small shed, by himself, in 26 hours. Alex can build the same small shed, but it will take him 2 days to do so. How long would it take them to build the shed working together? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 26 hours; t2 = 48 hours 1 1 ⋅t + ⋅t = 1 26 48 48t + 26t =1 26 ⋅ 48 74t = 1248 624 t= 37 32 t = 16 37
ANSWER: Jose and Alex can build the small shed in 16
together.
32 hours working 37
37.
PROBLEM: Allison can complete a sales route, by herself, in 6 hours. Working with an associate, they can complete the route in 4 hours. How long would it take her associate to complete the route by herself? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 6 hours; t = 4 hours 1 1 ⋅4+ ⋅4 =1 t2 6 4 4 = 1− t2 6 4 2 = t2 6 t2 =
4⋅6 = 12 2
ANSWER: The associate will take 12 hours to complete the sales route on her own.
38.
PROBLEM: James can prepare and paint a house by himself in 5 days. Working with his brother, Bryan, they can do it in 3 days. How long would it take Bryan to prepare and paint the house by himself? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 5 days; t = 3 days 1 1 ⋅3 + ⋅3 =1 5 t2 3 3 + =1 5 t2 3 3 = 1− 5 t2 3 2 = t2 5 t2 =
5 ⋅ 3 15 1 = =7 2 2 2
1 ANSWER: Bryan will take 7 days to prepare and paint the house himself. 2
39.
PROBLEM: Joe can assemble a computer, by himself, in 1 hour. Working with an assistant, he can assemble a computer in 40 minutes. How long would it take his assistant to assemble a computer working alone? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 1 hour = 60 minutes; t = 40 minutes 1 1 ⋅ 40 + ⋅ 40 = 1 60 t2 2 40 + =1 3 t2 40 2 = 1− 3 t2 40 1 = t2 3 t2 = 40 ⋅ 3 = 120
ANSWER: The assistant will take 120 minutes = 2 hours to assemble a computer working alone.
40.
PROBLEM: The teacher’s assistant can grade class homework assignment by herself in one hour. If the teacher helps, then the grading can be completed in 20 minutes. How long would it take the teacher to grade the papers working alone? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t2 = 1 hour = 60 minutes; t = 20 minutes 1 1 ⋅ 20 + ⋅ 20 = 1 60 t1 20 1 + =1 t1 3 20 1 = 1− 3 t1 20 2 = t1 3 t1 =
3 ⋅ 20 = 30 2
ANSWER: The teacher will take 30 minutes grading the paper alone.
41.
PROBLEM: A larger pipe can fill a water tank twice as fast as a smaller pipe. When both pipes are on, they can fill the tank in 5 hours. If the larger pipe is left off, then how long would it take the smaller pipe to fill the tank? SOLUTION: If we let x represent the time it takes for the small pipe to fill the 1 tank, then the small pipe has a work rate of . x 2 The large pipe has a work rate of . x When both the pipes are on, the tank is filled in 5 hours. Multiply the individual work rates by 5. The sum of these two products is equal to 1. 1 2 ⋅5 + ⋅5 = 1 x x We first note that x ≠ 0 . 1 2 ⋅5 + ⋅5 = 1 x x 5 + 10 =1 x x = 15 ANSWER: The smaller pipe takes 15 hours to fill the tank all alone.
42.
PROBLEM: A newer printer can print twice as fast as an older printer. If both printers, working together, can print a batch of flyers in 45 minutes, then how long would it take the newer printer to print the batch working alone? SOLUTION: If we let x represent the time it takes for the older printer to print a 1 batch of flyers, then the old printer has a work rate of . x x The new printer can print the same batch of flyers in . Hence the new printer 2 2 has a work rate of . x When both the printers are used the batch of flyers can be printed in 45 minutes. Multiply the individual work rates by 45. The sum of these two products is equal to 1. 1 2 ⋅ 45 + ⋅ 45 = 1 x x We first note that x ≠ 0 . 45 + 90 =1 x x = 135 ANSWER: The new printer can print the same batch of flyers in x 135 = = 67.5 minutes . 2 2
43.
PROBLEM: Working alone, Henry would take 9 hours longer than Mary to clean the carpets in the entire office. Working together they can clean the carpets in 6 hours. How long would it take Mary to clean the office carpets if Henry was not there to help? SOLUTION: If we let x represent the time it takes Mary to clean the carpets in 1 the entire office, then Mary has a work rate of . x Henry takes 9 hours longer than Mary, x + 9. Hence, Henry has a work rate of 1 . x+9 Mary and Henry can clean the carpets in 6 hours together. Multiply the individual work rates by 6. The sum of these two products is equal to 1. 1 1 ⋅6 + ⋅6 =1 x x+9 6 6 + =1 x x+9 We first note that x ≠ 0 and x ≠ −9 . Multiply both sides with the LCD, x ( x + 9 ) .
6 ⎞ ⎛6 ⎜ + ⎟ x ( x + 9 ) = 1⋅ x ( x + 9 ) ⎝ x x+9⎠ 6 x + 54 + 6 x = x 2 + 9 x
x 2 − 3 x − 54 = 0
( x − 9 )( x + 6 ) = 0 x = 9 or x = −6 The positive integer is the solution. ANSWER: Mary will take 9 hours to clean the carpets on her own.
44.
PROBLEM: Working alone, Monique would take 4 hours longer than Audrey to record the inventory of the entire shop. Working together they can take inventory in 1.5 hours. How long would it take Audrey to record the inventory working alone? SOLUTION: If we let x represent the time it takes for Audrey to record the 1 inventory of the entire shop, then Audrey has a work rate of . x Monique takes 4 hours longer than Audrey, x + 4 hours. Hence, Monique has a 1 . work rate of x+4 Working together they can take inventory in 1.5 hours. Multiply the individual work rates by 1.5. The sum of these two products is equal to 1. 1 1 ⋅1.5 + ⋅1.5 = 1 x x+4 We first note that x ≠ 0 and x ≠ −4 . 15 15 + = 10 x x+4 Multiply both the sides with the LCD, x ( x + 4 ) .
⎛ 15 15 ⎞ ⎜ + ⎟ x ( x + 4 ) = 10 ⋅ x ( x + 4 ) ⎝ x x+4⎠ 15 x + 60 + 15 x = 10 x 2 + 40 x 10 x 2 + 10 x − 60 = 0 2 ( x − 2 )( x + 3) = 0
x = 2 or x = −3 The positive integer is the solution. ANSWER: Audrey takes 2 hours to record the inventory of the entire shop.
45.
PROBLEM: Jerry can lay a tile floor in 3 hours less time than Jake. If they work together, the floor will take 2 hours. How long would it take Jerry to lay the floor by himself? SOLUTION: If we let x represent the time it takes for Jake to lay a tile floor, then 1 Jake has a work rate of . x Jerry takes 3 hours less than Jake, x – 3 hours. Hence, Jerry has a work rate of 1 . x−3 Working together they can tile the floor in 2 hours. Multiply the individual work rates by 2. The sum of these two products is equal to 1. 1 1 ⋅2+ ⋅2 =1 x x −3 2 2 + =1 x x −3 We first note that x ≠ 0 and x ≠ 3 . Multiply both the sides with the LCD, x ( x − 3) .
2 ⎞ ⎛2 ⎜ + ⎟ x ( x − 3) = 1 ⋅ x ( x − 3) ⎝ x x −3⎠ 2 x − 6 + 2 x = x 2 − 3x
x2 − 7 x + 6 = 0
( x − 6 )( x − 1) = 0 x = 6 or x = 1 x ≠ 1 (Jerry takes less than 3 hours.) ANSWER: Jerry takes x – 3 hours = 6 – 3 = 3 hours
46.
PROBLEM: Jeremy can build a model airplane in 5 hours less time than his brother could. Working together it took 6 hours. How long would it have taken Jeremy to build the model airplane working alone? SOLUTION: If we let x represent the time it takes for Jeremy to build the model 1 airplane, then Jeremy has a work rate of . x Jeremy’s brother takes 5 hours longer than Jeremy, x + 5 hours. Hence, Jeremy’s 1 . brother has a work rate of x+5 Working together they can build the model in 6 hours. Multiply the individual work rates by 6. The sum of these two products is equal to 1. 1 1 ⋅6 + ⋅6 =1 x x+5 We first note that x ≠ 0 and x ≠ −5 . 6 6 + =1 x x+5 Multiply both the sides with the LCD, x ( x + 5 ) .
6 ⎞ ⎛6 ⎜ + ⎟ x ( x + 5) = 1⋅ x ( x + 5) ⎝ x x+5⎠ 6 x + 30 + 6 x = x 2 + 5 x
x 2 − 7 x − 30 = 0
( x − 10 )( x + 3) = 0 x = 10 or x = −3 The positive integer is the solution. ANSWER: Jeremy takes 10 hours to build the model all alone.
47.
PROBLEM: Harry can paint a shed, by himself, in 6 hours and Jeremy can do it in 8 hours. How long will it take them to paint two sheds working together? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 6 hours; t2 = 8 hours 1 1 ⋅t + ⋅t = 1 6 8 6t + 8t =1 48 14t = 48 48 t= 14 24 t= 7
24 hours. 7 24 48 6 ANSWER: They can paint two sheds in , ⋅2 = = 6 hours. 7 7 7
Working together, Harry and Jeremy can paint a shed in
48.
PROBLEM: Joe can assemble a computer, by himself, in 1 hour. Working with an assistant, he can assemble 10 computers in 6 hours. How long would it take his assistant to assemble one computer working alone? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2
Joe and his assistant can assemble 10 computers in 6 hours. 6 hour. Joe and his assistant can assemble a computer in 10 6 t1 = 1 hour; t = hour 10 1 6 1 6 ⋅ + ⋅ =1 1 10 t2 10 6 ⎛ 1⎞ ⎜1 + ⎟ = 1 10 ⎝ t2 ⎠ 6 ⎛ t2 + 1 ⎞ ⎜ ⎟ =1 10 ⎝ t2 ⎠ 6t2 + 6 = 10t2 4t2 = 6 6 4 3 1 t2 = = 1 2 2 t2 =
1 ANSWER: Joe’s assistant will take 1 hours to assemble one computer. 2
49.
PROBLEM: Jerry can lay a tile floor in 3 hours and his assistant can do the same job in 4 hours. If Jerry starts the job and his assistant joins him 1 hour later, then how long will it take to lay the floor?
1 SOLUTION: Jerry can lay a tile floor in 3 hours. Hence, Jerry’s work rate is . 3 Jerry’s assistant can lay the same tile floor in 4 hours. Hence, assistant work rate 1 is . 4 1 1 Work done by Jerry in a hour is 1 ⋅ = 3 3 1 2 Remaining floor to be tiled: 1 − = 3 3 1 1 2 Using the formula, ⋅ t + ⋅ t = (Remaining floor to be tiled) t1 t2 3 t1 = 3 hours; t2 = 4 hours 1 1 2 ⋅t + ⋅t = 3 3 4 7t 2 = 12 3 2 12 t= ⋅ 3 7 8 t= 7 Jerry starts one hour earlier. ANSWER:
1+ Hence, the total time taken to lay the floor = 1 + t =
8 15 1 = = 2 hours. 7 7 7
50.
PROBLEM: Working alone, Monique would take 6 hours to record the inventory of the entire shop while it would take Audrey only 4 hours to do the same job. How long will it take them working together if Monique leaves 2 hours early? SOLUTION: Monique takes 6 hours to record the inventory of the entire shop. 1 Hence, Monique’s work rate is . 6 Audrey takes 4 hours to record the inventory of the entire shop. Hence, Audrey’s 1 work rate is . 4 Audrey leaves 2 hours early. Hence, Monique has to do Audrey’s 2 hour work. 1 1 Audrey’s work to be completed by Monique 2 ⋅ = . Hence, Monique takes 2 4 2 hours to complete Audrey’s work.
Work done together by Monique and Audrey: 1 −
1 1 = 2 2
1 1 1 Using the formula, ⋅ t + ⋅ t = (Work done by Audrey and Monique) t1 t2 2 t1 = 6 hours; t2 = 4 hours 1 1 1 ⋅t + ⋅t = 6 4 2 4t + 6t 1 = 24 2 1 24 t= ⋅ 2 10 6 t= 5 Total time taken = Time taken by Audrey and Monique + Time taken by Audrey 1 to complete of the work to complete Monique’s work 2 6 16 1 = +2= = 3 hours 5 5 5
ANSWER: 3
1 hours 5
7.7 Variation
Part A: Variation Problems Translate the following sentences into a mathematical formula.
1.
PROBLEM: The distance D an automobile can travel is directly proportional to the time t that it travels at a constant speed. ANSWER: D = kt
2.
PROBLEM: The extension of a hanging spring d is directly proportional to the weight w attached to it. ANSWER: d = kw
3.
PROBLEM: The breaking distance of an automobile d is directly proportional to the square of its speed v. ANSWER: d = kv2
4.
PROBLEM: The volume of a sphere V varies directly as the cube of its radius r. ANSWER: V = kr3
5.
6.
PROBLEM: The volume V of a given mass of gas is inversely proportional to the pressure p exerted on it. k ANSWER: V = p PROBLEM: The intensity of light from a light source I is inversely proportional to the square of the distance from the source. ANSWER: The distance from the source is represented by d. k I= 2 d
7.
PROBLEM: Every particle of matter in the universe attracts every other particle with a force F that is directly proportional to the product of the masses m1 and m2 of the particles and inversely proportional to the square of the distance d between them. ANSWER: F = k
8.
m1 ⋅ m2 d2
PROBLEM: Simple interest I is jointly proportional to the annual interest rate r and the time t, in years, a fixed amount of money is invested. ANSWER: I = krt
9.
PROBLEM: The period of a pendulum T is directly proportional to the square root of its length L. ANSWER: T = k L
10.
PROBLEM: The time t it takes an object to fall is directly proportional to the square root of the distance d it falls. ANSWER: t = k d
Construct a mathematical model given the following:
11.
PROBLEM: y varies directly as x and y = 30 when x = 6. SOLUTION: y = kx Substitute the values of x and y in the expression above. 30 = k6 k=5 ANSWER: The expression becomes y = 5x.
12.
PROBLEM: y varies directly as x and y = 52 when x = 4. SOLUTION: y = kx Substitute the values of x and y in the expression above. 52 = k4 k = 13 ANSWER: The expression becomes y = 13x.
13.
PROBLEM: y is directly proportional to x and y = 12 when x = 3. SOLUTION: y = kx Substitute the values of x and y in the expression above. 12 = k3 k=4 ANSWER: The expression becomes y = 4x.
14.
PROBLEM: y is directly proportional to x and y = 120 when x = 20. SOLUTION: y = kx Substitute the values of x and y in the expression above. 120 = k20 k=6 ANSWER: The expression becomes y = 6x.
15.
PROBLEM: y varies directly as x and y = 14 when x = 10. SOLUTION: y = kx Substitute the values of x and y in the expression above. 14 = k10 14 7 k= = 10 5 7 ANSWER: The expression becomes y = x. 5
16.
PROBLEM: y varies directly as x and y = 2 when x = 8. SOLUTION: y = kx Substitute the values of x and y in the expression above. 2 = k8 2 1 k= = 8 4 x ANSWER: The expression becomes y = . 4
17.
PROBLEM: y varies inversely as x and y = 5 when x = 7.
k x Substitute the values of x and y in the expression above. k 5= 7 k = 35 35 . ANSWER: The expression becomes y = x SOLUTION: y =
18.
PROBLEM: y varies inversely as x and y = 12 when x = 2.
k x Substitute the values of x and y in the expression above. k 12 = 2 k = 24 24 . ANSWER: The expression becomes y = x SOLUTION: y =
19.
PROBLEM: y is inversely proportional to x and y = 3 when x = 9.
k x Substitute the values of x and y in the expression above. k 3= 9 k = 27 27 ANSWER: The expression becomes y = . x SOLUTION: y =
20.
PROBLEM: y is inversely proportional to x and y = 21 when x = 3.
k x Substitute the values of x and y in the expression above. k 21 = 3 k = 63 63 ANSWER: The expression becomes y = . x SOLUTION: y =
21.
PROBLEM: y varies inversely as x and y = 2 when x = 1/8.
k x Substitute the values of x and y in the expression above. k 2= 1 8 1 k = 2⋅ 8 1 k= 4 1 ANSWER: The expression becomes y = . 4x SOLUTION: y =
22.
PROBLEM: y varies inversely as x and y = 3/2 when x = 1/9.
k x Substitute the values of x and y in the expression above. 3 k = 2 1 9 3 1 k= ⋅ 2 9 1 k= 6 1 ANSWER: The expression becomes y = . 6x SOLUTION: y =
23.
PROBLEM: y varies jointly as x and z where y = 8 when x = 4 and z = 1/2. SOLUTION: y = kxz Substitute the values of x, y, and z in the expression above. 1 8 = k ⋅4⋅ 2 k =4 ANSWER: The expression becomes y = 4xz.
24.
PROBLEM: y varies jointly as x and z where y = 24 when x = 1/3 and z = 9. SOLUTION: y = kxz Substitute the values of x, y, and z in the expression above. 1 24 = k ⋅ ⋅ 9 3 k =8 ANSWER: The expression becomes y = 8xz.
25.
PROBLEM: y is jointly proportional to x and z where y = 2 when x = 1 and z = 3. SOLUTION: y = kxz Substitute the values of x, y, and z in the expression above. 2 = k ⋅1 ⋅ 3
k=
2 3
ANSWER: The expression becomes y =
26.
2 xz . 3
PROBLEM: y is jointly proportional to x and z where y = 15 when x = 3 and z = 7. SOLUTION: y = kxz Substitute the values of x, y, and z in the expression above. 15 = k ⋅ 3 ⋅ 7
k=
5 7
ANSWER: The expression becomes y =
27.
5 xz. 7
PROBLEM: y varies jointly as x and z where y = 2/3 when x = 1/2 and z = 12. SOLUTION: y = kxz Substitute the values of x, y, and z in the expression above. 2 1 = k ⋅ ⋅12 3 2 2 1 k= = 3⋅ 6 9 xz . ANSWER: The expression becomes y = 9
28.
PROBLEM: y varies jointly as x and z where y = 5 when x = 3/2 and z = 2/9. SOLUTION: y = kxz Substitute the values of x, y, and z in the expression above. 3 2 5=k⋅ ⋅ 2 9 k = 5 ⋅ 3 = 15 ANSWER: The expression becomes y = 15xz.
29.
PROBLEM: y varies directly as the square of x where y = 45 when x = 3. SOLUTION: y = kx2 Substitute the values of x and y in the expression above. 45 = k 32 45 =5 9 ANSWER: The expression becomes y = 5 x 2 . k=
30.
PROBLEM: y varies directly as the square of x where y = 3 when x = 1/2. SOLUTION: y = kx2 Substitute the values of x and y in the expression above. 2 ⎛1⎞ 3= k⎜ ⎟ ⎝2⎠ k = 3 ⋅ 4 = 12 ANSWER: The expression becomes y = 12 x 2 .
31.
PROBLEM: y is inversely proportional to the square of x where y = 27 when x = 1/3.
k x2 Substitute the values of x and y in the expression above. k 27 = 2 ⎛1⎞ ⎜ ⎟ ⎝3⎠ 1 k = ⋅ 27 = 3 9 3 ANSWER: The expression becomes y = 2 . x SOLUTION: y =
32.
PROBLEM: y is inversely proportional to the square of x where y = 9 when x = 2/3.
k x2 Substitute the values of x and y in the expression above. k 9= 2 ⎛2⎞ ⎜ ⎟ ⎝3⎠ 4 k = ⋅9 = 4 9 4 ANSWER: The expression becomes y = 2 . x SOLUTION: y =
33.
PROBLEM: y varies jointly as x and the square of z where y = 54 when x = 2 and z = 3. SOLUTION: y = kxz 2 Substitute the values of x, y, and z in the expression above. 54 = k ⋅ 2 ⋅ 32 54 =3 18 ANSWER: The expression becomes y = 3xz 2 k=
34.
PROBLEM: y varies jointly as x and the square of z where y = 6 when x = 1/4 and z = 2/3. SOLUTION: y = kxz 2 Substitute the values of x, y, and z in the expression above. 2 1 ⎛2⎞ 6 = k ⋅ ⋅⎜ ⎟ 4 ⎝3⎠ k = 6 ⋅ 9 = 54 ANSWER: The expression becomes y = 54 xz 2
35.
PROBLEM: y varies jointly as x and z and inversely as the square of w, where y = 30 when x = 8, z = 3, and w = 2.
kxz w2 Substitute the values of x, y, and z in the expression above. k ⋅8⋅3 30 = 22 30 k= =5 6 5xz ANSWER: The expression becomes y = 2 w SOLUTION: y =
36.
PROBLEM: y varies jointly as x and z and inversely as the square of w, where y = 5 when x = 1, z = 3, and w = 1/2.
kxz w2 Substitute the values of x, y, and z in the expression above. k ⋅1 ⋅ 3 5= 2 ⎛1⎞ ⎜ ⎟ ⎝2⎠ 5 k= 12 5 xz ANSWER: The expression becomes y = 12w2 SOLUTION: y =
37.
PROBLEM: y varies directly as the square root of x and inversely as z where y = 12 when x = 9 and z = 5.
x z Substitute the values of x, y, and z in the expression above. 9 12 = k 5 5 ⋅12 k= = 20 3 20 x ANSWER: The expression becomes y = z SOLUTION: y = k
38.
PROBLEM: y varies directly as the square root of x and inversely as the square of z where y = 15 when x = 25 and z = 2. SOLUTION: y = k
x 2
z Substitute the values of x, y, and z in the expression above. 25 15 = k 2 2 15 ⋅ 4 k= = 12 5 12 x ANSWER: The expression becomes y = 2 z
39. PROBLEM: y varies directly as the square of x and inversely as z and the square of w where y = 14 when x = 4, w = 2, and z = 2. x2 SOLUTION: y = k 2 zw Substitute the values of x, y, z, and w in the expression above. 42 14 = k 2 ⋅ 22 14 =k 2 k =7 7x 2 ANSWER: The expression becomes y = 2 zw 40.
PROBLEM: y varies directly as the square root of x and inversely as z and the square of w where y = 27 when x = 9, w = 1/2, and z = 4.
x zw2 Substitute the values of x, y, z, and w in the expression above. 9 27 = k 2 ⎛1⎞ 4⋅⎜ ⎟ ⎝2⎠ 27 k= =9 3 9 x ANSWER: The expression becomes y = zw2
SOLUTION: y = k
Part B: Variation Problems Applications involving variation.
41.
PROBLEM: The revenue, in dollars, is directly proportional to the number of branded sweatshirts sold. If the revenue earned from selling 25 sweatshirts was $318.75, then determine the revenue if 30 sweatshirts are sold. SOLUTION: Revenue in dollars is represented by R. Number of sweatshirts sold is represented by x. R = kx Revenue earned from selling 25 sweatshirts: $318.75 R = kx 318.75 = k ⋅ 25 318.75 k= = 12.75 25 Revenue for 30 sweatshirts: R = kx R = 12.75 ⋅ 30 = 382.5 ANSWER: Revenue for 30 sweatshirts sold is $382.50.
42.
PROBLEM: The sales tax on the purchase of a new car varies directly as the price of the car. If a $18,000 new car is purchased, then the sales tax would be $1,350. How much sales tax would be required if the new car was priced at $22,000? SOLUTION: Sales tax in dollars is represented by S. Price of the car is represented by x. S = kx If a $18,000 new car is purchased, then the sales tax would be $1,350. S = kx 1350 = k ⋅18000 k = 0.075 Sales tax required if the new car was priced at $22,000: S = kx S = 0.075 ⋅ 22, 000 = 1650
ANSWER: Sales tax required if the new car was priced at $22,000 is $1,650.
43.
PROBLEM: The price of a share of common stock in a company is directly proportional to the earnings per share (EPS) of the previous 12 months. If the price of a share of common stock in a company is $22.55 and the EPS is published to be $1.10, then determine the value of the stock if the earnings per share increased by $0.20. SOLUTION: Price of a share of common stock in the company is represented by P. Earnings per share is represented by x. P = kx Price of a share of common stock in a company is $22.55 and the EPS is published to be $1.10. P = kx 22.55 = k ⋅1.10 k = 20.5 If the earnings per share increased by $0.20, x = 1.10 + 0.20 = 1.3. Value of the stock: P = kx P = 20.5 ⋅1.3 = 26.65
ANSWER: Value of the stock if the earnings per share increased by $0.20 is $26.65.
44.
PROBLEM: The distance traveled on a road trip varies directly as the time spent on the road. If a 126 mile trip can be made in 3 hours, then what distance could be traveled in 4 hours? SOLUTION: Distance traveled on a road trip D. Time spent on the road is represented by t. D = kt A 126 mile trip can be made in 3 hours. D = kt 126 = k ⋅ 3 k = 42 Distance travelled in 4 hours: D = kt D = 42 ⋅ 4 = 168
ANSWER: Distance travelled in 4 hours is 168 miles.
45.
PROBLEM: The circumference of a circle is directly proportional to its radius. If the circumference of a circle with radius 7 cm is measured to be 14π cm , then find the constant of proportionality. SOLUTION: Circumference of the circle is represented by C. Radius of the circle is represented by r. C = kr Circumference of a circle with radius 7 cm is measured to be 14π cm . C = kr 14π = k ⋅ 7 k = 2π
ANSWER: k = 2π
46.
PROBLEM: The area of circle varies directly as the square of its radius. If the area of a circle with radius 7 cm is determined to be 49π cm 2 , then find the constant of proportionality. SOLUTION: Area of the circle is represented by A. Radius of the circle is represented by r. A = kr 2 Area of a circle with radius 7 cm is measured to be 49π cm 2 .
A = kr 2 49π = k ⋅ 7 2 49π = k ⋅ 49 k =π ANSWER: k = π
47.
PROBLEM: The surface area of a sphere varies directly as the square of its radius. When the radius of a sphere measures 2 m, the surface area measures 16π m 2 . Find the surface area of a sphere with radius 3 m. SOLUTION: Surface area of the sphere is represented by S. Radius of the sphere is represented by r. S = kr 2 When the radius of a sphere measures 2 m, the surface area measures 16π m 2 .
S = kr 2 16π = k ⋅ 22 4k = 16π k = 4π Surface area of a sphere with radius 3 m: S = kr 2 S = 4π ⋅ 32 = 36π ANSWER: Surface area of a sphere with radius 3 m is 36π m2.
48.
PROBLEM: The volume of a sphere varies directly as the cube of its radius. When the radius of a sphere measures 3 m, the volume measures 36π m3 . Find the volume of a sphere with radius 1 m. SOLUTION: Volume of a sphere is represented by V. Radius of the sphere is represented by r. V = kr 3 When the radius of a sphere measures 3 m, the volume measures 36π m3 . V = kr 3
36π = k ⋅ 33 27k = 36π 4 k= π 3 Volume of a sphere with radius 1 m: V = kr 3 4 V = π ⋅13 3 4π V= 3 ANSWER: Volume of a sphere with radius 1 m is
4π 3 m. 3
49.
PROBLEM: With a fixed height, the volume of a cone is directly proportional to the square of the radius at the base. When the radius at the base measures 10 cm, the volume is determined to be 200 cubic centimeters. Determine the volume of the cone if the radius of the base is halved. SOLUTION: Volume of the cone is represented by V. Radius of the base of the cone is represented by r. V = kr 2 When the radius at the base measures 10 cm, the volume is determined to be 200 cubic centimeters. V = kr 2
200 = k ⋅102 k =2 Volume of the cone if the radius of the base is halved, r =
10 = 5 cm . 2
V = kr 2 V = 2 ⋅ 52 = 50 cm3 ANSWER: Volume of the cone if the radius of the base is halved is 50 cm3. Hooke’s law suggests that the extension of a hanging spring is directly proportional to the weight attached to it. The constant of variation is called the spring constant.
50.
PROBLEM: The distance d an object in freefall falls varies directly with the square of the time t that it has been falling. If an object in freefall falls 36 feet in 1.5 seconds, then how far will it have fallen in 3 seconds? SOLUTION: d = kt 2 An object in freefall falls 36 feet in 1.5 seconds. 36 = k ⋅1.52 k = 16 The object in freefall for 3 seconds falls: d = kt 2 d = 16 ⋅ 9 = 144 ANSWER: The object in freefall for 3 seconds falls 144 feet.
Hooke’s law suggests that the extension of a hanging spring is directly proportional to the weight attached to it. The constant of variation is called the spring constant.
51.
PROBLEM: If a hanging spring is stretched 5 inches when a 20 lb weight is attached to it, then determine its spring constant. SOLUTION: The extension of the hanging spring is represented by L. The weight attached to the hanging spring is represented by w. L = kw A hanging spring is stretched 5 inches when a 20 lb weight is attached to it. L = kw 5 = k ⋅ 20 1 k= 4 ANSWER: k =
52.
1 4
PROBLEM: If a hanging spring is stretched 3cm when a 2 kg weight is attached to it, then determine the spring constant. SOLUTION: The extension of the hanging spring is represented by L. The weight attached to the hanging spring is represented by w. L = kw A hanging spring is stretched 3cm when a 2 kg weight is attached to it. L = kw 3 = k ⋅2 3 k= 2 3 ANSWER: k = 2
53.
PROBLEM: If a hanging spring is stretched 3 inches when a 2 pound weight is attached, then how far will it stretch with a 5 lb weight attached? SOLUTION: The extension of the hanging spring is represented by L. The weight attached to the hanging spring is represented by w. L = kw A hanging spring is stretched 3 inches when a 2 pound weight is attached. L = kw 3 = k ⋅2 3 k= 2 If a 5 lb weight is attached: L = kw 3 L = ⋅5 2 15 1 L= =7 2 2 ANSWER: If a 5 lb weight is attached, the hanging spring will stretch by 7
1 2
inches. The breaking distance of an automobile is directly proportional to the square of its speed.
54.
PROBLEM: If a hanging spring is stretched 6 cm when a 4 kg weight is attached to it, then how far will it stretch with a 2 kg weight attached? SOLUTION: The extension of the hanging spring is represented by L. The weight attached to the hanging spring is represented by w. L = kw A hanging spring is stretched 6 cm when a 4 kg weight is attached. L = kw 6 = k ⋅4 6 3 k= = 4 2 If a 2 kg weight is attached: L = kw 3 L = ⋅2 2 L=3 ANSWER: If a 2 kg weight is attached, the hanging spring will stretch by 3 cm.
The breaking distance of an automobile is directly proportional to the square of its speed.
55.
PROBLEM: If it took 36 feet to stop a particular automobile moving at a speed of 30 miles per hour, then how much breaking distance is required if the speed is 35 miles per hour? SOLUTION: Breaking distance of the automobile is represented by d. Speed of the vehicle is represented by s. d = ks 2 It takes 36 feet to stop the automobile moving at a speed of 30 miles per hour. d = ks 2 36 = k ⋅ 302 36 1 = 900 25 If the speed is 35 miles per hour, breaking distance required: d = ks 2 1 d = ⋅ 352 = 49 25 k=
ANSWER: If the speed is 35 miles per hour, breaking distance required is 49 feet. Boyle's Law states that if the temperature remains constant, the volume V of a given mass of gas is inversely proportional to the pressure p exerted on it.
56.
PROBLEM: After an accident, it was determined that it took a driver 80 feet to stop his car. In an experiment, under similar conditions, it took 45 feet to stop the car moving at a speed of 30 miles per hour. Estimate how fast the driver was moving before the accident. SOLUTION: Breaking distance of the automobile is represented by d. Speed of the vehicle is represented by s. d = ks 2 In an experiment, under similar conditions, it took 45 feet to stop the car moving at a speed of 30 miles per hour. d = ks 2 45 = k ⋅ 302 1 20 After an accident, it was determined that it took a driver 80 feet to stop his car. d = ks 2 k=
1 2 s 20 s 2 = 1600
80 =
s = 40
ANSWER: The driver was moving at 40 miles/hour before the accident. Boyle's Law states that if the temperature remains constant, the volume V of a given mass of gas is inversely proportional to the pressure p exerted on it.
57.
PROBLEM: A balloon is filled to a volume of 216 cubic inches on a diving boat under 1 atmosphere (atm) of pressure. If the balloon is taken under water approximately 33 ft where the pressure measures 2 atm, then what is the volume of the balloon?
k p The balloon is filled to a volume of 216 cubic inches on a diving boat under 1 atmosphere (atm) of pressure. k V= p k 216 = 1 k = 216 The balloon is taken under water approximately 33 ft where the pressure measures 2 atm. k V= p 216 V= = 108 2 SOLUTION: Boyle’s law, V =
ANSWER: If the balloon is taken under water approximately 33 ft where the pressure measures 2 atm, the volume of the balloon is 108 in.3
58.
PROBLEM: If a balloon is filled to 216 cubic inches under a pressure of 3 atmospheres at a depth of 66 ft, then what would the volume be at the surface where the pressure is 1 atm?
k p The balloon is filled to a volume of 216 cubic inches under a pressure of 3 atmospheres at a depth of 66 ft. k V= p k 216 = 3 k = 648 The volume at the surface where the pressure is 1 atm. k V= p 648 V= 1 V = 648 SOLUTION: Boyle’s law, V =
ANSWER: The volume at the surface where the pressure is 1 atm is 648 in.3
59.
PROBLEM: To balance a seesaw, the distance from the fulcrum that a person must sit is inversely proportional to his weight. If a 72 lb boy is sitting 3 feet from the fulcrum, then how far from the fulcrum must a 54 lb boy sit to balance the seesaw? SOLUTION: The distance from the fulcrum that a person must sit is represented by d. Weight of the person is represented by w. k d= w A 72 lb boy is sitting 3 feet from the fulcrum. k d= w k 3= 72 k = 216
If a 54 lb boy sits on these saw, the distance from the fulcrum that a person must k sit: d = w 216 d= =4 54 ANSWER: If a 54 lb boy sits on these saw, the person must sit 4 feet away from the fulcrum.
60.
PROBLEM: The current ( I ) in an electrical conductor is inversely proportional to its resistance ( R ). If the current is 1/4 amp when the resistance is 100 ohms, then what is the current when the resistance is 150 ohms?
k R The current is 1/4 amp when the resistance is 100 ohms. 1 k = 4 100 k = 25 k When the resistance is 150 ohms: I = R 25 1 I= = 150 6 SOLUTION: I =
ANSWER: When the resistance is 150 ohms, the current is
1 amp. 6
61.
PROBLEM: The number of men y needed to lay a cobblestone driveway is directly proportional to the area A of the driveway and inversely proportional to the amount of time t allowed to complete the job. Typically, 3 men can lay 1200 square feet of cobblestone in 4 hours. How many men will be required to lay 2400 square feet of cobblestone given 6 hours?
kA t 3 men can lay 1200 square feet of cobblestone in 4 hours. k ⋅1200 3= 4 1 k= 100 SOLUTION: y =
Men required to lay 2400 square feet of cobblestone given 6 hours: y =
kA t
1 ⋅ 2400 100 y= =4 6 ANSWER: To lay 2400 square feet of cobblestone given 6 hours, 4 men are required.
62.
PROBLEM: The volume of a right circular cylinder varies jointly as the square of its radius and its height. A right circular with a 3 cm radius and a 4 cm height has a volume of 36π cm3 . Find a formula for the volume of a right circular cylinder in terms of its radius and height. SOLUTION: Volume of the right circular cylinder is represented by V. Radius of the right circular cylinder is represented by r. V = kr 2 h Right circular with a 3 cm radius and a 4 cm height has a volume of 36π cm3 .
36π = k 32 ⋅ 4 k =π ANSWER: The formula for a right circular cylinder in terms of its radius and height is V = π r 2 h
63.
PROBLEM: The period of a pendulum T is directly proportional to the square root of its length L. If the length of a pendulum is 1 meter, then the period is approximately 2 seconds. Approximate the period of a pendulum that is 0.5 meter in length. SOLUTION: T = k L When the length of a pendulum is 1 meter, the period is approximately 2 seconds. T =k L 2=k 1 k=2 The period of a pendulum that is 0.5 meter in length: T = k L T = 2 0.5 = 1.4
ANSWER: The period of a pendulum that is 0.5 meter in length is 1.4 seconds. Newton's universal law of gravitation states that every particle of matter in the universe attracts every other particle with a force F that is directly proportional to the product of the masses m1 and m2 of the particles and inversely proportional to the square of the distance d between them. The constant of proportionality is called the gravitational constant.
64.
PROBLEM: The time t it takes an object to fall is directly proportional to the square root of the distance d it falls. An object dropped from 4 feet will take 1/2 second to hit the ground. How long will it take an object dropped from 16 feet to hit the ground? SOLUTION: T = k d An object dropped from 4 feet will take 1/2 second to hit the ground. 1 =k 4 2 1 k= 4 If an object is dropped from 16 feet to hit the ground: T = k d 1 T= 16 = 1 4 ANSWER: It takes 1 second for an object dropped from 16 feet to hit the ground.
Newton's universal law of gravitation states that every particle of matter in the universe attracts every other particle with a force F that is directly proportional to the product of the masses m1 and m2 of the particles and inversely proportional to the square of the distance d between them. The constant of proportionality is called the gravitational constant.
65.
PROBLEM: If two objects with masses 50 kg and 100 kg are 1/2 meter apart, then they produce approximately 1.34 ×10−6 Newtons (N) of force. Calculate the gravitational constant.
km1m2 d2 Two objects with masses 50 kg and 100 kg 1/2 meter apart, produce approximately 1.34 ×10−6 Newtons (N) of force. km m F = 12 2 d k 50 ⋅100 1.34 ×10−6 = 2 ⎛1⎞ ⎜ ⎟ ⎝2⎠ k = 6.7 × 10−11 SOLUTION: F =
ANSWER: The gravitational constant is 6.7 × 10−11 N m 2 / kg 2 .
66.
PROBLEM: Use the gravitational constant from the previous exercise to write a formula that approximates the force F, in Newtons, between two masses m1 and m2 , in kilograms, given the distance d between them, in meters. ANSWER: F =
(
F = 6.7 × 10−11
)
km1m2 d2 m1m2 d2
67.
PROBLEM: Calculate the force, in Newtons, between the Earth and Moon given that the mass of the Moon is approximately 7.3 ×1022 kg , the mass of the Earth is approximately 6.0 ×1024 kg , and the distance between them is on average 1.5 × 1011 m .
(
SOLUTION: F = 6.7 × 10−11
) mdm 1
2
2
The mass of the Moon is approximately 7.3 ×1022 kg , the mass of the Earth is approximately 6.0 ×1024 kg , and the distance between them is on average 1.5 × 1011 m . 7.3 × 1022 × 6.0 ×1024 F = 6.7 × 10−11 2 1.5 × 1011
(
)
(
)
F = 1.3 ×1015 ANSWER: The force between the Earth and Moon is 1.3 × 1015 N.
68.
PROBLEM: Calculate the force, in Newtons, between the Earth and Sun given that the mass of the Sun is approximately 2.0 ×1030 kg , the mass of the Earth is approximately 6.0 ×1024 kg , and the distance between them is on average 3.85 ×108 m . SOLUTION: The mass of the Sun is approximately 2.0 ×1030 kg , the mass of the Earth is approximately 6.0 ×1024 kg , and the distance between them is on average 3.85 ×108 m . mm F = 6.7 × 10−11 1 2 2 d 2.0 × 1030 × 6.0 ×1024 F = 6.7 × 10−11 2 3.85 × 108
(
)
(
)
(
)
F = 5.4 ×1027 ANSWER: Force between the Earth and Sun is 5.4 × 1027 N.
69.
PROBLEM: If y varies directly as the square of x, then how does y change if x is doubled? SOLUTION: y = kx 2
If x is doubled, y = k ( 2 x ) = 4kx 2
ANSWER: y changes by a factor of 4
70.
PROBLEM: If y varies inversely as square of t, then how does y change if t is doubled? SOLUTION: y =
If t is doubled, y =
k t2 k
( 2t )
2
=
k 4t 2 1 4
ANSWER: y changes by a factor of
71.
PROBLEM: If y varies directly as the square of x and inversely as the square of t, then how does y change if both x and t are doubled? SOLUTION: y =
x2 t2
( 2x) y= 2 ( 2t )
2
If both x and t are doubled,
ANSWER: y remains unchanged
=
4 x2 x2 = 4t 2 t 2
7.8 Review Exercises and Sample Exam Chapter 7 Review Exercises 7.1 Simplifying Rational Expressions Evaluate for the given set of x‐values.
1.
PROBLEM: 25 ; {−5, 0, 5} 2x 2 ANSWER: Substitute the values in for x. x = −5
x=0
25 25 25 1 = = = 2 2 2x 2 ⋅ 25 2 2 ( −5 ) 2.
25 25 25 = = = undefined 2 2 2x 0 2 ( 0)
x=5
25 25 25 1 = = = 2 2 2x 2 ⋅ 25 2 2 ( 5)
PROBLEM: x−4 ; {1/2, 2, 4} 2x −1 ANSWER: Substitute the values in for x. x=
x=2
1 2
1 7 −4 − 2 = 2 = undefined 1 2 ⋅ −1 0 2 3.
2−4 2 =− 2 ( 2) −1 3
x=4
4−4 0 = =0 2 ( 4) −1 7
PROBLEM: 1 ; {−3, 0, 3} 2 x +9 ANSWER: Substitute the values in for x. x = −3
1
( −3)
2
+9
=
1 18
x=0
1 1 = 0 +9 9 2
x=3
1 1 = 3 + 9 18 2
4.
PROBLEM: x+3 ; {−3, 0, 3} x2 − 9 ANSWER: Substitute the values in for x. x = −3
−3 + 3
( −3)
2
−9
=−
x=0
0 =0 18
0+3 3 1 =− =− 0−9 9 3
x=3
3+3 = undefined 32 − 9
State the restrictions to the domain. 5. PROBLEM: 5 x SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x=0 ANSWER: The domain consists of any real number x where x ≠ 0 .
6.
PROBLEM: 1 x ( 3x + 1) SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: x ( 3x + 1) = 0
x = 0 or 3x + 1 = 0 1 x = 0 or x = − 3 1 ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ − . 3
7.
PROBLEM: x+2 x 2 − 25 SOLUTION: x+2 x+2 = 2 x − 25 ( x − 5)( x + 5 ) To find the restrictions to the domain, set the denominator equal to zero and solve: ( x − 5)( x + 5) = 0 x − 5 = 0 or x + 5 = 0 x=5
or x = −5
ANSWER: The domain consists of any real number x where x ≠ ±5 .
8.
PROBLEM: x −1 ( x − 1)( 2 x − 3) SOLUTION: To find the restrictions to the domain, set the denominator equal to zero and solve: ( x − 1)( 2 x − 3) = 0
x − 1 = 0 or 2 x − 3 = 0 3 x = 1 or x = 2 ANSWER: The domain consists of any real number x where x ≠ 1 and x ≠
3 . 2
State the restrictions and simplify.
9.
PROBLEM: x −8 x 2 − 64 SOLUTION: x −8 x −8 1 = = 2 x − 64 ( x − 8)( x + 8) x + 8 To find the restrictions to the domain, set the denominator equal to zero and solve: ( x − 8)( x + 8) = 0 x − 8 = 0 or x + 8 = 0 x=8
or x = −8
ANSWER: The domain consists of any real number x where x ≠ ±8 .
10.
PROBLEM: 3x 2 + 9 x 2 x 3 − 18 x SOLUTION: 3 x ( x + 3) 3x 2 + 9 x 3 = = 3 2 x − 18 x 2 x ( x + 3)( x − 3) 2 ( x − 3) To find the restrictions to the domain, set the denominator equal to zero and solve: 2 x ( x − 3)( x + 3) = 0 x = 0 or x − 3 = 0 or x + 3 = 0 x = 0 or x = 3
or x = −3
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ ±3 .
11.
PROBLEM: x 2 − 5 x − 24 x 2 − 3x − 40 SOLUTION: x 2 − 5 x − 24 ( x − 8 )( x + 3) x + 3 = = x 2 − 3x − 40 ( x − 8 )( x + 5 ) x + 5 To find the restrictions to the domain, set the denominator equal to zero and solve: ( x − 8)( x + 5) = 0 x − 8 = 0 or x + 5 = 0 x=8 or x = −5
ANSWER: The domain consists of any real number x where x ≠ 8 and x ≠ −5 .
12.
PROBLEM: 2x2 + 9x − 5 4x2 −1 SOLUTION: 2 x 2 + 9 x − 5 ( 2 x − 1)( x + 5 ) x+5 = = 2 4x −1 ( 2 x − 1)( 2 x + 1) 2 x + 1 To find the restrictions to the domain, set the denominator equal to zero and solve: ( 2 x − 1)( 2 x + 1) = 0
2 x − 1 = 0 or 2x + 1 = 0 1 1 x= or x = − 2 2 1 ANSWER: The domain consists of any real number x where x ≠ ± . 2
13.
PROBLEM: x 2 − 144 12 − x SOLUTION: x 2 − 144 ( x − 12 )( x + 12 ) = = − x − 12 − ( x − 12 ) 12 − x To find the restrictions to the domain, set the denominator equal to zero and solve: 12 − x = 0 x = 12
ANSWER: The domain consists of any real number x where x ≠ 12 .
14.
PROBLEM: 8 x 2 − 10 x − 3 9 − 4 x2 SOLUTION: 8 x 2 − 10 x − 3 ( 4 x + 1)( 2 x − 3) 4x +1 = =− 2 − ( 2 x − 3)( 2 x + 3) 9 − 4x 2x + 3 To find the restrictions to the domain, set the denominator equal to zero and solve: − ( 2 x − 3)( 2 x + 3)
2 x − 3 = 0 or 2x + 3 = 0 3 3 x= or x = − 2 2 3 ANSWER: The domain consists of any real number x where x ≠ ± . 2
15.
PROBLEM:
Given f ( x) =
x −3 find f ( −3) , f ( 0 ) , and f ( 3) . x2 + 9
ANSWER: Substitute the values in for x. f ( x ) = f ( −3 )
f ( x) = 16.
−3 − 3
( −3)
2
+9
=−
f ( x ) = f (0)
6 1 =− 18 3
f ( x) =
0−3
3 1 =− =− (0) + 9 9 3 2
f ( x ) = f ( 3)
f ( x) =
3−3
( 3)
2
+9
=
0 =0 9
PROBLEM:
Simplify g ( x ) =
x 2 − 2 x − 24 and state the restrictions. 2 x 2 − 9 x − 18
SOLUTION: ( x − 6 )( x + 4 ) = x + 4 x 2 − 2 x − 24 g ( x) = 2 = 2 x − 9 x − 18 ( 2 x + 3)( x − 6 ) 2 x + 3 To find the restrictions to the domain, set the denominator equal to zero and solve: ( 2 x + 3)( x − 6 ) = 0
2 x + 3 = 0 or x − 6 = 0 3 x=− or x = 6 2 ANSWER: The domain consists of any real number x where x ≠ −
3 and x ≠ 6 . 2
7.2 Multiplying and Dividing Rational Expressions Multiply. (Assume all denominators are nonzero.)
17.
PROBLEM: 3x5 x − 3 ⋅ x − 3 9 x2 SOLUTION: 3x5 x − 3 3x5 x3 ⋅ = = x − 3 9 x2 9 x2 3 ANSWER:
18.
x3 3
PROBLEM: ( 2 y − 1) 12 y 2 ⋅ 3 3y y ( 2 y − 1) SOLUTION: ( 2 y − 1) = 12 y 2 = 4 12 y 2 ⋅ y 3 ( 2 y − 1) 3y 3y4 y2 ANSWER:
19.
4 y2
PROBLEM: 3x 2 x 2 − 4 x + 4 ⋅ x−2 5 x3 SOLUTION:
3( x − 2) 3x 2 x 2 − 4 x + 4 3x 2 ( x − 2 ) ⋅ = ⋅ = 3 3 x−2 x − 2 5x 5x 5x 2
ANSWER:
3( x − 2) 5x
20.
PROBLEM: x 2 − 8 x + 15 12 x 2 ⋅ 9 x5 x−3 SOLUTION: x 2 − 8 x + 15 12 x 2 ( x − 3)( x − 5 ) 12 x 2 4 ( x − 5 ) ⋅ = ⋅ = 9 x5 x−3 9 x5 x−3 3x3 ANSWER:
21.
4 ( x − 5) 3x3
PROBLEM: x 2 − 36 2 x 2 + 10 x ⋅ x 2 − x − 30 x 2 + 5 x − 6 SOLUTION: x 2 − 36 2 x 2 + 10 x ( x + 6 )( x − 6 ) 2 x ( x + 5 ) ⋅ = ⋅ x 2 − x − 30 x 2 + 5 x − 6 ( x − 6 )( x + 5 ) ( x + 6 )( x − 1)
= ANSWER:
22.
2x x −1
2x x −1
PROBLEM: 9 x 2 + 11x + 2 9 x − 2 ⋅ 2 4 − 81x 2 ( x + 1) SOLUTION: ( 9 x + 2 )( x + 1) ⋅ 9 x − 2 9 x 2 + 11x + 2 9 x − 2 ⋅ = 2 2 2 4 − 81x ( x + 1) − ( 9 x − 2 )( 9 x + 2 ) ( x + 1) =−
ANSWER: −
1 x +1
1 x +1
Divide. (Assume all denominators are nonzero.)
23.
PROBLEM: 9 x 2 − 25 3 x + 5 ÷ 5 x3 15 x 4 SOLUTION: 9 x 2 − 25 3x + 5 9 x 2 − 25 15 x 4 ( 3x + 5 )( 3x − 5 ) 15 x 4 ÷ = ⋅ = ⋅ 5 x3 15 x 4 5 x3 3x + 5 5 x3 3x + 5 = 3x ( 3x − 5) ANSWER: 3 x ( 3 x − 5 )
24.
PROBLEM: 4x2 2x ÷ 2 4x −1 2x −1 SOLUTION: 4x2 2x 4 x2 2 x −1 4 x2 2x −1 ÷ = ⋅ = ⋅ 2 2 4x −1 2x −1 4x −1 2x ( 2 x − 1)( 2 x + 1) 2 x
=
ANSWER:
25.
2x 2x +1
2x 2x +1
PROBLEM: 3x 2 − 13 x − 10 9 x 2 + 12 x + 4 ÷ 2 x 2 − x − 20 x + 8 x + 16 SOLUTION: 3x 2 − 13x − 10 9 x 2 + 12 x + 4 3 x 2 − 13 x − 10 x 2 + 8 x + 16 ÷ 2 = 2 ⋅ x 2 − x − 20 x + 8 x + 16 x − x − 20 9 x 2 + 12 x + 4
( 3x + 2 )( x − 5) ⋅ ( x + 4 ) = ( x − 5)( x + 4 ) ( 3x + 2 )2 2
= ANSWER:
x+4 3x + 2
x+4 3x + 2
26.
PROBLEM: 2 x 2 + xy − y 2 4x2 − y 2 ÷ x 2 + 2 xy + y 2 3x 2 + 2 xy − y 2 SOLUTION: 2 x 2 + xy − y 2 4 x2 − y 2 2 x 2 + xy − y 2 3x 2 + 2 xy − y 2 ÷ = ⋅ x 2 + 2 xy + y 2 3 x 2 + 2 xy − y 2 x 2 + 2 xy + y 2 4 x2 − y 2
ANSWER:
27.
=
( 2 x − y )( x + y ) ⋅ ( 3x − y )( x + y ) 2 ( 2 x − y )( 2 x + y ) ( x + y)
=
3x − y 2x + y
3x − y 2x + y
PROBLEM: 2 x 2 − 6 x − 20 ÷ 8 x 2 + 39 x − 5 2 8 x + 17 x + 2
(
)
SOLUTION: 2 x 2 − 6 x − 20 2 x 2 − 6 x − 20 1 2 ÷ 8 x + 39 x − 5 = ⋅ 2 2 2 8 x + 17 x + 2 8 x + 17 x + 2 8 x + 39 x − 5
(
)
(
=
ANSWER:
2 ( x − 5)
2 ( x − 5 )( x + 2 )
⋅
1 (8 x − 1)( x + 5)
(8 x + 1)( x + 2 ) 2 ( x − 5) = (8 x + 1)(8 x − 1)( x + 5)
(8 x + 1)(8 x − 1)( x + 5)
)
28.
PROBLEM: 12 x 2 − 27 x 4 ÷ 3x 2 + x − 2 4 3 15 x + 10 x
(
)
SOLUTION: 12 x 2 − 27 x 4 12 x 2 − 27 x 4 1 2 ÷ 3x + x − 2 = ⋅ 4 3 4 3 2 15 x + 10 x 15 x + 10 x 3x + x − 2
(
)
(
3x ( 2 + 3 x )( 2 − 3 x ) 2
=
5 x ( 3x + 2 ) 3
=− ANSWER: −
29.
⋅
)
1 ( 3x − 2 )( x + 1)
3 5 x ( x + 1)
3 5 x ( x + 1)
PROBLEM: 25 y 2 − 1 1 10 y 2 ⋅ ÷ 5 y 4 ( y − 2 ) 5 y − 1 ( y − 2 )2 SOLUTION:
( y − 2 ) = ( 5 y − 1)( 5 y + 1) ⋅ 1 ⋅ ( y − 2 ) 25 y 2 − 1 1 10 y 2 25 y 2 − 1 1 ⋅ ÷ = ⋅ ⋅ 2 4 4 5 y ( y − 2) 5 y −1 ( y − 2) 5 y ( y − 2 ) 5 y − 1 10 y 2 5 y4 ( y − 2) 5 y − 1 10 y 2 2
=
ANSWER:
( 5 y + 1)( y − 2 ) 50 y 6
( 5 y + 1)( y − 2 ) 50 y 6
2
30.
PROBLEM: 10 x 4 5x2 x −1 ÷ ⋅ 2 2 1 − 36 x 6 x − 7 x + 1 2 x SOLUTION: x −1 10 x 4 5x2 10 x 4 6 x 2 − 7 x + 1 x − 1 ÷ ⋅ = ⋅ ⋅ 1 − 36 x 2 6 x 2 − 7 x + 1 2 x 1 − 36 x 2 5x2 2x 4 ( 6 x − 1)( x − 1) ⋅ x − 1 10 x = ⋅ − (1 + 6 x )( 6 x − 1) 5x2 2x
=− ANSWER: −
31.
x ( x − 1)
x ( x − 1)
2
1+ 6x
2
1+ 6x
PROBLEM:
16 x 2 − 9 x 2 + 3x − 10 and g ( x ) = 2 , calculate ( f ⋅ g )( x ) and state x+5 4x + 5x − 6 the restrictions. Given f ( x ) =
SOLUTION: 16 x 2 − 9 x 2 + 3x − 10 f ( x) = and g ( x ) = 2 x+5 4x + 5x − 6 2 2 16 x − 9 x + 3x − 10 ( 4 x + 3)( 4 x − 3) ( x + 5 )( x − 2 ) ⋅ = ⋅ ( f ⋅ g ) ( x) = x + 5 4x2 + 5x − 6 ( x + 5) ( 4 x − 3)( x + 2 )
=
( 4 x + 3)( x − 2 )
x+2 In this case, the domain of f ( x) consists of all real numbers except –5, and the 3 domain of g ( x) consists all real numbers except and –2. 4 ANSWER: ( f ⋅ g ) ( x) =
( 4 x + 3)( x − 2 ) x+2
3 The domain consists of any real number x where x ≠ −5, x = , and x ≠ −2. 4
32.
PROBLEM:
Given f ( x ) =
x+7 x 2 − 49 and g ( x ) = , calculate ( f / g )( x ) and state the 5x − 1 25 x 2 − 5 x
restrictions. SOLUTION: x+7 x 2 − 49 f ( x) = and g ( x ) = 5x − 1 25 x 2 − 5 x x+7 x + 7 25 x 2 − 5 x x + 7 5 x ( 5 x − 1) ⋅ = ⋅ ( f / g ) ( x) = 52x − 1 = x − 49 5 x − 1 x 2 − 49 5 x − 1 ( x + 7 )( x − 7 ) 25 x 2 − 5 x 5x = x−7
In this case, the domain of f ( x) consists of all real numbers except domain of g ( x) consists all real numbers except 0 and reciprocal of g ( x) has a restriction of ±7 . ANSWER: ( f / g ) ( x) =
1 . In addition, the 5
5x x−7
The domain consists of any real number x where x ≠ 0 , x ≠ 7.3 Adding and Subtracting Rational Expressions Simplify. (Assume all denominators are nonzero.)
33.
PROBLEM: 5x 3 − y y SOLUTION: 5x 3 5x − 3 − = y y y ANSWER:
5x − 3 y
1 , and the 5
1 , and x ≠ ±7 . 5
34.
PROBLEM: x 3 − 2 2 x − x−6 x − x−6 SOLUTION: x 3 x −3 x−3 − 2 = 2 = 2 x − x − 6 x − x − 6 x − x − 6 ( x − 3)( x + 2 )
= ANSWER:
35.
1 x+2
1 x+2
PROBLEM: 2x 1 + 2x + 1 x − 5 SOLUTION: LCD = ( 2 x + 1)( x − 5 )
To obtain equivalent terms with this common denominator, we will multiply the x−5 2x +1 first term by . and the second term by x−5 2x +1 2x 1 2x x − 5 1 2 x + 1 2 x 2 − 10 x + 2 x + 1 + = ⋅ + ⋅ = 2x +1 x − 5 2x +1 x − 5 x − 5 2x +1 ( 2 x + 1)( x − 5)
2 x2 − 8x + 1 = ( 2 x + 1)( x − 5) 2 x2 − 8x + 1 ANSWER: ( 2 x + 1)( x − 5)
36.
PROBLEM: 3 1 − 2x + 2 x−7 x SOLUTION: LCD = x 2 ( x − 7 )
To obtain equivalent terms with this common denominator, we will multiply the x2 x−7 . first term by 2 and the second term by x−7 x 3 1− 2x 3 x 2 1 − 2 x x − 7 3x 2 − 2 x 2 + 15 x − 7 + 2 = ⋅ + 2 ⋅ = x−7 x x − 7 x2 x x−7 x2 ( x − 7 )
= ANSWER:
37.
x 2 + 15 x − 7 x2 ( x − 7 )
x 2 + 15 x − 7 x2 ( x − 7 )
PROBLEM: 7x 2 − 2 4x − 9x + 2 x − 2 SOLUTION: 7x 2 7x 2 − = − 2 4 x − 9 x + 2 x − 2 ( 4 x − 1)( x − 2 ) x − 2
LCD = ( 4 x − 1)( x − 2 ) To obtain equivalent terms with this common denominator, we will multiply the 1 4x −1 first term by and the second term by . 1 4x −1 7x 2 7x 1 2 4x −1 − = ⋅ − ⋅ ( 4 x − 1)( x − 2 ) x − 2 ( 4 x − 1)( x − 2 ) 1 x − 2 4 x − 1 =
ANSWER: −
1 4x −1
7 x − 8x + 2 ( 4 x − 1)( x − 2 )
=−
x−2 ( 4 x − 1)( x − 2 )
=−
1 4x −1
38.
PROBLEM: 5 20 − 9 x + 2 x − 5 2 x − 15 x + 25 SOLUTION: 5 20 − 9 x 5 20 − 9 x + 2 = + x − 5 2 x − 15 x + 25 x − 5 ( 2 x − 5 )( x − 5 )
LCD = ( 2 x − 1)( x − 5 ) To obtain equivalent terms with this common denominator, we will multiply the 2x − 5 1 first term by and the second term by . 2x − 5 1 5 20 − 9 x 5 2x − 5 20 − 9 x 1 + = ⋅ + ⋅ x − 5 ( 2 x − 5 )( x − 5 ) x − 5 2 x − 5 ( 2 x − 5 )( x − 5 ) 1
ANSWER:
1 2x − 5
=
10 x − 25 + 20 − 9 x ( 2 x − 5)( x − 5)
=
x −5 ( 2 x − 5)( x − 5)
=
1 2x − 5
39.
PROBLEM: 5 ( x − 3) x 2 − − 2 x − 5 x − 3 x − 8 x + 15 SOLUTION: 5 ( x − 3) 5 ( x − 3) x 2 x 2 − − 2 = − − x − 5 x − 3 x − 8 x + 15 x − 5 x − 3 ( x − 3)( x − 5 )
LCD = ( x + 3)( x − 5 )( x − 3) To obtain equivalent terms with this common denominator, we will multiply the x−3 x −5 1 first term by , and the third term by . , the second term by x−3 x −5 1 5 ( x − 3) 5 ( x − 3) x 2 x x−3 2 x−5 1 − − = ⋅ − ⋅ − ⋅ x − 5 x − 3 ( x + 3)( x − 5 ) x − 5 x − 3 x − 3 x − 5 ( x − 3)( x − 5 ) 1 = =
x 2 − 3x − 2 x + 10 − 5 x + 15 ( x − 3)( x − 5) x 2 − 10 x + 25 ( x − 3)( x − 5)
( x − 5) = ( x − 3)( x − 5) 2
=
ANSWER:
x−5 x−3
x −5 x−3
40.
PROBLEM: x − 4 12 ( 2 − x ) 3x − + 2 x − 1 x + 4 2 x2 + 7 x − 4 SOLUTION: 12 ( 2 − x ) 3x x − 4 12 ( 2 − x ) 3x x−4 − + 2 = − + 2 x − 1 x + 4 2 x + 7 x − 4 2 x − 1 x + 4 ( 2 x − 1)( x + 4 )
LCD = ( 2 x − 1)( x + 4 ) To obtain equivalent terms with this common denominator, we will multiply the x+4 2x −1 1 first term by , and the third term by . , the second term by x+4 2x −1 1 12 ( 2 − x ) 12 ( 2 − x ) 3x x−4 3x x + 4 x − 4 2 x − 1 1 − + = ⋅ − ⋅ + ⋅ 2 x − 1 x + 4 ( 2 x − 1)( x + 4 ) 2 x − 1 x + 4 x + 4 2 x − 1 ( 2 x − 1)( x + 4 ) 1 3x 2 + 12 x − 2 x 2 + 9 x − 4 + 24 − 12 x = ( 2 x − 1)( x + 4 )
ANSWER:
x+5 2x −1
=
x 2 + 9 x + 20 ( 2 x − 1)( x + 4 )
=
( x + 5)( x + 4 ) ( 2 x − 1)( x + 4 )
=
x+5 2x −1
41.
PROBLEM: 1 1 − 2 2 x + 8 x − 9 x + 11x + 18 SOLUTION: 1 1 1 1 − 2 = − 2 x + 8 x − 9 x + 11x + 18 ( x + 9 )( x − 1) ( x + 9 )( x + 2 )
LCD = ( x + 9 )( x + 2 )( x − 1) To obtain equivalent terms with this common denominator, we will multiply the x+2 x −1 first term by . and the second term by x+2 x −1 1 1 1 x+2 1 x −1 − = ⋅ − ⋅ ( x + 9 )( x − 1) ( x + 9 )( x + 2 ) ( x + 9 )( x − 1) x + 2 ( x + 9 )( x + 2 ) x − 1
ANSWER:
=
x + 2 − x +1 ( x + 9 )( x + 2 )( x − 1)
=
3 ( x + 9 )( x + 2 )( x − 1)
3 ( x + 9 )( x + 2 )( x − 1)
42.
PROBLEM: 4 3 + 2 2 x + 13x + 36 x + 6 x − 27 SOLUTION: 4 3 4 3 + 2 = + 2 x + 13x + 36 x + 6 x − 27 ( x + 4 )( x + 9 ) ( x + 9 )( x − 3)
LCD = ( x + 9 )( x − 3)( x + 4 ) To obtain equivalent terms with this common denominator, we will multiply the x−3 x+4 first term by . and the second term by x−3 x+4 4 3 4 3 x−3 x+4 + = ⋅ + ⋅ ( x + 4 )( x + 9 ) ( x + 9 )( x − 3) ( x + 4 )( x + 9 ) x − 3 ( x + 9 )( x − 3) x + 4
ANSWER:
=
4 x − 12 + 3x + 12 ( x + 9 )( x − 3)( x + 4 )
=
7x ( x + 9 )( x − 3)( x + 4 )
7x ( x + 9 )( x − 3)( x + 4 )
43.
PROBLEM: y +1 1 2y − + 2 y+2 2− y y −4 SOLUTION: y +1 1 2y y +1 1 2y − + 2 = + + y + 2 2 − y y − 4 y + 2 y − 2 ( y + 2 )( y − 2 )
LCD = ( y + 2 )( y − 2 ) To obtain equivalent terms with this common denominator, we will multiply the 1 y−2 y+2 first term by and the third term by . , the second term by 1 y−2 y+2 y +1 1 2y y +1 y − 2 1 y+2 2y 1 + + = ⋅ + ⋅ + ⋅ y + 2 y − 2 ( y + 2 )( y − 2 ) y + 2 y − 2 y − 2 y + 2 ( y + 2 )( y − 2 ) 1 =
y2 − y − 2 + y + 2 + 2 y ( y + 2 )( y − 2 )
y2 + 2 y = ( y + 2 )( y − 2 ) = =
ANSWER:
y y−2
y ( y + 2)
( y + 2 )( y − 2 ) y y−2
44.
PROBLEM: 1 1 2 − − 2 y 1 − y y −1 SOLUTION: 1 1 2 1 1 2 − − 2 = + − y 1 − y y − 1 y y − 1 ( y + 1)( y − 1)
LCD = y ( y + 1)( y − 1) To obtain equivalent terms with this common denominator, we will multiply the ( y + 1)( y − 1) , the second term by y ( y + 1) and the third term by first term by y ( y + 1) ( y + 1)( y − 1) y . y 1 1 2 1 ( y + 1)( y − 1) 1 y ( y + 1) 2 y + − = ⋅ + ⋅ − ⋅ y y − 1 ( y + 1)( y − 1) y ( y + 1)( y − 1) y − 1 y ( y + 1) ( y + 1)( y − 1) y y2 −1 + y2 + y − 2 y = y ( y + 1)( y − 1)
ANSWER:
2 y +1 y ( y + 1)
=
2 y2 − y −1 y ( y + 1)( y − 1)
=
( 2 y + 1)( y − 1) y ( y + 1)( y − 1)
=
2 y +1 y ( y + 1)
45.
PROBLEM:
Given f ( x ) =
x +1 x and g ( x ) = , calculate ( f + g )( x ) and state the x +1 2x − 5
restrictions. SOLUTION:
x x +1 + 2x − 5 x + 1 LCD = ( 2 x − 5 )( x + 1) To obtain equivalent terms with this common denominator, we will multiply the x +1 2x − 5 first term by . and the second term by x +1 2x − 5 x +1 x x +1 x +1 x 2x − 5 + + = ⋅ ⋅ 2x − 5 x +1 2x − 5 x +1 x +1 2x − 5 x2 + 2 x + 1 + 2 x2 − 5x = ( 2 x − 5)( x + 1)
( f + g ) ( x) =
=
3x 2 − 3x + 1 ( 2 x − 5)( x + 1)
5 In this case, the domain of f ( x) consists of all real numbers except , and the 2 g ( x ) consists all real numbers except –1. domain of ANSWER: ( f + g ) ( x) =
3x 2 − 3x + 1 ( 2 x − 5)( x + 1)
The domain consists of any real number x where x ≠
5 and x ≠ −1. 2
46.
PROBLEM:
Given f ( x ) =
x +1 2 and g ( x ) = , calculate ( f − g )( x ) and state the x −8 3x
restrictions. SOLUTION: 2 x +1 − ( f − g ) ( x) = 3x x − 8 LCD = 3x ( x − 8)
To obtain equivalent terms with this common denominator, we will multiply the x+8 3x first term by and the second term by . x+8 3x x +1 2 x +1 2 − = ⋅ 3x ( x − 8) − ⋅ 3x ( x − 8) 3x x − 8 3x x −8 x2 − 7 x − 8 − 6 x = 3x ( x − 8) x 2 − 13 x − 8 3x ( x − 8) In this case, the domain of f ( x) consists of all real numbers except 0 , and the domain of g ( x) consists all real numbers except 8. =
x 2 − 13x − 8 3x ( x − 8 ) The domain consists of any real number x where x ≠ 0 and x ≠ 8. ANSWER: ( f − g ) ( x) =
7.4 Complex Fractions Simplify.
47.
PROBLEM: 2 4− x 2x −1 3x SOLUTION: 2 4x − 2 4− 4 x − 2 3x x = x = ⋅ 2x −1 2x −1 2x −1 x 3x 3x 6 ( 2 x − 1) = 2x −1 =6 ANSWER: 6
48.
PROBLEM: 1 1 − 3 3y 1 1 − 5 5y SOLUTION: 1 1 y −1 − y −1 5 y 5 3 3y 3y = = ⋅ = 1 1 y −1 3 y y −1 3 − 5 5y 5y ANSWER:
5 3
49.
PROBLEM: 1 1 + 6 x 1 1 − 36 x 2 SOLUTION: 1 1 x+6 + 2 6 x = 6 x = x + 6 ⋅ 36 x 2 1 1 6 x x 2 − 36 − 2 x − 36 36 x 36 x 2 36 x 2 x+6 = ⋅ 6 x ( x + 6 )( x − 6 ) =
6x x−6
ANSWER:
50.
6x x−6
PROBLEM: 1 1 − 2 100 x 1 1 − 10 x SOLUTION: 1 1 x 2 − 100 − 2 2 100 x = 100 x 2 = x − 100 ⋅ 10 x 1 1 x − 10 100 x 2 x − 10 − 10 x 10 x ( x − 10 )( x + 10 ) ⋅ 10 x = 100 x 2 x − 10 x + 10 = 10 x ANSWER:
x + 10 10 x
51.
PROBLEM: x 2 − x + 3 x +1 x 1 + x+4 x+3 SOLUTION:
x2 + x − 2x − 6 x 2 x x +1 2 x+3 − ⋅ − x + 3 x + 1 = x + 3 x + 1 x + 1 x + 3 = ( x + 3)( x + 1) x 1 x x+3 1 x + 4 x 2 + 3x + x + 4 + + ⋅ x+4 x+3 x + 4 x + 3 x + 3 x + 4 ( x + 4 )( x + 3) x2 − x − 6 ( x + 3)( x + 1) = x2 + 4 x + 4 ( x + 4 )( x + 3) =
x 2 − x − 6 ( x + 4 )( x + 3) ⋅ ( x + 3)( x + 1) x 2 + 4 x + 4
( x − 3)( x + 2 ) ⋅ ( x + 4 )( x + 3) ( x + 3)( x + 1) ( x + 2 )2 ( x + 4 )( x − 3) = ( x + 1)( x + 2 ) =
ANSWER:
( x + 4 )( x − 3) ( x + 1)( x + 2 )
52.
PROBLEM: 3 1 − x x −5 5 2 − x+2 x SOLUTION:
3 ( x − 5) 1 x 3 x − 15 − x 3 1 ⋅ − ⋅ − x x − 5 = x ( x − 5) x − 5 x = x ( x − 5) 5 2 5 x 2 x+2 5x − 2 x − 4 − ⋅ − ⋅ x+2 x x+2 x x x+2 x ( x + 2) 2 x − 15 x ( x − 5) = 3x − 4 x ( x + 2) =
2 x − 15 x ( x + 2 ) ⋅ x ( x − 5) 3x − 4
=
( 2 x − 15)( x + 2 ) ( x − 5)( 3x − 4 )
ANSWER:
( 2 x − 15)( x + 2 ) ( x − 5)( 3x − 4 )
53.
PROBLEM: 12 35 1− + 2 x x 25 1− 2 x SOLUTION: ⎛ 12 35 ⎞ 2 12 35 1− + 2 ⎜1 − + 2 ⎟ ⋅ x x 2 − 12 x + 35 x x ⎠ x x = ⎝ = 25 x 2 − 25 ⎛ 25 ⎞ 2 1− 2 1 − ⋅ x ⎜ 2 ⎟ x ⎝ x ⎠ ( x − 7 )( x − 5) = ( x + 5)( x − 5) =
ANSWER:
54.
x−7 x+5
x−7 x+5
PROBLEM: 15 25 2− + 2 x x 2x − 5 SOLUTION: 15 25 ⎛ 15 25 ⎞ 2 2 − + 2 ⎜2 − + 2 ⎟⋅ x 2 x 2 − 15 x + 25 x x ⎠ x x =⎝ = 2x − 5 x 2 ( 2 x − 5) ( 2 x − 5) ⋅ x2
=
( 2 x − 5)( x − 5) x 2 ( 2 x − 5)
=
x −5 x2
ANSWER:
x−5 x2
7.5 Solving Rational Equations Solve.
55.
PROBLEM: 6 2 = x − 6 2x −1
1 . 2 Multiply both sides with the LCD, ( x − 6 )( 2 x − 1) .
SOLUTION: We first note that x ≠ 6 and x ≠
6 2 ⋅ ( x − 6 )( 2 x − 1) = ⋅ ( x − 6 )( 2 x − 1) x−6 2x −1 12 x − 6 = 2 x − 12 10 x = −6 6 x=− 10 3 x=− 5 3 We can check our answer by substituting − in for x to see if we obtain a true 5 statement. 6 2 = 3 3 − − 6 2 ⋅ ⎛⎜ − ⎞⎟ − 1 5 ⎝ 5⎠ 10 10 − =− 11 11 3 ANSWER: The solution is − . 5
56.
PROBLEM: x x+2 = x−6 x−2 SOLUTION: We first note that x ≠ 6 and x ≠ 2 . Multiply both sides with the LCD, ( x − 6 )( x − 2 ) .
x x+2 ⋅ ( x − 6 )( x − 2 ) = ⋅ ( x − 6 )( x − 2 ) x−6 x−2 x 2 − 2 x = x 2 − 4 x − 12 2 x = −12 x = −6 We can check our answer by substituting −6 in for x to see if we obtain a true statement. −6 −6 + 2 = −6 − 6 −6 − 2 −6 −4 = −12 −8 1 1 = 2 2 ANSWER: The solution is −6 .
57.
PROBLEM: 1 2 1 − = 3x 9 x SOLUTION: We first note that x ≠ 0 . Multiply both sides with the LCD, 9x . 1 ⎛ 1 2⎞ ⎜ − ⎟ 9x = ⋅ 9x x ⎝ 3x 9 ⎠ 3 − 2x = 9
2 x = −6 x = −3 We can check our answer by substituting −3 in for x to see if we obtain a true statement. 1 2 1 − =− 3 ( −3) 9 3 3 1 − =− 9 3 1 1 − =− 3 3 ANSWER: The solution is −3 .
58.
PROBLEM: 2 3 1 + = x −5 5 x −5 SOLUTION: We first note that x ≠ 5 . Multiply both sides with the LCD, 5 ( x − 5 ) .
3⎞ 1 ⎛ 2 + ⎟ 5 ( x − 5) = ⋅ 5 ( x − 5) ⎜ x −5 ⎝ x−5 5⎠ 10 + 3x − 15 = 5 3x = 10 10 x= 3 We can check our answer by substituting statement. 2 3 1 + = 10 10 −5 5 −5 3 3 6 3 3 − + =− 5 5 5 3 3 − =− 5 5 ANSWER: The solution is
10 . 3
10 in for x to see if we obtain a true 3
59.
PROBLEM: x 4 10 + =− 2 x −5 x +5 x − 25 SOLUTION: x 4 10 + =− 2 x −5 x +5 x − 25 x 4 10 + =− x −5 x +5 ( x − 5)( x + 5)
We first note that x ≠ ±5 . Multiply both sides with the LCD, ( x − 5 )( x + 5 ) .
4 ⎞ 10 ⎛ x + ⋅ ( x − 5)( x + 5) ⎜ ⎟ ( x − 5)( x + 5) = − ( x − 5)( x + 5) ⎝ x−5 x +5⎠ x 2 + 5 x + 4 x − 20 = −10 x 2 + 9 x − 10 = 0
( x − 1)( x + 10) = 0 x = 1 or x = −10 Check x = 1 1 4 10 + =− 1− 5 1+ 5 (1 − 5)(1 + 5) 4 1 5 − = 6 4 12 8−3 5 = 12 12 5 5 = 12 12 ANSWER: The solution is 1 or –10.
Check x = −10 −10 4 10 + =− −10 − 5 −10 + 5 ( −10 − 5 )( −10 + 5 ) 2 4 2 − =− 3 5 15 2 2 − =− 15 15
60.
PROBLEM: 2 12 2 − 3x 2 − = 2 x 2 x + 3 2 x + 3x SOLUTION: 2 12 2 − 3x 2 − = 2 x 2 x + 3 2 x + 3x 2 12 2 − 3x 2 − = x 2 x + 3 x ( 2 x + 3)
3 We first note that x ≠ 0 and x ≠ − . 2 Multiply both sides with the LCD, x ( 2 x + 3) .
12 ⎞ 2 − 3x 2 ⎛2 ⋅ x ( 2 x + 3) ⎜ − ⎟ x ( 2 x + 3) = x ( 2 x + 3) ⎝ x 2x + 3 ⎠ 4 x + 6 − 12 x = 2 − 3x 2 3x 2 − 8 x + 4 = 0
( 3x − 2 )( x − 2) = 0 x=
2 or x = 2 3 Check x =
2 3 2
⎛2⎞ 2 − 3⎜ ⎟ 2 12 ⎝3⎠ − = 2 2 2⎛ 2 ⎞ 2⋅ + 3 2 ⋅ + 3⎟ ⎜ 3 3 3⎝ 3 ⎠ 36 1 3− = 13 13 3 3 3 = 13 13
ANSWER: The solution is
2 or 2. 3
Check x = 2 2 12 2 − 3 ⋅ 22 − = 2 2 ⋅ 2 + 3 2 ( 2 ⋅ 2 + 3) 12 10 =− 7 14 5 5 − =− 7 7
1−
61.
PROBLEM: x +1 x−6 + =1 2 ( x − 2) x SOLUTION: We first note that x ≠ 0 and x ≠ 2 . Multiply both sides with the LCD, 2 x ( x − 2 ) .
⎛ x +1 x−6⎞ + ⎜⎜ ⎟ 2 x ( x − 2 ) = 1⋅ 2 x ( x − 2 ) x ⎟⎠ ⎝ 2 ( x − 2) x 2 + x + 2 x 2 − 16 x + 24 = 2 x 2 − 4 x x 2 − 11x + 24 = 0
( x − 8)( x − 3) = 0 x = 8 or x = 3 Check x = 8 8 +1 8−6 + =1 2 (8 − 2 ) 8 9 1 + =1 12 4 1=1
ANSWER: The solution is 8 or 3.
Check x = 3 3 +1 3−6 + =1 2 (3 − 2) 3 2 −1 = 1 1=1
62.
PROBLEM: x 5x + 2 − =4 x +1 x + 4 SOLUTION: We first note that x ≠ −1 and x ≠ −4 . Multiply both sides with the LCD, ( x + 1)( x + 4 ) .
x ⎞ ⎛ 5x + 2 − ⎜ ⎟ ( x + 1)( x + 4 ) = 4 ⋅ ( x + 1)( x + 4 ) ⎝ x +1 x + 4 ⎠ 5 x 2 + 22 x + 8 − x 2 − x = 4 x 2 + 20 x + 16 x =8 We can check our answer by substituting 8 in for x to see if we obtain a true statement. 5⋅8 + 2 8 − =4 8 +1 8 + 4 14 2 − =4 3 3 4=4 ANSWER: The solution is 8 .
63.
PROBLEM: x 1 4x − 7 + = 2 x + 5 x − 4 x + x − 20 SOLUTION: x 1 4x − 7 + = x + 5 x − 4 ( x + 5)( x − 4 ) We first note that x ≠ −5 and x ≠ 4 . Multiply both sides with the LCD, ( x + 5 )( x − 4 ) .
1 ⎞ 4x − 7 ⎛ x + ⋅ ( x + 5)( x − 4 ) ⎜ ⎟ ⋅ ( x + 5)( x − 4 ) = ( x + 5)( x − 4) ⎝ x+5 x−4⎠ x2 − 4x + x + 5 = 4 x − 7 x 2 − 7 x + 12 = 0
( x − 3)( x + 4) = 0 x = 3 or x = −4 x ≠ 4. ANSWER: Hence the only solution is 3.
We can check our answer by substituting 3 in for x to see if we obtain a true statement. 3 1 4⋅3 − 7 + = 3 + 5 3 − 4 ( 3 + 5 )( 3 − 4 )
3 5 −1 = − 8 8 5 5 − =− 8 8
64.
PROBLEM: 2 (3 − 4x ) x 2 + = 2 3x − 1 2 x + 1 6 x + x − 1 SOLUTION: 2 (3 − 4x ) 2 x + = 2 3x − 1 2 x + 1 6 x + x − 1 2 (3 − 4x ) 2 x + = 3x − 1 2 x + 1 ( 3 x − 1)( 2 x + 1)
1 1 and x ≠ − . 3 2 Multiply both sides with the LCD, ( 3 x − 1)( 2 x + 1) . We first note that x ≠
2 (3 − 4x ) x ⎞ ⎛ 2 + ( 3x − 1)( 2 x + 1) ⎜ ⎟ ( 3x − 1)( 2 x + 1) = ( 3x − 1)( 2 x + 1) ⎝ 3x − 1 2 x + 1 ⎠ 4 x + 2 + 3x 2 − x = 6 − 8 x
3x 2 + 11x − 4 = 0
( x + 4)( 3x − 1) = 0 x = −4 or x =
1 3
1 x≠ . 3 ANSWER: Hence the only solution is –4. We can check our answer by substituting −4 in for x to see if we obtain a true statement. 2 ( 3 − 4 ( −4 ) ) ( −4 ) = 2 + 3 ( −4 ) − 1 2 ( −4 ) + 1 ( 3 ( −4 ) − 1) ( 2 ( −4 ) + 1)
2 4 2 ⋅19 + = 13 7 13 ⋅ 7 38 38 = 91 91
−
65.
PROBLEM: x 1 2x + = 2 x −1 x + 1 x −1 SOLUTION: x 1 2x + = 2 x −1 x + 1 x −1 1 2x x + = x − 1 x + 1 ( x + 1)( x − 1)
We first note that x ≠ ±1 . Multiply both sides with the LCD, ( x + 1)( x − 1) .
1 ⎞ 2x ⎛ x + ( x + 1)( x − 1) ⎜ ⎟ ( x + 1)( x − 1) = ( x + 1)( x − 1) ⎝ x −1 x +1 ⎠ x2 + x + x −1 = 2x x2 −1 = 0
( x + 1)( x − 1) = 0 x = −1 or x = 1 x ≠ ±1 . ANSWER: Hence, the solution is∅.
66.
PROBLEM: 2x 1 4 − 7x − = 2 x + 5 2 x − 3 2 x + 7 x − 15 SOLUTION: 2x 1 4 − 7x − = 2 x + 5 2 x − 3 2 x + 7 x − 15 2x 1 4 − 7x − = x + 5 2 x − 3 ( x + 5 )( 2 x − 3)
3 . 2 Multiply both sides with the LCD, ( x + 5 )( 2 x − 3) . We first note that x ≠ −5 and x ≠
1 ⎞ 4 − 7x ⎛ 2x − ⋅ ( x + 5)( 2 x − 3) ⎜ ⎟ ( x + 5)( 2 x − 3) = ( x + 5)( 2 x − 3) ⎝ x + 5 2x − 3 ⎠ 4x2 − 6x − x − 5 = 4 − 7 x 4x2 − 9 = 0
( 2 x + 3)( 2 x − 3) = 0 x=± x≠
3 2
3 . 2
3 ANSWER: Hence the only solution is − . 2 We can check our answer by substituting −
3 in for x to see if we obtain a true 2
statement. ⎛ 3⎞ ⎛ 3⎞ 2⎜ − ⎟ 4 − 7⎜− ⎟ 1 ⎝ 2⎠ − ⎝ 2⎠ = 3 3 − + 5 2 ⎛⎜ − ⎞⎟ − 3 ⎜⎛ ⎛⎜ − 3 ⎞⎟ + 5 ⎟⎞ ⎜⎛ 2 ⎛⎜ − 3 ⎞⎟ − 3 ⎟⎞ 2 ⎝ 2⎠ ⎝⎝ 2 ⎠ ⎠⎝ ⎝ 2 ⎠ ⎠ 29 6 1 − + =− 2 7 6 21 29 29 − =− 42 42
67.
PROBLEM: 1 1 1 Solve for a: = + a b c SOLUTION: Multiply both the sides with the LCD, abc. 1 ⎛1 1⎞ ⋅ abc = ⎜ + ⎟ ⋅ abc a ⎝b c⎠ bc = ac + ab
bc = a ( c + b ) a=
bc c+b
ANSWER: a =
68.
PROBLEM:
Solve for y: x =
bc c+b 2 y −1 3y
SOLUTION: 2 y −1 x= 3y 3xy = 2 y − 1 2 y − 3xy = 1 y ( 2 − 3x ) = 1 y=
1 2 − 3x
ANSWER: y =
1 2 − 3x
7.6 Applications of Rational Equations Use algebra to solve the following applications.
69.
PROBLEM: A positive integer is twice that of another. The sum of the reciprocals of the two positive integers is 1/4. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: 2n 1 1 1 + = n 2n 4 We first note that n ≠ 0. Multiply both sides with the LCD, 2n. 1 ⎛1 1 ⎞ ⎜ + ⎟ 2n = ⋅ 2n 4 ⎝ n 2n ⎠ n 2 +1 = 2 n=6 Larger integer: 2n = 2 ⋅ 6 = 12 ANSWER: n = 6 , 2n = 12
70.
PROBLEM: If the reciprocal of the smaller of two consecutive integers is subtracted from three times the reciprocal of the larger, the result is 3/10. Find the integers. SOLUTION: Smaller integer is represented by n. Larger integer: n + 1 3 1 3 − = n + 1 n 10 We first note that n ≠ 0 and n ≠ 1 . Multiply both the sides with the LCD, 10n ( n + 1) .
1⎞ 3 ⎛ 3 − ⎟10n ( n + 1) = ⋅10n ( n + 1) ⎜ 10 ⎝ n +1 n ⎠ 2 30n − 10n − 10 = 3n + 3n 3n 2 − 17 n + 10 = 0
( 3n − 2 )( n − 5) = 0 2 or n = 5 3 The problem speaks about integers. Hence, the solution is 5. Larger integer: n + 1 = 5 +1 = 6 n=
ANSWER: The two integers are 5 and 6.
71.
PROBLEM: Mary can jog, on average, 2 mph faster than her husband James. James can jog 6.6 miles in the same amount of time it takes Mary to jog 9 miles. How fast, on average, can Mary jog? SOLUTION: James’ speed is represented by x. Mary’ speed: x + 2 D 9 Time taken by Mary to cover 9 miles: t = = r x+2 D 6.6 Time taken by James to cover 6.6 miles: t = = r x
Time taken by Mary to cover 9 miles = Time taken by James to cover 6.6 miles 9 6.6 = x+2 x We first note that x ≠ 0 and x ≠ −2 . Multiply both the sides with the LCD, x ( x + 2 ) . 9 x = 6.6 ( x + 2 ) 9 x = 6.6 x + 13.2 90 x = 66 x + 132 24 x = 132 x = 5.5
ANSWER: Mary’s speed: 5.5 + 2 = 7.5 miles/hour
72.
PROBLEM: Billy traveled 140 miles to visit his grandmother on the bus and then drove the 140 miles back in a rental car. On average, the bus was 14 mph slower than the car. If the total time spent traveling was 4.5 hours, then what was the average speed of the bus? SOLUTION: Speed of the rental car is represented by x. Speed of the bus: x – 14 D 140 Time taken by the bus to cover 140 miles: t = = r x − 14 D 140 Time taken by the rental car to cover 140 miles: t = = r x
Time taken by the bus + Time taken by the rental car to to cover 140 miles cover 140 miles 140 140 + = 4.5 x − 14 x We first note that x ≠ 14 and x ≠ 0 . Multiply both the sides with the LCD, x ( x − 14 ) .
= 4.5 hours
⎛ 140 140 ⎞ + ⎜ ⎟ x ( x − 14 ) = 4.5 ⋅ x ( x − 14 ) x ⎠ ⎝ x − 14 140 x + 140 x − 196 = 4.5 x 2 − 63 x 4.5 x 2 − 343 x + 1960 = 0 45 x 2 − 3430 x + 19600 = 0
( x − 70 )( 9 x − 56 ) = 0 56 9 Consider the whole number to be the solution. x = 70 or x =
ANSWER: Speed of the bus: x – 14 = 70 – 14 = 56 miles/hour
73.
PROBLEM: Jerry takes twice as long as Manny to assemble a skateboard. If they work together, they can assemble a skateboard in 6 minutes. How long would it take Manny to assemble the skateboard without Jerry’s help? SOLUTION: If Manny can assemble a skateboard in x minutes, Manny’s work 1 rate is . x 1 Jerry can assemble the skateboard in 2x minutes. Hence, Jerry’s work rate is . 2x If they work together, they can assemble the skateboard in 6 minutes. Multiply the individual work rates by 6. The sum of these two products is equal to 1. 1 1 ⋅6 + ⋅6 =1 x 2x We first note that x ≠ 0. 1 1 ⋅6 + ⋅6 =1 x 2x 6 3 + =1 x x 9 =1 x x=9 ANSWER: Manny can assemble the skateboard in 9 minutes.
74.
PROBLEM: Working alone, Joe can complete the yard work in 30 minutes. It takes Mike 45 minutes to complete the same yard. How long would it take them working together? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 30 minutes; t2 = 45 minutes 1 1 ⋅t + ⋅t = 1 30 45 t t + =1 30 45 t ⎞ ⎛ t ⎜ + ⎟ ⋅ 90 = 1⋅ 90 ⎝ 30 45 ⎠ 3t + 2t = 90 5t = 90 t = 18
ANSWER: It would take Joe and Manny to build the yard together in 18 minutes. 7.7 Variation Construct a mathematical model given the following:
75.
PROBLEM: y varies directly x and y = 12 when x = 4. SOLUTION: y = kx Substitute the values of x and y in the expression above. y = kx 12 = k 4 k =3 ANSWER: The expression becomes y = 3x.
76.
PROBLEM: y varies inversely as x and y = 2 when x = 5.
k x Substitute the values of x and y in the expression above. k y= x k 2= 5 k = 10 SOLUTION: y =
ANSWER: The expression becomes y =
77.
10 . x
PROBLEM: y is jointly proportional to x and z where y = 36 when x = 3 and z = 4. SOLUTION: y = kxz Substitute the values of x, y, and z in the expression above. y = kxz 36 = k ⋅ 3 ⋅ 4 k =3 ANSWER: The expression becomes y = 3xz .
78.
PROBLEM: y is directly proportional to the square of x and inversely proportional to z, where y = 20 when x = 2 and z = 5.
x2 z Substitute the values of x, y, and z in the expression above. x2 y=k z 22 20 = k 5 20 ⋅ 5 k= = 25 4 25x 2 ANSWER: The expression becomes y = . z SOLUTION: y = k
79.
PROBLEM: The distance an object in freefall falls varies directly with the time that it has been falling. It is observed that an object falls 16 feet in 1 second, find an equation that models the distance an object will fall and use it to determine how far it will fall in 2 seconds. SOLUTION: Distance an object falls in freefall is represented by d. Time of fall is represented by t. d = kt It is observed that an object falls 16 feet in 1 second. d = kt 16 = k ⋅1 k = 16 The formula is d = 16t . ANSWER: In 2 seconds the object falls at a distance of , d = 16 ⋅ 2 = 32 feet.
80.
PROBLEM: The weight of an object varies inversely as the square of its distance from the center of the Earth. If an object weighs 180 lbs on the surface of the Earth (approximately 4000 miles from the center), then how much will it weigh at 2000 miles above the Earth’s surface? SOLUTION: The weight of the object is represented by w. The distance from the center of the earth is represented by d. k w= 2 d An object weighs 180 lbs on the surface of the Earth (approximately 4000 miles from the center). k w= 2 d k 180 = 40002 k = 288 × 107 Weight of the object 2000 miles above the Earth’s surface, i.e. 4000 + 2000 = k 6000 miles: w = 2 d 7 288 ×10 w= = 80 60002 ANSWER: Weight of the object 2000 miles above the Earth’s surface is 80 lb.
Chapter 7 Sample Exam Simplify and state the restrictions.
1.
PROBLEM: 2 15 x3 ( 3x − 1)
3x ( 3x − 1)
SOLUTION: 2 15 x3 ( 3x − 1)
= 5 x 2 ( 3x − 1)
3x ( 3x − 1) To find the restrictions to the domain, set the denominator equal to zero and solve: 3x ( 3x − 1) = 0 x = 0 or 3x − 1 = 0 1 x = 0 or x = 3
ANSWER: 5 x 2 ( 3 x − 1)
The domain consists of any real number x where x ≠ 0 and x ≠ 2.
1 . 3
PROBLEM: x 2 − 144 x 2 + 12 x SOLUTION: x 2 − 144 ( x + 12 )( x − 12 ) x − 12 = = x 2 + 12 x x ( x + 12 ) x To find the restrictions to the domain, set the denominator equal to zero and solve: x ( x + 12 ) = 0 x = 0 or x + 12 = 0 x = 0 or x = −12
ANSWER: The domain consists of any real number x where x ≠ 0 and x ≠ −12 .
3.
PROBLEM: x 2 + x − 12 2x2 + 7 x − 4 SOLUTION: ( x + 4 )( x − 3) = x − 3 x 2 + x − 12 = 2 2 x + 7 x − 4 ( 2 x − 1)( x + 4 ) 2 x − 1 To find the restrictions to the domain, set the denominator equal to zero and solve: ( 2 x − 1)( x + 4 ) = 0
2 x − 1 = 0 or x + 4 = 0 1 x = or x = −4 2 ANSWER:
x −3 2x −1
The domain consists of any real number x where x ≠ 4.
1 and x ≠ −4 . 2
PROBLEM: 9 − x2
( x − 3)
2
SOLUTION: ( x − 3)( x + 3) = − x + 3 9 − x2 =− 2 2 x−3 ( x − 3) ( x − 3) To find the restrictions to the domain, set the denominator equal to zero and solve:
( x − 3)
2
=0
x −3 = 0 x=3
ANSWER: The domain consists of any real number x where x ≠ 3 .
Simplify. (Assume all variables in the denominator are positive.)
5.
PROBLEM: 5x x −5 ⋅ 2 x − 25 25 x 2 SOLUTION: 5x x −5 5x x−5 1 ⋅ = ⋅ = 2 2 2 x − 25 25 x ( x − 5)( x + 5) 25 x 5 x ( x + 5) ANSWER:
6.
1 5x ( x + 5)
PROBLEM: x 2 + x − 6 3x 2 − 5 x − 2 ⋅ x2 − 4x + 4 x2 − 9 SOLUTION: x 2 + x − 6 3x 2 − 5 x − 2 ( x + 3)( x − 2 ) ( 3 x + 1)( x − 2 ) 3 x + 1 ⋅ = ⋅ = 2 x2 − 4x + 4 x2 − 9 ( x + 3)( x − 3) x − 3 ( x − 2) ANSWER:
7.
3x + 1 x −3
PROBLEM: x 2 − 4 x − 12 x − 6 ÷ 12 x 2 6x SOLUTION: ( x − 6 )( x + 2 ) ⋅ 6 x x 2 − 4 x − 12 x − 6 x 2 − 4 x − 12 6 x ÷ = ⋅ = 2 2 12 x 6x 12 x 12 x 2 x−6 x−6 x+2 = 2x x+2 ANSWER: 2x
8.
PROBLEM: 2 x2 − 7 x − 4 2 x2 + 7 x + 3 ÷ 6 x 2 − 24 x 10 x 2 + 30 x SOLUTION: 2 x 2 − 7 x − 4 2 x 2 + 7 x + 3 2 x 2 − 7 x − 4 10 x 2 + 30 x ÷ = ⋅ 6 x 2 − 24 x 10 x 2 + 30 x 6 x 2 − 24 x 2 x 2 + 7 x + 3 ( 2 x + 1)( x − 4 ) ⋅ 10 x ( x + 3) = 6x ( x − 4) ( 2 x + 1)( x + 3)
= ANSWER:
9.
5 3
5 3
PROBLEM: 1 1 + x −5 x +5 SOLUTION: LCD = ( x − 5 )( x + 5 )
To obtain equivalent terms with this common denominator, we will multiply the x+5 x−5 and the second term by first term by . x+5 x−5 1 1 1 x+5 1 x −5 + = ⋅ + ⋅ x −5 x +5 x −5 x +5 x +5 x −5 x +5+ x −5 = ( x − 5)( x + 5)
=
ANSWER:
2x ( x − 5)( x + 5)
2x ( x − 5)( x + 5)
10.
PROBLEM: x 8 12 x − − 2 x +1 2 − x x − x − 2 SOLUTION: x 8 12 x x 8 12 x − − 2 = + − x + 1 2 − x x − x − 2 x + 1 x − 2 ( x − 2 )( x + 1) LCD = ( x − 2 )( x + 1)
To obtain equivalent terms with this common denominator, we will multiply the x−2 x +1 1 first term by , and the third term by . , the second term by x−2 1 x +1 x 8 12 x x x−2 8 x +1 12 x 1 + − = ⋅ + ⋅ − ⋅ x + 1 x − 2 ( x − 2 )( x + 1) x + 1 x − 2 x − 2 x + 1 ( x − 2 )( x + 1) 1
=
x 2 − 2 x + 8 x + 8 − 12 x ( x − 2 )( x + 1)
x2 − 6 x + 8 = ( x − 2 )( x + 1)
ANSWER:
x−4 x +1
=
( x − 2 )( x − 4 ) ( x − 2 )( x + 1)
=
x−4 x +1
11.
PROBLEM: 1 1 + y x 1 1 − 2 2 y x SOLUTION: 1 1 1 x 1 y x+ y + ⋅ + ⋅ y x y x x y xy = = 2 2 2 1 1 1 x 1 y x − y2 − ⋅ − ⋅ y 2 x2 y 2 x2 x2 y 2 x2 y2 x+ y xy = ( x + y )( x − y ) x2 y2 =
x+ y x2 y2 ⋅ xy ( x + y )( x − y )
=
xy x− y
ANSWER:
xy x− y
12.
PROBLEM: 6 9 1− + 2 x x 5 3 2− − 2 x x SOLUTION: ⎛ 6 9 ⎞ 2 6 9 1 − + 2 ⎜1 − + 2 ⎟ x 2 x x = ⎝ x x ⎠ = x − 6x + 9 2 5 3 5 3 2 − − 2 ⎜⎛ 2 − − 2 ⎟⎞ x 2 2 x − 5 x − 3 x x x x ⎠ ⎝
( x − 3) = ( 2 x + 1)( x − 3) 2
=
x −3 2x +1
ANSWER:
x −3 2x +1
13.
PROBLEM:
Given f ( x ) =
x 2 − 81
( 4 x − 3)
2
and g ( x ) =
4x − 3 , calculate ( f ⋅ g )( x ) and state the x −9
restrictions. SOLUTION: x 2 − 81 4 x − 3 ( x − 9 )( x + 9 ) 4 x − 3 ( ) ⋅ f g x ⋅ = ⋅ = ( ) 2 2 x −9 ( 4 x − 3) x − 9 ( 4 x − 3)
=
x+9 4x − 3
3 In this case, the domain of f ( x) consists of all real numbers except , and the 4 domain of g ( x) consists all real numbers except 9. ANSWER: ( f ⋅ g ) ( x) =
x+9 4x − 3
The domain consists of any real number x where x ≠
3 and x ≠ 9. 4
14.
PROBLEM:
Given f ( x ) =
x 1 and g ( x ) = , calculate ( f − g )( x ) and state the x −5 3x − 5
restrictions. SOLUTION:
x 1 − x − 5 3x − 5 LCD = ( x − 5)( 3x − 5) To obtain equivalent terms with this common denominator, we will multiply the 3x − 5 x−5 first term by . and the second term by 3x − 5 x−5 x x 3x − 5 x−5 1 1 − − = ⋅ ⋅ x − 5 3x − 5 x − 5 3x − 5 3x − 5 x − 5 3x 2 − 5 x − x + 5 = ( x − 5 )( 3x − 5)
( f − g ) ( x) =
3x 2 − 6 x + 5 ( x − 5)( 3x − 5) In this case, the domain of f ( x) consists of all real numbers except 5 , and the 5 domain of g ( x) consists all real numbers except . 3 5 ANSWER: The domain consists of any real number x where x ≠ 5 and x ≠ . 3 =
Solve.
15.
PROBLEM: 1 1 + =2 3 x SOLUTION: We first note that x ≠ 0. Multiplying both the side with the LCD, 3x. ⎛1 1⎞ ⎜ + ⎟ 3x = 2 ⋅ 3x ⎝3 x⎠ x + 3 = 6x
5x = 3 3 x= 5 We can check our answer by substituting statement. 1 1 + =2 3 3 5 1 5 + =2 3 3 1+ 5 =2 3 2=2 3 ANSWER: The solution is . 5
3 in for x to see if we obtain a true 5
16.
PROBLEM: 1 3 = x − 5 2x − 3
3 SOLUTION: We first note that x ≠ 5 and x ≠ . 2 Multiplying both the sides with the LCD, ( x − 5 )( 2 x − 3) .
⎛ 1 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ( x − 5 )( 2 x − 3) = ⎜ ⎟ ( x − 5 )( 2 x − 3) ⎝ x−5⎠ ⎝ 2x − 3 ⎠ 2 x − 3 = 3x − 15 x = 12 We can check our answer by substituting 12 in for x to see if we obtain a true statement. 1 3 = 12 − 5 2 ⋅12 − 3 1 3 = 7 21 1 1 = 7 7 ANSWER: The solution is 12 .
17.
PROBLEM: 9 20 1− + 2 = 0 x x SOLUTION: We first note that x ≠ 0 . Multiply both the sides with the LCD, x2. ⎛ 9 20 ⎞ 2 2 ⎜1 − + 2 ⎟ x = 0 ⋅ x ⎝ x x ⎠ x 2 − 9 x + 20 = 0
( x − 4)( x − 5) = 0 x = 4 or x = 5 Check x = 4 9 20 1− + 2 = 0 4 4 9 20 1− + =0 4 16 16 − 36 + 20 =0 16 0=0 ANSWER: The solution is 4 or 5.
Check x = 5 9 20 1− + 2 = 0 5 5 9 20 =0 1− + 5 25 25 − 45 + 20 =0 25 0=0
18.
PROBLEM: 4 ( x + 1) x+2 1 + = 2 x−2 x+2 x −4 SOLUTION: 4 ( x + 1) x+2 1 + = 2 x−2 x+2 x −4 4 ( x + 1) x+2 1 + = x − 2 x + 2 ( x + 2 )( x − 2 ) We first note that x ≠ ±2 . Multiply both the sides with the LCD, ( x − 2 )( x + 2 ) .
4 ( x + 1) 1 ⎞ ⎛ x+2 + ⋅ ( x + 2 )( x − 2 ) ⎜ ⎟ ( x + 2 )( x − 2 ) = ( x + 2 )( x − 2 ) ⎝ x−2 x+2⎠ x2 + 4x + 4 + x − 2 = 4 x + 4 x2 + x − 2 = 0
( x + 2 )( x − 1) = 0 x = −2 or x = 1 x ≠ ±2 . ANSWER: Hence, the only solution is 1. We can check our answer by substituting 1 in for x to see if we obtain a true statement. 4 (1 + 1) 1+ 2 1 + = 1 − 2 1 + 2 (1 + 2 )(1 − 2 )
1 8 −3 + = − 3 3 8 8 − =− 3 3
19.
PROBLEM: x 1 3x − 10 − = 2 x − 2 x − 3 x − 5x + 6 SOLUTION: x 1 3 x − 10 − = 2 x − 2 x − 3 x − 5x + 6 x 1 3x − 10 − = x − 2 x − 3 ( x − 2 )( x − 3)
We first note that x ≠ 2 and x ≠ 3 . Multiply both the sides with the LCD, ( x − 2 )( x − 3) .
1 ⎞ 3x − 10 ⎛ x − ⋅ ( x − 2 )( x − 3) ⎜ ⎟ ( x − 2 )( x − 3) = ( x − 2)( x − 3) ⎝ x −2 x −3⎠ x 2 − 3x − x + 2 = 3x − 10 x 2 − 7 x + 12 = 0
( x − 3)( x − 4) = 0 x = 3 or x = 4 x ≠ 3. ANSWER: Hence, the only solution is 4. We can check our answer by substituting 4 in for x to see if we obtain a true statement. 4 1 3 ⋅ 4 − 10 − = 4 − 2 4 − 3 ( 4 − 2 )( 4 − 3)
2 −1 = 1 1=1
20.
PROBLEM: 5 x 9x − 4 − = 2 x + 4 4 − x x − 16 SOLUTION: x 5 9x − 4 − = 2 x + 4 4 − x x − 16 x 5 9x − 4 + = x + 4 x − 4 ( x + 4 )( x − 4 )
We first note that x ≠ ±4 . Multiply both the sides with the LCD, ( x − 4 )( x + 4 ) .
x ⎞ 9x − 4 ⎛ 5 + ⋅ ( x − 4 )( x + 4 ) ⎜ ⎟ ( x − 4 )( x + 4 ) = ( x + 4)( x − 4) ⎝ x+4 x−4⎠
5 x − 20 + x 2 + 4 x = 9 x − 4 x 2 − 16 = 0
( x − 4)( x + 4 ) = 0 x = 4 or x = −4 x ≠ ±4 . ANSWER: Hence, the solution is∅.
21.
PROBLEM:
Solve for r: P =
120 1 + 3r
SOLUTION:
1 We first note that r ≠ − . 3 120 P= 1 + 3r 120 1 + 3r = P 120 3r = −1 P 120 1 r= − 3P 3 40 1 r= − P 3 ANSWER: r =
40 1 − P 3
Set up an algebraic equation and then solve:
22.
PROBLEM: An integer is three times that of another. The sum of the reciprocals of the two integers is 1/3. Find the two integers. SOLUTION: Smaller integer is represented by n. Larger integer: 3n 1 1 1 + = n 3n 3 We first note that n ≠ 0. Multiply both sides by the LCD, 3n. 1 ⎛1 1 ⎞ ⎜ + ⎟ 3n = ⋅ 3n 3 ⎝ n 3n ⎠ 3 +1 = n n=4 Larger integer: 3n = 3 ⋅ 4 = 12 ANSWER: The two integers are 4 and 12.
23.
PROBLEM: Working alone, Joe can paint the room in 6 hours. If Manny helps, then together they can paint the room in 2 hours. How long would it take Manny to paint the room by himself? 1 1 SOLUTION: Using the formula, ⋅ t + ⋅ t = 1 t1 t2 t1 = 6 hours; t = 2 hours 1 1 ⋅t + ⋅t =1 t1 t2 1 1 ⋅2+ ⋅2 =1 6 t2 2 2 + =1 6 t2
Multiplying both sides by the LCD, 6t2. ⎛2 2⎞ ⎜ + ⎟ 6t2 =1 ⋅ 6t2 ⎝ 6 t2 ⎠ 2t2 + 12 = 6t2 4t2 = 12 t2 = 3
ANSWER: It would take Manny to paint the room in 3 hours.
24.
PROBLEM: A river tour boat can average 6 mph in still water. With the current, the boat can travel 17 miles in the same time it can travel 7 miles against it. What is the speed of the current? SOLUTION: The speed of the current is represented by x. Speed of the boat downstream: 6 + x Speed of the boat upstream: 6 – x
D 17 = t 6+ x D 7 Time taken by the boat to cover 7 miles upstream: t = = t 6− x Time taken by the boat to cover = Time taken by the boat to cover 17 miles downstream 7 miles upstream 17 7 = 6+ x 6− x Multiply both the sides by the LCD, (6 + x)(6 – x) 17 7 ⋅ ( 6 + x )( 6 − x ) = ⋅ ( 6 + x )( 6 − x ) 6+ x 6− x 102 − 17 x = 42 + 7 x 24 x = 60 5 1 x= =2 2 2 1 ANSWER: The speed of the current is 2 miles/hour. 2
Time taken by the boat to cover 17 miles downstream: t =
25.
PROBLEM: The breaking distance of an automobile is directly proportional to the square of its speed. Under optimal conditions, a certain automobile moving at 35 mph can break in 25 feet. Find an equation that models the breaking distance under optimal conditions and use it to determine the breaking distance if the automobile is moving 28 mph. SOLUTION: Breaking distance of the automobile is represented by d. Speed of the automobile is represented by s. d = ks 2 Under optimal conditions, the automobile moving at 35 mph can break in 25 feet. d = ks 2
25 = k 352 25 1 k= 2 = 35 49 1 2 d= s 49 If the automobile is moving at 28 mph: d = ks 2 25 d = 2 282 = 16 35 ANSWER: If the automobile is moving at 28 mph, the breaking distance is 16 feet.
