4.1 Solving Linear Systems by Graphing
4.1 Solving Linear Systems by Graphing Part A: Solutions to Linear Systems 1.
PROBLEM: x y 1 (3, 2 ); 2 x 2 y 2 SOLUTION: Check: (3, 2) Equation 1: x y 1
Equation 2: 2 x 2 y 2
x y 1
2 x 2 y 2
3 2 1
2 3 2 2 2
3 2 1 1 1
6 4 2 6 4 2 2 2
ANSWER: (3, -2) does not satisfy both equations, hence, it is not a solution.
2.
PROBLEM: x y 1 (5, 0); 2 x 2 y 2 SOLUTION:
Check: (5, 0) Equation 1: x y 1
Equation 2: 2 x 2 y 2
x y 1
2 x 2 y 2
5 0 1 5 1
2 5 2 0 2 10 0 2 10 2
ANSWER: Since (5, 0) does not satisfy both equations, hence, it is not a solution
3.
PROBLEM: x y 4 (2, 6); 3x y 12 SOLUTION:
Check (2, 6) Equation 1: x y 4
Equation 2: 3x y 12 3x y 12 3 2 6 12
x y 4 2 6 4
6 6 12 0 12
2 6 4 4 4
ANSWER: Since (3, 4) does not satisfy both equations, hence, it is not a solution
4.
PROBLEM: 3 x 2 y 8 (2, 7); 5 x 3 y 11 SOLUTION:
Check: (2, 7) Equation 1: 3x 2 y 8
Equation 2: 5 x 3 y 11
3x 2 y 8
5 x 3 y 11
3 2 2 7 8
5 2 3 7 11
6 14 8
10 (21) 11 10 21 11 11 11
8 8
ANSWER: Since (2, 7) satisfy both equations, hence, it is a solution
5.
PROBLEM: 5 x 5 y 15 (0, 3); 13x 2 y 6 SOLUTION:
Check (0, 3) Equation 1: 5 x 5 y 15
Equation 2: 13 x 2 y 6
5 x 5 y 15
13x 2 y 6
5 0 5 3 15
13 0 2 3 6
0 (15) 15 15 15
0 6 6 6 6
ANSWER: Since (0, 3) satisfy both equations, hence, it is a solution
6.
PROBLEM: 1 x y 1 1 ( 2 , 4 ) ; 4 2 x 4 y 0 SOLUTION:
Check: ( 12 , 14 ) Equation 1: x y x y
1 4
1 1 1 2 4 4 1 1 4 4
1 4
Equation 2: 2 x 4 y 0 2 x 4 y 0 2 12 4 14 0 1 1 0 00
ANSWER: Since ( 12 , 14 ) satisfy both equations, hence, it is a solution
7.
PROBLEM: ( 34 , 14 ) ; x y 1 4 x 8 y 5 SOLUTION:
Check: ( 34 , 14 ) Equation 1: x y 1
Equation 2: 4 x 8 y 5
x y 1
4 34 8 14 5
34 14 1
3 2 5 5 5
1 1
ANSWER: Since ( 34 , 14 ) does not satisfy both equations, hence, it is not a solution.
8.
PROBLEM: 1 1 3 x 2 y 1 (3, 4); 2 x 3 y 8 3 2 SOLUTION:
Check: (3, 4)
1 1 x y 1 3 2 1 1 3 4 1 3 2 1 2 1 11
Equation 1:
2 3 x y 8 3 2 2 3 x y 8 3 2 2 3 3 4 8 3 2 2 6 8 8 8
Equation 2:
ANSWER: Since (3, 4) satisfy both equations, hence, it is a solution
9.
PROBLEM: y 3 (5, 3); 5 x 10 y 5 SOLUTION:
Check: (5, 3) Equation 1: y 3
Equation 2: 5 x 10 y 5
y 3
5 x 10 y 5
3 3
5 5 10 3 5 25 30 5 25 30 5 55
ANSWER: Since (5, 3) satisfy both equations, hence, it is a solution
10. .
PROBLEM: x4 (4, 2); 7 x 4 y 8 SOLUTION:
Check: (4, 2) Equation 1: x 4 x4 44
Equation 2: 7 x 4 y 8 7x 4 y 8 7 4 4 2 8 28 8 8 20 8
ANSWER: Since (4, 2) does not satisfy both equations, hence, it is not a solution
11.
PROBLEM:
SOLUTION: In the above graph, the point of intersection appears to be (5, 0). ANSWER: Hence, solution is (5, 0).
12.
PROBLEM:
SOLUTION: In the above graph, the point of intersection appears to be ( 2, -2). ANSWER: So, solution is (2, -2).
13.
PROBLEM:
SOLUTION: In the above graph, the point of intersection appears to be (2, 1). ANSWER: Hence, solution is (2, 1).
14.
PROBLEM:
SOLUTION:
In the above graph, the point of intersection appears to be ( 5, -2). ANSWER: So, solution is ( 5, -2).
15.
PROBLEM:
SOLUTION: In the above graph, the point of intersection appears to be (0, 0). ANSWER: Hence, solution is (0, 0).
16.
PROBLEM:
SOLUTION: In the above graph, the point of intersection is not there. ANSWER: So, there is no solution.
17.
PROBLEM:
ANSWER: In the above graph, y depends on x and so we express all the ordered pair solutions ( x, y ) in the form ( x, 2 x 2) where x is any real number.
18.
PROBLEM:
SOLUTION: In the above graph, the point of intersection appears to be ( -1, -1). ANSWER: So, solution is (-1, -1).
19.
PROBLEM:
SOLUTION: In the above graph, the point of intersection is not there. ANSWER: Hence there is no solution.. Part B: Solving Linear Systems
20.
PROBLEM:
ANSWER: In the above graph, y depends on x and so we may express all the ordered pair solutions ( x, y ) in the form ( x, 3x 6) where x is any real number
21.
PROBLEM: 3 y x 6 2 y x 1 SOLUTION:
Equation 1: y
3 x6 2
Equation 2: y x 1
If x = 0, then y=0+6 y=6
If x = 0, then y=0+1 y=1
If y = 0, then 0=3/2x + 6 - 6 = 3/2x x = -4
If y = 0, then -x = -1 x=1
ANSWER: The solution is (-2, 3)
22.
PROBLEM: 3 y 4 x 2 y 1 x 2 4 SOLUTION:
Equation 1: y
3 x2 4
1 Equation 2: y x 2 4
If x=0, then y=0+2 y=2
If x=0, then y=0-2 y= -2
If y=0, then 3 y x2 4 3 0 x2 4 3 x2 4 3x 2* 4 x 8 / 3
If y=0, then 1 0 x2 4 1 x 2 4 1 x 2 4 x 8
23.
PROBLEM: y x 4 y x 2 SOLUTION:
Equation 1: y x 4
Equation 2: y x 2
If x = 0, then y=0-4 y = -4
If x = 0, then y=0+2 y=2
If y =0, then 0=x-4
If y = 0, then 0= -x +2
x=4
ANSWER: The solution is (3,-1)
24.
PROBLEM: y 5 x 4 y 4x 5 SOLUTION:
x=2
Equation 1: y 5 x 4
Equation 2: y 4 x 5
If x=0, then y 5 0 4
If x=0, then y 4 0 5
y4 If y=0, then 0 5 x 4
y 5 If y=0, then 0 4x 5
5x 4
4x 5
x .8
x 1.25
ANSWER:
25.
PROBLEM: 2 y 5 x 1 y 3 x 5
SOLUTION:
Equation 1: y If x=0, then 2 y 0 1 5 y 0 1
y 1 If y=0, then 2 y x 1 5 2 0 x 1 5 2 x 1 5 5 x 2
ANSWER:
2 x 1 5
Equation 2: y If x=0, then 3 y x 5 3 y 0 5 y0 If y=0, then 3 y x 5 3 0 x 5 x0
3 x 5
26.
PROBLEM: 2 y 5 x 6 y 2 x 10 5
SOLUTION:
2 Equation 1: y x 6 5
Equation 2: y
2 x 10 5
If x=0, then y 06 y6 If y=0, then 2 y x6 5 2 0 x6 5 2 x6 5 x 15 ANSWER:
If x=0, then y=0+10 y= 10 If y=0, then 2 0 x 10 5 2 10 x 5 x 25
27.
PROBLEM: y 2 y x 1 SOLUTION:
Equation 1: y 2 If every real value of x, y= -2
Equation 2: y x 1 If x=0, then y=0+1 y=1 If y=0, then y x 1 0 x 1 x 1
ANSWER:
28.
PROBLEM: y 3 x 3 SOLUTION:
Equation 1: y 3 For x = any real number, y=3
ANSWER:
Equation 2: x 3 For y = any real number x= -3
29.
PROBLEM: y 0 2 y 5 x 4 SOLUTION: Equation 1: y 0
For every real value of x, y=0
Equation 2: y If x=0, then y=0-4 y=-4 If y=0, then 2 0 x4 5 2 4 x 5 x 10
ANSWER:
2 x4 5
30.
PROBLEM: x 2 y 3x SOLUTION:
Equation 1: x 2 For y = any real number x=2
Equation 2: y 3x If x=0, then y= 0 If y= 0, then x=0
ANSWER:
31.
PROBLEM: 3 y 5 x 6 y 3 x 3 5 SOLUTION:
Equation 1: y If x=0, then y= 0-6 y= -6 If y=0, then 3 0 x 6 5 3 x6 5 x 10 ANSWER:
3 x6 5
Equation 2: y If x=0, then y=0-3 y=-3 If y=0, then 3 0 x 3 5 3 3 x 5 x5
3 x3 5
32.
PROBLEM: 1 y 2 x 1 y 1 x 1 2
SOLUTION:
1 Equation 1: y x 1 2 If x=0, then 1 y 0 1 2 y 1 If y= 0, then 1 0 x 1 2 1 x 1 2 x2 ANSWER:
1 Equation 2: y x 1 2 If x=0, then 1 y 0 1 2 y 1 If y= 0, then 1 0 x 1 2 1 x 1 2 x2
33.
PROBLEM: 2 x 3 y 18 6 x 3 y 6 SOLUTION:
Equation 1: 2 x 3 y 18
Equation 2: 6 x 3 y 6
If x = 0, then 2 0 3 y 18
If x = 0, then 3 y 6 6 x
0 3 y 18 y6 If y = 0, then 2 x 3 0 18
3 y 6 0 y 2 If y = 0, then 6 x 3 0 6
2 x 0 18 x9
ANSWER:
6 x 0 6 6 x 6 x 1
34.
PROBLEM: 3x 4 y 20 2x 8 y 8 SOLUTION:
Equation 1: 3x 4 y 20
Equation 2: 2 x 8 y 8
If x=0, then 3 0 4 y 20
If x=0, then 2 0 8 y 8
0 4 y 20 y5
0 8y 8 y 1
If y= 0, then 3 x 4 0 20
If y= 0, then 2 x 8 0 8
3 x 0 20
2x 0 8
x 6.67
x4
ANSWER:
35.
PROBLEM: 2 x y 1 2x 3y 9 SOLUTION:
Equation 1: 2 x y 1
Equation 2: 2 x 3 y 9
If x = 0, then 2 0 y 1
If x = 0, then 2 0 3 y 9
0 y 1 y 1 If y = 0, then 2 x 0 1 2 x 1 x 0.5
0 3y 9 y 3 If y = 0, then 2x 3 0 9
ANSWER:
2x 0 9 x 4.5
36.
PROBLEM: x 2 y 8 5 x 4 y 4 SOLUTION:
Equation 1: x 2 y 8
Equation 2: 5 x 4 y 4
If x=0, then 0 2 y 8 2 y 8
If x=0, then 5 0 4 y 4
y 4
y 1
0 4 y 4
If y= 0, then x 2 0 8
If y= 0, then 5 x 4 0 4
x 0 8
5 x 0 4
x 8
x .8
ANSWER:
37.
PROBLEM: 4 x 6 y 36 2x 3 y 6 SOLUTION:
Equation 1: 4 x 6 y 36
Equation 2: 2 x 3 y 6
If x = 0, then 4 0 6 y 36
If x = 0, then 2 0 3 y 6
0 6 y 36
0 3y 6 y 2 If y = 0, then 2x 3 0 6
6 y 36 y6 If y = 0, then 4 x 6 0 36 4 x 0 36 x9
ANSWER:
2x 0 6 x3
38.
PROBLEM: 2 x 3 y 18 6 x 3 y 6 SOLUTION:
Equation 1: 2 x 3 y 18
Equation 2: 6 x 3 y 6
If x=0, then 2 0 3 y 18
If x=0, then 6 0 3 y 6
0 3 y 18 3 y 18
0 3 y 6 3y 6
y 6 If y= 0, then
y2 If y= 0, then
2 x 3 0 18
6 x 3 0 6
2 x 0 18
6 x 0 6 6 x 6
x9
x 1 ANSWER:
39.
PROBLEM: 3 x 5 y 30 6 x 10 y 10 SOLUTION:
Equation 1: 3 x 5 y 30
Equation 2: 6 x 10 y 10
If x = 0, then 3 0 5 y 30
If x = 0, then 6 0 10 y 10
0 5 y 30
0 10 y 10
y6 If y = 0, then 3 x 5 0 30
10 y 10
3 x 0 30 x 10
y 1 If y = 0, then 6 x 10 0 10
6 x 0 10 6 x 10 x 1.67
ANSWER:
40.
PROBLEM: x 3y 3 5 x 15 y 15 SOLUTION:
Equation 1: x 3 y 3
Equation 2: 5 x 15 y 15
If x=0, then 0 3 y 3
If x=0, then 5 0 15 y 15
0 3y 3 y 1 If y= 0, then
0 15 y 15 15 y 15 y 1
x 3 0 3 x 0 3 x 3
If y= 0, then 5 x 15 0 15 5 x 0 15 5 x 15 x 3
ANSWER:
41.
PROBLEM: x y 0 x y 0 SOLUTION:
Equation 1: x y 0
Equation 2: x y 0
If x = 0, then 0 y 0
If x = 0, then 0 y 0
y 0 y0 If y = 0, then x 0 0
y0 If y = 0, then x 0 0
x0 ANSWER:
x 0 x0
42.
PROBLEM: yx y x 1 SOLUTION:
Equation 1: y x X= Y
Equation 2: y x 1 If x=0, then y 0 1 y 1 If y= 0, then
0 x 1 x 1 ANSWER:
43.
PROBLEM: 3x 2 y 0 x2 SOLUTION:
Equation 1: 3x 2 y 0 If x=0, then 3 0 2 y 0 0 2y 0 2y 0 y0 If y=0, then 3x 2 0 0 3x 0 0 x0
ANSWER:
Equation 2: x 2 For every value of y, x will remain 2
44.
PROBLEM: 1 2 2 x 3 y 3 3 x 1 y 2 2 SOLUTION:
Equation 1: 2 x If x=0, then 1 2 2 0 y 3 3 1 2 0 y 3 3 1 2 y 3 3 y2 If y= 0, then 1 2 0 3 3 2 2x 3 1 x 3 x .33
2x
ANSWER:
1 2 y 3 3
Equation 2: 3x If x=0, then 1 3 0 y 2 2 1 y 2 2 y 4 If y= 0, then 1 3 x 0 2 2 3 x 0 2 3x 2 x .67
1 y 2 2
45.
PROBLEM: 1 1 10 x 5 y 2 1 x 1 y 1 5 5
SOLUTION:
Equation 1 :
1 1 x y 2 10 5
If x = 0, then 1 1 0 y 2 10 5 1 0 y 2 5 y 10 If y = 0, then 1 1 x 0 2 10 5 1 x0 2 10 1 x2 10 x 20
ANSWER:
1 1 Equation 2: x y 1 5 5 If x = 0, then 1 1 0 y 1 5 5 1 y 1 5 y 5 If y = 0, then 1 1 x 0 1 5 5 1 x 1 5 x5
46.
PROBLEM: 1 1 3 x 2 y 1 1 x 1 y 1 3 5 SOLUTION:
Equation 1: If x=0, then
1 1 x y 1 3 2
Equation 2: If x=0, then
1 1 x y 1 3 5
1 1 0 y 1 3 2 1 y 1 2 y 2
1 1 0 y 1 3 5 1 y 1 5 y5
If y= 0, then 1 1 x 0 1 3 2 1 x 1 3 x3
If y= 0, then 1 1 x 0 1 3 5 1 x 1 3 x3
ANSWER:
47.
PROBLEM: 1 1 9 x 6 y 0 1 x 1 y 1 9 4 2 SOLUTION:
1 1 Equation 1 : x y 0 9 6 If x = 0, then 1 1 0 y 0 9 6 1 y0 6 y0 If y = 0, then 1 1 x 0 0 9 6 1 x0 9 x0 ANSWER:
Equation 2: If x = 0, then 1 1 1 0 y 9 4 2 1 1 y 4 2 y2 If y = 0, then 1 1 1 x 0 9 4 2 1 1 x 9 2 x 4.5
1 1 1 x y 9 4 2
48.
PROBLEM: 1 5 x y 5 16 2 5 x 1 y 5 16 2 2 SOLUTION:
Equation 1: If x=0, then
5 1 x y 5 16 2
Equation 2: If x=0, then
5 1 5 x y 16 2 2
5 1 0 y 5 16 2 1 y5 2 y 10 If y= 0, then 5 1 x 0 5 16 2 5 x5 16 x 16 ANSWER:
5 1 5 0 y 16 2 2 1 5 y 2 2 y5
If y= 0, then 5 1 5 x 0 16 2 2 5 5 x 16 2 x 8
49.
PROBLEM: 1 9 1 6 x 2 y 2 1 x 1 y 3 18 6 2 SOLUTION:
1 1 9 Equation 1 : x y 6 2 2
Equation 2:
If x = 0, then 1 1 9 0 y 6 2 2 1 9 y 2 2 y 9 If y = 0, then 1 1 9 x 0 6 2 2 1 9 x 6 2 x 27
If x = 0, then 1 1 3 0 y 18 6 2 1 3 y 6 2 y 9 If y = 0, then 1 1 3 x 0 18 6 2 1 3 x 18 2 x 27
ANSWER:
1 1 3 x y 18 6 2
50.
PROBLEM: 1 1 1 2 x 4 y 2 1 x 1 y 3 3 2 SOLUTION:
Equation 1:
1 1 1 x y 2 4 2
Equation 2:
If x=0, then 1 1 1 0 y 2 4 2 1 1 y 4 2 y2
If x=0, then 1 1 0 y 3 3 2 1 y3 2 y 6
If y= 0, then 1 1 1 x 0 2 4 2 1 1 x 2 2 x 1
If y= 0, then 1 1 x 0 3 3 2 1 x3 3 x9
ANSWER:
1 1 x y 3 3 2
51.
PROBLEM: y 4 x 5 SOLUTION:
Equation 1 : y 4 For every value of x, y will remain 4
Equation 2: x 5
For every value of y, x will remain -5
ANSWER:
52.
PROBLEM: y 3 x 2 SOLUTION:
Equation 1 : y 3 For every value of x, y will remain -3
Equation 2: x 2
For every value of y, x will remain 2
ANSWER:
53.
PROBLEM: y 0 x 0 SOLUTION:
Equation 1 : y 0 For every value of x, y will remain 0
Equation 2: x 0 For every value of y, x will remain 0
ANSWER:
54.
PROBLEM: y 2 y 3 SOLUTION:
Equation 1 : y 2 For every value of x, y will remain -2
Equation 2: y 3 For every value of x, y will remain 3
ANSWER:
55.
PROBLEM: y 5 y 5 SOLUTION:
Equation 1 : y 5 For every value of x, y will remain 5
Equation 2: y 5 For every value of x, y will remain -5
ANSWER:
56.
PROBLEM: y2 y 2 0 SOLUTION:
Equation 1 : y 2 For every value of x, y will remain 2
Equation 2: y 2 0 For every value of x, y will remain 2
ANSWER:
57.
PROBLEM: x 5 x 1 SOLUTION:
Equation 1 : x 5 For every value of y, x will remain -5
Equation 2: x 1 For every value of y, x will remain 1
ANSWER:
58.
PROBLEM: y x x 0 SOLUTION:
Equation 1 : y x For every value of x, y will remain 2
Equation 2: x 0 For every value of x, y will remain 2
ANSWER:
59.
PROBLEM: 4 x 6 y 3 x y 2 SOLUTION:
Equation 1 : 4 x 6 y 3 If x=0, then
Equation 2: x y 2 If x=0, then
4 0 6 y 3
0 y 2
6y 3 y .5 If y=0, then 4x 6 0 3
y 2 If y=0, then x 0 2
4x 3 x .75
ANSWER:
60.
PROBLEM: 2 x 20 y 20 3x 10 y 10
x 2 x2
SOLUTION:
Equation 1: 2 x 20 y 20
Equation 2: 3x 10 y 10
If x=0, then 2 0 20 y 20
If x=0, then 3 0 10 y 10
20 y 20 y 1
10 y 10 y 1
If y=0, then 2 x 20 0 20
If y=0, then 3x 10 0 10
2 x 20
3x 10 x 3.33
x 10
ANSWER:
61.
PROBLEM: The sum of two numbers is 20. The larger number is 10 less than five times the smaller. SOLUTION:
Let the two numbers be x and y Then two equations are x+ y = 20 5x- y = 10 Equation 1 : 4 x 6 y 3
Equation 2: x y 2
If x = 0, then 0 y 20
If x = 0, then 5 0 y 10
y 20
y 10
If y = 0, then x 0 20
y 10 If y = 0, then 5x 0 10
x 20
ANSWER:
5x 10 x2
62.
PROBLEM: The difference between two numbers is 12 and the sum is 4. SOLUTION: Let the two numbers be x and y Then two equations are x - y = 12 x+y=4
Equation 1: x y 12 If x=0, then
Equation 2: x y 4 If x=0, then 0 y 4 y4
0 y
12
y 12 y 12 If y=0, then x 0 12 x 12 ANSWER:
63.
PROBLEM: 3x 2 y 6 SOLUTION:
If y=0, then x 0 4 x4
x 1 2 3 4 5 6
ANSWER:
y -3/2 0 3/2 3 9/2 6
3x 6 2
64.
PROBLEM: 5 x 2 y 30 SOLUTION:
x 0 -2 -4 -6 -8 -10
y 15 10 5 0 -5 -10 ANSWER:
30 5 x 2
65.
PROBLEM: Determine the price at which the quantity demanded is equal to the quantity supplied. SOLUTION:
The price at which the quantity demanded is equal to the quantity supplied is = $1.25. ANSWER:
66.
PROBLEM: If production of bottled water slips to 20 tons, then what price does the demand curve predict for a bottle of water? SOLUTION:
Production of bottled water slips to 20 tons, the price demand curve predict for a bottle of water is $2.00 ANSWER:
67.
PROBLEM: If production of bottled water increases to 40 tons, then what price does the demand curve predict for a bottle of water? SOLUTION: Production of bottled water increases to 40 tons, the predicted price by the demand curve predict for a bottle of water is $1.00. ANSWER:
68.
PROBLEM: If the price of bottled water is set at $2.50 dollars per bottle, what quantity does the demand curve predict? SOLUTION:
The price of bottled water is set at $2.50 dollars per bottle. The quantity predicted by the demand curve is 10 tons. ANSWER:
4.2 Solving Linear Systems by Substitution Part A: Substitution Method 1.
PROBLEM: y 4x 1 3 x y 1 SOLUTION: Step 1: Put the value of y of first equation in second equation 3 x 4 x 1 1 3 x 4 x 1 1 x2
Step 2: Put the value of x in first equation
y 4 2 1 y 8 1 y7 ANSWER: (x, y) = (2,7)
2.
PROBLEM: y 3x 8 4 x y 2 SOLUTION: Step 1: Put the value of y of first equation in second equation 4 x 3x 8 2 4 x 3x 8 2 x 6 Step 2: Put the value of x in first equation
y 3 6 8 y 18 8 y 26 ANSWER: (x, y) = (-6,-26)
3.
PROBLEM: x 2y 3 x 3 y 8 SOLUTION: Step 1: Put the value of x of first equation in second equation
2 y 3 3 y 8 5 y 3 8 5 y 5 y 1
Step 2: Put the value of y in first equation
x 2 1 3 x 5 ANSWER: (x, y) = (-5,-1)
4.
PROBLEM: x 4 y 1 2 x 3 y 12 SOLUTION: Step 1: Put the value of x of first equation in second equation 2 4 y 1 3 y 12 8 y 2 3 y 12 5 y 10 y 2 Step 2: Put the value of y in first equation x 4 2 1 x 8 1 x9 ANSWER: (x, y) = (9,-2)
5.
PROBLEM: y 3x 5 x 2 y 2 SOLUTION: Step 1: Put the value of y of first equation in second equation 5 x 2 3 x 2 5 x 6 x 2 x2
Step 2: Put the value of x in first equation y 3 2 y6
ANSWER: (x, y) = (2,6)
6.
PROBLEM: yx 2 x 3 y 10 SOLUTION: Step 1: Put the value of y of first equation in second equation 2 x 3 x 10 5 x 10 x2 Step 2: Put the value of x in first equation y 2 y2
ANSWER: (x, y) = (2,2)
7.
PROBLEM: y 4x 1 4 x y 2 SOLUTION: Step 1: Put the value of y of first equation in second equation
4 x 4 x 1 2 1 2 ANSWER: Hence there is no solution.
8.
PROBLEM: y 3x 5 3x y 5 SOLUTION: Step 1: Put the value of y of first equation in second equation 3 x 3 x 5 5
55 ANSWER: (x, y) = (x,-3x +5)
9.
PROBLEM: y 2x 3 2 x y 3 SOLUTION: Step 1: Put the value of y of first equation in second equation
2 x 2 x 3 3 3 3 ANSWER: (x, y) = (x, 2x+3)
10.
PROBLEM: y 5x 1 x 2 y 5 SOLUTION: Step 1: Put the value of y of first equation in second equation x 2 5 x 1 5
x 10 x 2 5 9 x 3 1 x 3 Step 2: Put the value of x in first equation
1 y 5 1 3 8 y 3 ANSWER: (x, y) = (-1/3,-8/3)
11.
PROBLEM: y 7 x 1 3x y 4 SOLUTION: Step 1: Put the value of y of first equation in second equation 3x 7 x 1 4
10 x 1 4 10 x 5 x
1 2
Step 2: Put the value of x in first equation
1 y 7 1 2 7 y 1 2 5 y 2 ANSWER: (x, y) = (1/2, - 5/2)
12.
PROBLEM: x 6 y 2 5 x 2 y 0
SOLUTION: Step 1: Put the value of x of first equation in second equation 5 6 y 2 2 y 0
30 y 10 2 y 0 28 y 10 10 y 28 5 y 14 Step 2: Put the value of y in first equation 5 x 6 2 14 15 x 2 7 1 x 7 ANSWER: (x, y) = (-1/7, -5/14)
13.
PROBLEM: y 2 2 x y 6 SOLUTION: Step 1: Put the value of y = -2 of first equation in second equation
2 x 2 6 2 x 2 6 2 x 8 x4 ANSWER: (x, y) = (4, -2)
14.
PROBLEM: x 3 x 4 y 3 SOLUTION: Step 1: Put the value of x =-3 of first equation in second equation
3 4 y 3 4 y 0 y0 ANSWER: (x, y) = (-3, 0)
15.
PROBLEM: 1 y x 3 5 7 x 5 y 9 SOLUTION: Step 1: Put the value of y of first equation in second equation 1 7x 5 x 3 9 5 7 x x 15 9 7 x x 15 9 8 x 24 x3
Step 2: Put the value of x in first equation
1 3 3 5 3 y 3 5 12 y 5 y
ANSWER: (x, y) = (3, 12/5)
16.
PROBLEM: 2 y x 1 3 6 x 9 y 0 SOLUTION: Step 1: Put the value of y of first equation in second equation 2 6 x 9 x 1 0 3 6x 6x 9 0 90
ANSWER: So, there is no solution
17.
PROBLEM: 1 1 y x 2 3 x 6 y 4 SOLUTION: Step 1: Put the value of y of first equation in second equation 1 1 x 6 x 4 3 2 x 3x 2 4 2 x 6 x 3
Step 2: Put the value of x in first equation 1 1 y 3 2 3 3 1 y 2 3 7 y 6 ANSWER: (x, y) = (-3, -7/6)
18.
PROBLEM: 3 1 y x 8 2 2 x 4 y 1 SOLUTION: Step 1: Put the value of y of first equation in second equation
1 3 2x 4 x 1 2 8 3 2x x 2 1 2 1 x 1 2 x 2 Step 2: Put the value of x in first equation 3 1 2 8 2 3 1 y 4 2 5 y 4 ANSWER: (x, y) = (-2, 5/4) y
19.
PROBLEM: x y 6 2 x 3 y 16 SOLUTION: Step 1: Write the first equation in the form of y
x y 6 y 6 x Step 2: Put the value of y of first equation in second equation 2 x 3 6 x 16
2 x 18 3x 16 x 2 x2 Step 3: Put the value of x in first equation 2 y 6
y 62 y4 ANSWER: (x, y) = (2, 4)
20.
PROBLEM: x y 3 2 x 3 y 2 SOLUTION: Step 1: Write the first equation in the form of y x y 3
y x3
Step 2: Put the value of y of first equation in second equation 2 x 3 x 3 2 2 x 3 x 9 2 x7
Step 3: Put the value of x in first equation 7 3 y 4 y
ANSWER: (x, y) = (7, 4)
21.
PROBLEM: 2x y 2 3x 2 y 17 SOLUTION: Step 1: Write the first equation in the form of y 2x y 2
2x 2 y y 2 2x Step 2: Put the value of y of first equation in second equation 3x 2 2 2 x 17
3x 4 4 x 17 7 x 21 x3 Step 3: Put the value of x in first equation
2 3 y 2 6 2 y y 4 ANSWER: (x, y) = (3, -4)
22.
PROBLEM: x 3 y 11 3x 5 y 5 SOLUTION: Step 1: Write the first equation in the form of x x 3 y 11 x 11 3 y Step 2: Put the value of x of first equation in second equation 3 11 3 y 5 y 5 33 9 y 5 y 5 14 y 28 y2
Step 3: Put the value of y in first equation x 3 2 11
x 5 ANSWER: (x, y) = (-5, 2)
23.
PROBLEM: x 2 y 3 3x 4 y 2 SOLUTION: Step 1: Write the first equation in the form of x
x 2 y 3 x 3 2 y Step 2: Put the value of x of first equation in second equation 3 3 2 y 4 y 2
9 6 y 4 y 2 10 y 7 7 y 10 Step 3: Put the value of y in first equation 7 x 2 3 10 7 x 3 5 8 x 5 ANSWER: (x, y) = (-8/5, -7/10)
24.
PROBLEM: 5 x y 12 9 x y 10
SOLUTION: Step 1: Write the first equation in the form of y 5 x y 12
5 x 12 y Step 2: Put the value of y of first equation in second equation
9 x 5 x 12 10 9 x 5 x 12 10 4 x 2 1 x 2 Step 3: Put the value of x in first equation 1 5 y 12 2 5 12 y 2 29 y 2 ANSWER: (x, y) = (-1/2, -29/2)
25.
PROBLEM: x 2 y 6 4 x 8 y 24
SOLUTION: Step 1: Write the first equation in the form of x
x 2 y 6 x 6 2 y Step 2: Put the value of x of first equation in second equation 4 6 2 y 8 y 24
24 8 y 8 y 24 24 24 ANSWER: (x, y) = (x, -x/2-3)
26.
PROBLEM: x 3 y 6 2 x 6 y 12
SOLUTION: Step 1: Write the first equation in the form of x x 3 y 6 x 6 3 y Step 2: Put the value of x of first equation in second equation 2 6 3 y 6 y 12
12 6 y 6 y 12 12 12 ANSWER: So, there is no solution.
27.
PROBLEM: 3x y 4 6 x 2 y 2 SOLUTION: Step 1: Write the first equation in the form of y 3x y 4
y 4 3 x Step 2: Put the value of y of first equation in second equation 6 x 2 4 3 x 2 6 x 8 6 x 2 8 2
ANSWER: Hence there is no solution.
28.
PROBLEM: x 5 y 10 2 x 10 y 20 SOLUTION: Step 1: Write the first equation in the form of x
x 5 y 10 x 5 y 10 Step 1: Put the value of x of first equation in second equation 2 5 y 10 10 y 20 10 y 20 10 y 20 20 20
ANSWER: (x, y) = (x, x/5+2)
29.
PROBLEM: 3x y 9 4 x 3 y 1 SOLUTION: Step 1: Write the first equation in the form of y 3x y 9
3x 9 y Step 2: Put the value of y of first equation in second equation 4 x 3 3x 9 1
4 x 9 x 27 1 13 x 26 x2 Step 2: Put the value of x in first equation 3 2 y 9
69 y y 3 ANSWER: (x, y) = (2, -3)
30.
PROBLEM: 2x y 5 4 x 2 y 2 SOLUTION: Step 1: Write the first equation in the form of y 2x y 5
2x 5 y Step 2: Put the value of y of first equation in second equation 4 x 2 2 x 5 2
4 x 4 x 10 2 8x 8 x 1 Step 3: Put the value of x in first equation
2 1 y 5 25 y y 3 ANSWER: (x, y) = (1, -3)
31.
PROBLEM: x 4 y 0 2 x 5 y 6 SOLUTION: Step 1: Write the first equation in the form of x x 4 y 0
x 4 y x 4y Step 2: Put the value of x of first equation in second equation
2 4 y 5 y 6 8 y 5 y 6 3 y 6 y 2
Step 3: Put the value of y in first equation x 4 2 0 x 8 x 8
ANSWER: (x, y) = (-8, -2)
32.
PROBLEM: 3y x 5 5 x 2 y 8 SOLUTION: Step 1: Write the first equation in the form of x 3y x 5 3y 5 x Step 2: Put the value of x of first equation in second equation 5 3 y 5 2 y 8 15 y 25 2 y 8 17 y 17 y 1
Step 3: Put the value of y in first equation 3 1 x 5 35 x x 2
ANSWER: (x, y) = (-2, 1)
33.
PROBLEM: 2x 5 y 1 4 x 10 y 2
SOLUTION: Step 1: Write the first equation in the form of x 2x 5 y 1 2x 1 5 y 1 5y x 2 Step 2: Put the value of x of first equation in second equation
1 5y 4 10 y 2 2 2 10 y 10 y 2 20 y 0 y0 Step 3: Put the value of y in first equation
2x 5 0 1 2x 1 0 1 x 2 ANSWER: (x, y) = (1/2, 0)
34.
PROBLEM: 3x 7 y 3 6 x 14 y 0 SOLUTION: Step 1: Write the first equation in the form of y 3x 7 y 3 7 y 3 3x 7 y 3 3x 3 3 y x 7 7 Step 2: Put the value of y of first equation in second equation 3 3 6 x 14 x 0 7 7 6x 6 6x 0 12 x 6 1 x 2 Step 3: Put the value of x in first equation 1 3 7 y 3 2 3 7 y 3 2 3 7 y 2 3 y 14 ANSWER: (x, y) = (-1/2, 3/14)
35.
PROBLEM: 10 x y 3 1 5 x 2 y 1 SOLUTION: Step 1: Write the first equation in the form of y 10 x y 3
10 x 3 y Step 2: Put the value of y of first equation in second equation 1 5 x 10 x 3 1 2 3 5 x 5 x 1 2 3 1 2 ANSWER: Hence there is no solution.
36.
PROBLEM: 1 2 1 3 x 6 y 3 1 x1 y 3 2 3 2
SOLUTION: Step 1: Write the first equation in the form of y
1 1 2 x y 3 6 3 1 1 2 y x 6 3 3 x2 1 y 6 3 y 2x 4
Step 2: Put the value of y of first equation in second equation 1 1 3 x 2x 4 2 3 2 1 2 4 3 x x 2 3 3 2 3x 4 x 4 3 6 3 2 x 8 9 6 6 x 1 6 6 x 1 Step 3: Put the value of x in first equation 1 1 2 1 y 3 6 3 1 1 2 y 6 3 3 1 y 1 6 y6 ANSWER: (x, y) = (1, 6)
37.
PROBLEM: 2 1 3 x 3 y 1 1 x 1 y 1 4 3 12 SOLUTION: Step 1: Write the first equation in the form of y
1 2 x y 1 3 3 1 2 x 1 y 3 3 2 x3 y 3 3 x 3 y 2 3 1 x y 2 2 Step 2: Put the value of y of first equation in second equation 1 1 3 1 1 x x 4 3 2 2 12 1 1 1 1 x x 4 2 6 12 6x 4x 1 1 24 2 12 10 x 6 1 24 12 x 1
Step 3: Put the value of x in first equation 1 2 1 y 1 3 3 2 1 y 1 3 3 2 2 y 3 3 y 1 ANSWER: (x, y) = (1, 1)
38.
PROBLEM: 1 1 7 x y 2 1 x 1 y 2 4 2 SOLUTION: Step 1: Write the first equation in the form of y
1 1 x y 7 2 1 1 x y 7 2 Step 2: Put the value of y of first equation in second equation 1 11 1 x x 2 4 27 2 1 1 1 x x 2 4 14 4 7x 2x 1 2 28 4 9x 1 2 28 4 9x 9 28 4 x7
Step 3: Put the value of x in first equation
1 1 7 y 7 2 1 1 y 2 1 y 2 ANSWER: (x, y) = (7, 1/2)
39.
PROBLEM: 2 1 3 5 x 5 y 2 1 x 1 y 1 3 12 3 SOLUTION: Step 1: Write the first equation in the form of y 3 2 1 x y 5 5 2 2 1 3 y x 5 2 5 2 5 6x y 5 10 5 6x y 4 Step 2: Put the value of y of first equation in second equation 1 1 5 6x 1 x 3 12 4 3 1 5 1 1 x x 3 48 8 3 8x 3x 1 5 24 3 48 5 x 5 16 24 48 11 x 10 Step 3: Put the value of x in first equation 3 11 2 1 y 5 10 5 2 2 1 33 y 5 2 50 2 25 33 y 5 50 2 y 5 ANSWER: (x, y) = (-11/10, -2/5)
40.
PROBLEM: 1 2 x y 2 3 x 2 y 2 3 SOLUTION: Step 1: Write the first equation in the form of y
1 2 x y 2 3 3 x y 4 Step 2: Put the value of y of first equation in second equation 23 x x 2 34 1 x x 2 2 1 x2 2 x4 Step 3: 1 4 2 3 4 4 y3
Put the value of x in first equation 2 y 3
y
ANSWER: (x, y) = (4, 3)
41.
PROBLEM: 1 5 1 2 x 2 y 8 1 x 1 y 1 4 2 4 SOLUTION: Step 1: Write the first equation in the form of y
1 1 5 x y 2 2 8 1 5 1 y x 2 8 2 5 y x 4 Step 2: Put the value of y of first equation in second equation 1 15 1 x x 4 24 4 1 5 1 1 x x 4 8 2 4 3 1 5 x 4 4 8 3 25 x 4 8 3 x 6 1 x 2 Step 3: Put the value of x in first equation 1 1 1 5 y 2 2 2 8 1 5 1 y x 2 8 4 52 y 4 3 y 4 ANSWER: (x, y) = (-1/2, 3/4)
42.
PROBLEM: x y 0 x 2 y 3 SOLUTION: Step 1: Write the first equation in the form of y
x y 0 x y Step 2: Put the value of y of first equation in second equation x 2 x 3
x3 Step 3: Put the value of x in first equation x 3 0
x3 ANSWER: (x, y) = (3, 3)
43.
PROBLEM: y 3x 2 x 3 y 0
SOLUTION: Step 1: Put the value of y = 3x of first equation in second equation 2x 3y 0
2 x 3 3x 0 2x 9x 0 7 x 0 x0 Step 2: Put the value of x in first equation y 3 0 y0
ANSWER: (x, y) = (0,0)
44.
PROBLEM: 2 x 3 y 18 6 x 3 y 6
SOLUTION: Step 1: Write the first equation in the form of y
2 x 3 y 18 3 y 18 2 x 2 y 6 x 3 Step 2: Put the value of y of first equation in second equation 2 6 x 3 6 x 6 3 6 x 18 2 x 6 8 x 24 x3 Step 3: Put the value of x in first equation 2 3 3 y 18 3 y 18 6 3 y 12 y4
ANSWER: (x, y) = (3, 4)
45.
PROBLEM: 3 x 4 y 20 2x 8 y 8 SOLUTION: Step 1: Write the first equation in the form of y 3 x 4 y 20 4 y 20 3x
y 5
3 x 4
Step 2: Put the value of y of first equation in second equation 3 2x 8 5 x 8 4 2 x 40 6 x 8 8 x 8 40 8 x 32 x 4
Step 3: Put the value of x in first equation 3 4 4 y 20
4 y 20 12 8 y 4 y2 ANSWER: (x, y) = (-4, 2)
46.
PROBLEM: 5 x 3 y 1 3x 2 y 7
SOLUTION: Step 1: Write the first equation in the form of y 5 x 3 y 1
5x 1 3 y 5 1 y x 3 3 Step 2: Put the value of y of first equation in second equation
1 5 3x 2 x 7 3 3 10 2 3x x 7 3 3 19 2 x 7 3 3 19 19 x 3 3 x 1 Step 3: Put the value of x in first equation 5 1 3 y 1
5 1 3y y2 ANSWER: (x, y) = (1, 2)
47.
PROBLEM: 3 x 7 y 2 2x 7 y 1 SOLUTION: Step 1: Write the first equation in the form of y 3 x 7 y 2 7 y 2 3x
y
2 3 x 7 7
Step 2: Put the value of y of first equation in second equation 2 3 2x 7 x 1 7 7 2 x 2 3x 1
5 x 1 1 x 5 Step 3: Put the value of x in first equation
1 3 7 y 2 5 3 7y 2 5 2 3 y 7 35 10 3 y 35 1 y 5
ANSWER: (x, y) = (-1/5, 1/5)
48.
PROBLEM: y3 y 3 ANSWER: As the values of y in both the equations are not equal, the solution for the two equations is not defined.
49.
PROBLEM: x5 x 2 ANSWER: Since the values of x in both the equations are not equal, the solution for the two equations is not defined
50.
PROBLEM: y 4 y 4 ANSWER: As the values of y in both the equations are equal, the solution is (x, y) = (x, 4)
51.
PROBLEM: The sum of two numbers is 19. The larger number is 1 less than three times the smaller. SOLUTION:
Let the two numbers be x and y x + y = 19 3x - y = 1 Step 1: Write the first equation in the form of y x y 19
y 19 x Step 2: Put the value of y of first equation in second equation 3x 19 x 1
3x 19 x 1 4x=20 x=5 Step 3: Put the value of x in first equation
5
y 19
y 14 ANSWER: (x, y) = (5, 14)
52.
PROBLEM: The sum of two numbers is 15. The larger is 3 more than twice the smaller. SOLUTION:
Let the two numbers be x and y x + y = 15 y - 2x = 3 Step 1: Write the first equation in the form of y x y 15
y 15-x Step 2: Put the value of y of first equation in second equation
15-x 2x
3
15- x- 2x=3 3x=12 x=4 Step 3: Put the value of x in first equation y 15- 4
y 11 ANSWER: (x, y) = (4, 11)
53.
PROBLEM: The difference of two numbers is 7 and the sum is 1. SOLUTION:
Let the two numbers be x and y x-y=7 x+y=1 Step 1: Write the first equation in the form of y x y 7
x 7 y Step 2: Put the value of y of first equation in second equation x x 7 1
2x = 8 x=4 Step 3: Put the value of x in first equation 4 y 7
y 47 y 3 ANSWER: (x, y) = (4, -3)
54.
PROBLEM: The difference of two numbers is 3 and the sum is -7.
Let the two numbers be x and y x-y=3 x + y = -7 SOLUTION: Step 1: Write the first equation in the form of y x y 3
y x3 Step 2: Put the value of y of first equation in second equation
x x 3 7 2x-3= -7 2x = -4 x=-2 Step 3: Put the value of x in first equation 2 y 3
2 3 y y 5 ANSWER: (x, y) = (-2, -5)
55.
PROBLEM: 5 x 3 y 30 SOLUTION: x
0 -3 -6 -9 -12 -15
5 y 10 x 3 10 5 0 -5 -10 -15
ANSWER: So the solution is (-15, -15)
56.
PROBLEM: 1 1 x y 1 2 3 SOLUTION:
X 0 1 2 3 4 5 6
y
3x 6 2
-3 -3/2 0 3/2 3 9/2 6
ANSWER: So the solution is (6, 6)
58.
PROBLEM: Merits and drawbacks of the substitution method ANSWER: Merits: The substitution method is the simplest method when compared to the other methods available. It is most helpful when the question deals with just two unknown variables.
Drawbacks: Substitution is a time consuming process, since we have to move the equation around. First, we need to take out equation in the form of one variable then substitute this variable in the next equation. This method becomes quite difficult when it comes to dealing with square roots. It is long and complicated when any question consists of more than two equations. There is always an element of confusion involved as to which would be the best way to go ahead when we are working with three or four variables.
Part B: Discussion Board Topics
1.
PROBLEM:
Describe what drives the choice of variable to solve for when beginning the process of solving by substitution. ANSWER: The most important drive that makes any student to go for the choice of any variable in the equations is that variable which is there in the equation with coefficient 1. This variable is expressed in the form of rest part of the equation. When all the variables are available with the coefficient 1 then out of which the variable which is positive is generally chosen by students and is expressed in the rest part of the equation. These perceptions can include the tendency to convert negative signs to positive by shifting the value to the other side of the equation. In the substitution method, the student will try to make the substitution as simple as possible, which will make it easier to solve the value of the other variable.
4. 3 Solving Linear Systems by Elimination Part A : Elimination Method 1.
PROBLEM: x y 3 2 x y 9 SOLUTION: Step 1: Add both the equations to eliminate y, we get x y 3 2x y 9
3x
12
x4 Step 2 Substitute the value of x in first equation 4 y 3 y 3 4 y 1 ANSWER: (x, y) = (4, -1)
2.
PROBLEM: x y 6 5 x y 18 SOLUTION: Step 1: Add both the equations to eliminate y, we get x y 6 5 x y 18
6 x 24 x 4 Step 2 Substitute the value of x in first equation
4 y 6 4 6 y y2 ANSWER: (x, y) = (-4, 2)
3.
PROBLEM: x 3y 5 x 2 y 0 SOLUTION: Step 1: Add both the equations to eliminate x, we get x 3y 5 x 2y 0
y
5
y5 Step 2 Substitute the value of y in first equation x 3 5 5
x 15 5 x 5 15 x 10 ANSWER: (x, y) = (-10, 5)
4.
PROBLEM: x 4 y 4 x y 7
SOLUTION: Step 1: Add both the equations to eliminate x , we get x 4y 4 x y 7
3 y 3 y 1 Step 2 Substitute the value of y in first equation x 4 1 4 x 8 x 8
ANSWER: (x, y) = (-8, -1)
5.
PROBLEM: x y 2 x y 3 SOLUTION: Step 1: Add both the equations , we get
x y 2 x y 3 0
1
ANSWER: Hence there is no solution..
6.
PROBLEM: 3x y 2 6 x 4 y 2 SOLUTION: Step 1: Multiply first equation by 4 and then add both the equations to eliminate y 4
3 x y 2 6 x 4 y 2
12 x 4 y 8 6x 4 y 2
12 x 4 y 8 6x 4 y 2 18 x 1 x 3
6
Step 2: Substitute the value of x in first equation 1 3 y 2 3 1 y 2
y 2 1 y 1 y 1 ANSWER: (x, y) = (-1/3, 1)
7.
PROBLEM: 5 x 2 y 3 10 x y 4 SOLUTION: Step 1: Multiply second equation by 2 then add both the equations to eliminate y
5 x 2 y 3 2 10 x y 4
5 x 2 y 3 20 x 2 y 8
5 x 2 y 3 20 x 2 y 8 25 x x
5
1 5
Step 2 Substitute the value of x in first equation 1 5 2 y 3 5 1 2 y 3
2 y 4 y 2
ANSWER: (x, y) = (1/5, -2)
8.
PROBLEM: 2 x 14 y 28 x 7 y 21
SOLUTION: Step 1: Multiply second equation by 2 and then add both the equations
2 x 14 y 28 x 7 y 21
2
2 x 14 y 28 2 x 14 y 42 0
70
ANSWER: So, there is no solution
2 x 14 y 28 2 x 14 y 42
9.
PROBLEM: 2 x y 4 12 x 6 y 24 SOLUTION: Step 1: Multiply first equation by 6 then add both the equations 6
2 x y 4 12 x 6 y 24
12 x 6 y 24 12 x 6 y 24
12 x 6 y 24 12 x 6 y 24 0
0
ANSWER: (x, y) = (x, 2x+4)
10.
PROBLEM: x 8y 3 3x 12 y 6 SOLUTION: Step 1: Multiply first equation by 3 and then subtract second equation from first equation. x 8y 3 3 x 12 y 6
3
3 x 24 y 9 3 x 12 y 6
3 x 24 y 9 3 x 12 y 6 12 y
3
1 4 Step 2 Substitute the value of y in first equation y
1 x 8 3 4 x23 x 1
ANSWER: (x, y) = (1, 1/4)
11.
PROBLEM: 2 x 3 y 15 4 x 10 y 14
SOLUTION: Step 1: Multiply first equation by 2 then subtract second equation from first equation
2 x 3 y 15 4 x 10 y 14
2
4 x 6 y 30 4 x 10 y 14
4 x 6 y 30 4 x 10 y 14 16 y y 1
16
Step 2 Substitute the value of x in first equation 2 x 3 1 15 2 x 15 3 x6
ANSWER: (x, y) = (6, -1)
12.
PROBLEM: 4 x 3 y 10 3x 9 y 15 SOLUTION:
Step 1: Multiply first equation by 3 and then add both the equations to eliminate y
4 x 3 y 10 3 x 9 y 15
3
12 x 9 y 30 3 x 9 y 15
12 x 9 y 30 3 x 9 y 15 15 x 15 x 1 Step 2 Substitute the value of x in first equation 4 1 3 y 10 3 y 10 4 3 y 6 y 2
ANSWER: (x, y) = (-1, -2)
13.
PROBLEM: 4 x 5 y 3 8 x 3 y 15 SOLUTION: Step 1: Multiply first equation by 2 then add both the equations to eliminate x
4 x 5 y 3 8 x 3 y 15
2
8 x 10 y 6 8 x 3 y 15
8 x 10 y 6 8 x 3 y 15 7 y 21 y3 Step 2 Substitute the value of x in first equation 4 x 5 3 3 4 x 12 x 3 ANSWER: (x, y) = (-3, 3)
14.
PROBLEM: 2 x 7 y 56 4 x 2 y 112 SOLUTION: Step 1: Multiply first equation by 2 then add both the equations to eliminate
x
2 x 7 y 56 4 x 2 y 112
2
4 x 14 y 112 4 x 2 y 112
4 x 14 y 112 4 x 2 y 112 12 y
0
y0 Step 2 Substitute the value of y in first equation 2 x 7 0 56 2 x 56 x 28
ANSWER: (x, y) = (-28, 0)
15.
PROBLEM: 9 x 15 y 15 3 x 5 y 10 SOLUTION: Step 1: Multiply second equation by 3 then add both the equations to eliminate y
9 x 15 y 15 3 3x 5 y 10
9 x 15 y 15 9 x 15 y 30
9 x 15 y 15 9 x 15 y 30 0 45 ANSWER: Hence there is no solution..
16.
PROBLEM: 6 x 7 y 4 2 x 6 y 7 SOLUTION: Step 1: Multiply second equation by 3 and then subtract second equation from first equation
6 x 7 y 4 3 2 x 6 y 7
6x 7 y 4 6 x 18 y 21
6x 7 y 4 6 x 18 y 21 25 y y 1
25
Step 2 Substitute the value of y in first equation 6 x 7 1 4
6x 4 7 3 x 6 1 x 2 ANSWER: (x, y) = (-1/2, -1)
17.
PROBLEM: 4x 2 y 4 5 x 3 y 7
SOLUTION: Step 1: Multiply second equation by 2 and multiply first equation by 3 then add both the equations to eliminate y
3
4x 2 y 4 5 x 3 y 7 2
12 x 6 y 12 10 x 6 y 14
12 x 6 y 12 10 x 6 y 14 2
2x x 1
Step 2 Substitute the value of x in first equation 4 1 2 y 4
2y 8 y4 ANSWER: (x, y) = ( -1, 4)
18.
PROBLEM: 5 x 3 y 1 3x 2 y 7 SOLUTION: Step 1: Multiply first equation by 2 and multiply second equation by 3 then add both the equations to eliminate y 2
5 x 3 y 1 3 3x 2 y 7
10 x 6 y 2 9 x 6 y 21
10 x 6 y 2 9 x 6 y 21 19 x
19
x 1 Step 2: Substitute the value of x in first equation
5 1 3 y 1 3 y 5 1 3 y 6 y2
ANSWER: (x, y) = (1, 2)
19.
PROBLEM: 7 x 3 y 9 2 x 5 y 14 SOLUTION: Step 1: Multiply second equation by 2 then add both the equations to eliminate y 5
7 x 3 y 9 2 x 5 y 14 3
35 x 15 y 45 6 x 15 y 42
35 x 15 y 45 6 x 15 y 42 29 x
87
x3
Step 2 Substitute the value of x in first equation 7 3 3 y 9
3 y 12 y 4
ANSWER: (x, y) = (3, -4)
20.
PROBLEM: 9x 3y 3 7 x 2 y 15 SOLUTION: Step 1: Multiply first equation by 2 and multiply second equation by 3 then add both the equations to eliminate y 2
9x 3y 3 7 x 2 y 15 3
18 x 6 y 6 21x 6 y 45
18 x 6 y 6 21x 6 y 45 x
39 x 1
39
Step 2: Substitute the value of y in first equation 9 1 3 y 3 3 y 3 9 3 y 12 y 4
ANSWER: (x, y) = (-1, -4) 21.
PROBLEM: 5 x 3 y 7 7 x 6 y 11 SOLUTION: Step 1: Multiply first equation by 2 then add both the equations to eliminate y
2
5 x 3 y 7 7 x 6 y 11
10 x 6 y 14 7 x 6 y 11
10 x 6 y 14 7 x 6 y 11 3 x 3 x 1 Step 2 Substitute the value of x in first equation 5 1 3 y 7
3 y 2 2 y 3 ANSWER: (x, y) = ( -1, 2/3)
22.
PROBLEM: 2 x 9 y 8 3x 7 y 1 SOLUTION: Step 1: Multiply first equation by 3 and multiply second equation by 2 then subtract second equation from first equation 3
2 x 9 y 8 3 x 7 y 1 2
6 x 27 y 24 6 x 14 y 2
6 x 27 y 24 6 x 14 y 2 13 y y
26
2
Step 2: Substitute the value of y in first equation
2x 9 2 8 2 x 8 18 2 x 10 x 5 ANSWER: (x, y) = (-5, 2)
23.
PROBLEM: 2 x 2 y 5 3x 3 y 5
SOLUTION: Step 1: Multiply first equation by 3 and multiply second equation by 2 then subtract second equation from first equation 3
2 x 2 y 5 3 x 3 y 5 2
6 x 6 y 15 6 x 6 y 10
6 x 6 y 15 6 x 6 y 10 0 25
ANSWER: Hence there is no solution..
24.
PROBLEM: 3x 6 y 12 2x 4 y 8 SOLUTION: Step 1: Multiply first equation by 2 and multiply second equation by 3 then add both the equations 2
3 x 6 y 12 2x 4 y 8
3
6 x 12 y 24 6 x 12 y 24
6 x 12 y 24 6 x 12 y 24 0 0
ANSWER: (x, y) = (x, 1/2 x-2)
25.
PROBLEM: 25 x 15 y 1 15 x 10 y 1
SOLUTION: Step 1: Multiply first equation by 6 and multiply second equation by 10 then subtract second equation from first equation 6
25 x 15 y 1 15 x 10 y 1 10 150 x 90 y 6 150 x 100 y 10 10 y 4 2 y 5
150 x 90 y 6 150 x 100 y 10
Step 2 Substitute the value of y in first equation
25 x 15 y 1 2 25 x 15 1 5 25 x 6 1 25 x 5 1 x 5 ANSWER: (x, y) = (1/5, -2/5)
26.
PROBLEM: 2x 3y 2 18 x 12 y 5 SOLUTION: Step 1: Multiply first equation by 9 and then subtract second equation from first equation
2x 3y 2 18 x 12 y 5
9
18 x 27 y 18 18 x 12 y 5
18 x 27 y 18 18 x 12 y 5 15 y 13 13 y 15 Step 2 Substitute the value of y in first equation
13 2x 3 2 15 13 2x 2 5 10 13 2x 5 3 2x 5 3 x 10
ANSWER: (x, y) = (-3/10, -13/15)
27.
PROBLEM: y 2 x 3 3 x 2 y 4
SOLUTION: Step 1: Rewrite the first equation as y 2 x 3
2 x y 3 Step 2: Multiply first equation by 2 then add both the equations to eliminate y
2 x y 3 3 x 2 y 4
2
4 x 2 y 6 3 x 2 y 4
4 x 2 y 6 3x 2 y 4 x 2 Step 3: Substitute the value of x in first equation y 2 2 3 y 1 ANSWER: (x, y) = (-2, 1)
28.
PROBLEM: 28 x 6 y 9 6 y 4 x 15 SOLUTION: Step 1: Rewrite the second equation as 6 y 4 x 15
4 x 6 y 15 Step 2: Subtract second equation from the first equation 28 x 6 y 9 4 x 6 y 15 32 x 24 x
3 4
Step 3: Substitute the value of x in first equation
3 28 6 y 9 4 21 6 y 9 6 y 12 y 2 ANSWER: (x, y) = (3/4, -2) 29.
PROBLEM: y 5 x 15 y 5 x 5
SOLUTION: Step 1: Add both the equations to eliminate x
y 5 x 15 y 5 x 5 2 y 20 y 10
Step 2 Substitute the value of y in first equation
10 5 x 15 5 x 5 x 1
ANSWER: (x, y) = ( -1,10)
30.
PROBLEM: 2 x 3 y 9 5 x 8 y 16 SOLUTION: Step 1: Multiply first equation by 5 and multiply second equation by 2 then subtract second equation from first equation 5
2 x 3 y 9 5 x 8 y 16 2
10 x 15 y 45 10 x 16 y 32
10 x 15 y 45 10 x 16 y 32 y 77
Step 2 Substitute the value of y in first equation 2 x 3 77 9
2 x 231 9 2 x 240 x 120 ANSWER: (x, y) = (120, 77)
31.
PROBLEM: 1 1 1 2 x 3 y 6 5 x y 7 2 2
SOLUTION: Step 1: Multiply first and second equations by 6 1 1 6 1 2 x 3 y 6 3x 2 y 1 6 15 x 6 y 21 5 x y 7 2 2
3x 2 y 1 15 x 6 y 21 Step 2: Multiply first equation by 5 then subtract second equation from first equation 5
3x 2 y 1 15 x 6 y 21
15 x 10 y 5 15 x 6 y 21
15 x 10 y 5 15 x 6 y 21 16 y 16 y 1
Step 3: Substitute the value of y in first equation 1 1 1 x 1 2 3 6 1 1 1 x 2 6 3 1 1 x 2 2 x 1 ANSWER: (x, y) = (1, 1)
32.
PROBLEM: 1 1 4 x 9 y 1 3 x y 4 SOLUTION: Step 1: Rewrite the first equation as 1 1 x y 1 4 9 9x 4 y 1 36 9 x 4 y 36 Step 2: Multiply second equation by 4 then add both the equations to eliminate y 9 x 4 y 36 9 x 4 y 36 3 4 4 x 4 y 3 x y 4
9 x 4 y 36 4x 4 y 3 13 x 39 x3 Step 3: Substitute the value of x in first equation 1 1 3 y 1 4 9 3 1 1 y 4 9 1 1 y 4 9 9 y 4 ANSWER: (x, y) = (3, -9/4)
33.
PROBLEM: 1 1 1 2 x 4 y 3 1 x 1 y 19 4 2 6
SOLUTION: Step 1: Multiply first and second equations by 12 1 1 1 12 6 x 3 y 4 2 x 4 y 3 12 3x 6 y 38 1 x 1 y 19 6 2 4
6x 3y 4 3x 6 y 38 Step 2: Multiply first equation by 2 then subtract second equation from first equation
6 x 3 y 4 3 x 6 y 38
2
12 x 6 y 8 3 x 6 y 38
12 x 6 y 8 3 x 6 y 38 15 x 30 x 2
Step 3: Substitute the value of x in first equation 1 1 1 2 y 2 4 3 1 4 y 4 3 16 y 3 ANSWER: (x, y) = (-2, -16/3)
34.
PROBLEM: 14 3 x 2 y 4 1 x 1 y 4 3 7 21 SOLUTION: Step 1: Rewrite the first and second equations as 14 x 2y 4 3 14 x 6 y 4 3 14 x 6 y 12 1 1 4 x y 3 7 21 7 x 3 y 4 21 21 7 x 3 y 4
Step 2: Multiply second equation by 2 then subtract second equation from first equation
14 x 6 y 12 7 x 3 y 4
2
14 x 6 y 12 14 x 6 y 8 0
12
ANSWER: So, there is no solution.
14 x 6 y 12 14 x 6 y 8
35.
PROBLEM: 0.025 x 0.1 y 0.5 0.11x 0.04 y 0.2 SOLUTION: Step 1: Multiply first equation by 1000 and multiply second equation by 100 to get equations in the form 1000
0.025 x 0.1 y 0.5 0.11x 0.04 y 0.2
100
25 x 100 y 500 11x 4 y 20
25 x 100 y 500 11x 4 y 20
Step 2: Multiply second equation by 25 then subtract second equation from first equation
25 x 100 y 500 11x 4 y 20
25
25 x 100 y 500 275 x 100 y 500
25 x 100 y 500 275 x 100 y 500 250 x x 4
1000
Step 3: Substitute the value of x in first equation 0.025 4 0.1 y 0.5 .1 .1 y .5 .1 y .6 y6
ANSWER: (x, y) = (-4, 6)
36.
PROBLEM: 1.3x 0.1y 0.35 0.5 x y 2.75
SOLUTION: Step 1: Multiply first equation by 100 and multiply second equation by 100 to rewrite equations in the form 100
1.3 x 0.1 y 0.35 0.5 x y 2.75 100
130 x 10 y 35 50 x 100 y 275
130 x 10 y 35 50 x 100 y 275
Step 2: Multiply first equation by 10 then subtract second equation from first equation
130 x 10 y 35 50 x 100 y 275
10
1300 x 100 y 350 50 x 100 y 275
1300 x 100 y 350 50 x 100 y 275 1250 x 625 1 2 x .5 x
Step 3: Substitute the value of x in first equation 1.3 .5 0.1 y 0.35 .65 0.1 y 0.35 0.1 y .30 y 3 ANSWER: (x, y) = ( .5, -3)
37.
PROBLEM: x y 5 0.02 x 0.03 y 0.125 SOLUTION: Step 1: Multiply second equation by 1000 to get equations in the form as
x y 5 1000 0.02 x 0.03 y 0.125 x y 5
x y 5 20 x 30 y 125
20 x 30 y 125 Step 2: Multiply first equation by 20 then subtract second equation from first equation x y 5 20 x 30 y 125
20
20 x 20 y 100 20 x 30 y 125
20 x 20 y 100 20 x 30 y 125 10 y 25 5 2 y 2.5 y
Step 3: Substitute the value of y in first equation x 2.5 5
x 2.5 ANSWER: (x, y) = (2.5, 2.5)
38.
PROBLEM: x y 30 0.05 x 0.1y 2.4
SOLUTION:
Step 1: Multiply second equation by 100 to rewrite equations in the form
x y 30 0.05 x 0.1y 2.4
100
x y 30 5 x 10 y 240
x y 30 5 x 10 y 240 Step 2: Multiply first equation by 5 then subtract second equation from first equation 5
x y 30 5 x 10 y 240
5 x 5 y 150 5 x 10 y 240
5 x 5 y 150 5 x 10 y 240 5y y 18
90
Step 3: Substitute the value of y in first equation
x 18 3 x 12 ANSWER: (x, y) = (12, 18)
39.
PROBLEM: The sum of two numbers is 14. The larger number is 1 less than two times the smaller. SOLUTION:
Let the two numbers be x and y Given: x + y =14 2x - y =1
Step 1: Multiply first equation by 2 then subtract second equation from first equation
x y 14 2 x y 1
2
2 x 2 y 28 2x y 1
2 x 2 y 28 2x y 1 3 y 27 y9 Step 2 Substitute the value of y in first equation x 9 14
x 5 ANSWER: (x, y) = (5, 9)
40.
PROBLEM: The sum of two numbers is 30. The larger is 2 more than three times the smaller. SOLUTION: Given: x + y = 30 - 3x + y = 2 Step 1: Subtract second equation from first equation x y 30 3 x y 2
4 x 28 x7 Step 2: Substitute the value of x in first equation 7 y 30 y 23
41.
ANSWER: (x, y) = (7, 23) PROBLEM: The difference of two numbers is 13 and the sum is 11.
SOLUTION: Given: x - y = 13 x + y =11 Step 1: Add both the equations to eliminate y
x y 13 x y 11 2 x 24 x 12 Step 2 Substitute the value of y in first equation 12 y 13 y = -1 ANSWER: (x, y) = (12, -1)
42.
PROBLEM: The difference of two numbers is 2 and the sum is -12. SOLUTION: Given: x + y = -12 x-y=2 Step 1: Subtract second equation from first equation x y 12 x y 2
2 x 10 x 5 Step 2: Substitute the value of x in first equation 5 y 12 y 7
ANSWER: (x, y) = (-5,-7) Part B: Mixed Exercises
43.
PROBLEM: y 2x 3 3x y 12 SOLUTION: Step 1: Rewrite the first equation in the form y 2x 3
2 x y 3 Step 2: Subtract second equation from first equation 2 x y 3 3x y 12
5 x 15 x3 Step 3: Substitute the value of x in first equation 2 3 y 3 y3
ANSWER: (x, y) = (3, 3)
44.
PROBLEM: x 3 y 5 1 y 3 x 5
SOLUTION: Step 1: Rewrite the second equation as 1 y x5 3 3 y x 15 x 3 y 15 Step 2: Add both the equations to eliminate x
x 3 y 5 x 3 y 15 6 y 10 y
5 3
Step 3: Substitute the value of y in first equation 5 x 3 5 3 x 10 ANSWER: (x, y) = (-10, 5/3)
45.
PROBLEM: x 1 y3 ANSWER: From above equations it is clear that the solution is (x, y) = (-1, 3)
46.
PROBLEM: 1 y 2 x 9 0 SOLUTION: Step 1: Rewrite the second equation as x9 0 x 9 ANSWER: (x, y) = (-9, 1/2)
47.
PROBLEM: yx x y 1 SOLUTION: Step 1: Rewrite the second equation as
x y 1 y 1 x Step 2: yx y 1 x ANSWER: Hence there is no solution..
48.
PROBLEM: y 5x y 10
SOLUTION: Step 1: Put the value of y of second equation in the first equation, we get
10 5 x 5 x 10 x 2
ANSWER: (x, y) = (-2, -10)
49.
PROBLEM: 3 y 2 x 24 3x 4 y 2
SOLUTION: Step 1: Rewrite the first equation in the form 3 y 2 x 24
2 x 3 y 24 Step 2: Multiply first equation by 3 and multiply second equation by 2 then add both the equations 2 x 3 y 24 3x 4 y 2
3
2
6 x 9 y 72 6x 8 y 4
6 x 9 y 72 6x 8 y 4 17 y 68 y 4
Step 2 Substitute the value of y in first equation 3 4 2 x 24 3 4 2 x 24 2 x 12 x6
ANSWER: (x, y) = (6, -4)
50.
PROBLEM: 3 y x 1 2 2 y 2 3 x
SOLUTION: Step 1: Rewrite the first and second equations in the form 3 y x 1 2 3 x y 1 2 3x 2 y 2
2 y 2 3 x 3x 2 y 2 Step 2: Subtract second equation from first equation 3x 2 y 2 3x 2 y 2
0 0 ANSWER: (x, y) = (x, -3/2 x +1)
51.
PROBLEM: 7 y 2 x 1 7 x 2 y 23 SOLUTION: Step 1: Rewrite the first and second equations in the form
7 y 2 x 1 2 x 7 y 1 7 x 2 y 23 7 x 2 y 23 2 x 7 y 1 7 x 2 y 23 Step 2: Multiply first equation by 3 and multiply second equation by 2 then add both the equations 2
2 x 7 y 1 7 x 2 y 23 7
4 x 14 y 2 49 x 14 y 161
4 x 14 y 2 49 x 14 y 161 53x 159 x3
Step 3: Substitute the value of x in first equation
7 y 2 3 1 7 y 7 y 1 ANSWER: (x, y) = (3, -1)
52.
PROBLEM: 5 x 9 y 14 0 3x 2 y 5 0 SOLUTION: Step 1: Rewrite the first and second equations in the form 5 x 9 y 14 0 5 x 9 y 14
3x 2 y 5 0 3x 2 y 5 Step 2: Multiply first equation by 3 and multiply second equation by 5 and then subtract second equation from first equation 5 x 9 y 14 3 x 2 y 5
3
5
15 x 27 y 42 15 x 10 y 25
15 x 27 y 42 15 x 10 y 25 17 y 17 y 1
Step 3: Substitute the value of y in first equation 5 x 9 1 14
5 x 14 9 5x 5 x 1 ANSWER: (x, y) = (1, 1)
53 .
PROBLEM: 5 y 16 x 10 y 5 x 10 16
Step 1: Rewrite the first and second equations in the form. 5 y x 10 16 5 x y 10 16 5 x 16 y 10 16 5 x 16 y 160
5 x 10 16 5 x 160 y 16 16 y 5 x 160 5 x 16 y 160 y
5 x 16 y 160 5 x 16 y 160 Step 2: Add both the equations 5 x 16 y 160 5 x 16 y 160
10 x 320 x 32 Step 3: Substitute the value of x in first equation 5 y 32 10 16 y0 ANSWER: (x, y) = (32, 0)
54.
PROBLEM: 6 y x 12 5 x6 SOLUTION:
Given x=6 Step 1: Substitute the value of x in first equation 6 y 6 12 5 36 60 y 5 24 y 5 ANSWER: (x, y) = (6, 24/5)
55.
PROBLEM: 2 x 3 y 0 3 2 x y 1 15 SOLUTION: Step 1: Rewrite the first and second equations in the form 2 x 3 y 0
2x 6 y 0 2x y 6 3 2 x y 1 15 2x y 1 5 2x y 6 Step 2: Subtract second equation from first equation 2x y 6
2x y 6 0 0 ANSWER: (x, y) = (x, -2x+6)
56.
PROBLEM: 3 2 x y 3 4 x 3 y 1 8 SOLUTION: Step 1: Rewrite the first and second equations in the form 3 2 x y 3
3 2 x 2 y 3 2 x 2 y 6 2x 2 y 6 x y 3 4 x 3 y 1 8 4x 3y 3 8 4 x 3 y 11 Step 2: Multiply first equation by 4 and then subtract second equation from first equation
x y 3 4 x 3 y 11
4
4 x 4 y 12 4 x 3 y 11
4 x 4 y 12 4 x 3 y 11 y 1 y 1 Step 3: Substitute the value of y in first equation x 1 3 x 1 3 x2
ANSWER: (x, y) = (2, -1)
57.
PROBLEM: 2 x 1 3 2 y 1 21 3 x 2 1 (3 y 2)
SOLUTION: Step 1: Rewrite the first and second equations in the form
2 x 1 3 2 y 1 21 2 x 2 6 y 3 21 2 x 6 y 26 x 3 y 13 3 x 2 1 (3 y 2) 3x 6 1 3 y 2 3x 3 y 3 x y 1 Step 2: Subtract second equation from first equation x 3 y 13 x y 1
4 y 12 y 3 Step 3: Substitute the value of y in first equation
2 x 2 6 3 3 21 2 x 6 2 2 x 8 x 4 ANSWER: (x, y) = (-4, 3)
58.
PROBLEM: x y 2 3 7 x y 8 3 2
SOLUTION: Step 1: Rewrite the first and second equations in the form 100
0.15 x 1.25 y 0.4 0.03x 0.25 y 0.08 100
15 x 125 y 40 3 x 25 y 80
15 x 27 y 42 15 x 10 y 25 17 y 17 y 1 0.15 x 1.25 y 0.4 0.03x 0.25 y 0.08 Step 2: Multiply first equation by 2 and multiply second equation by 3 and then subtract second equation from first equation 3 x 2 y 42 2 x 3 y 48
2
3
6 x 4 y 84 6 x 9 y 144
6 x 4 y 84 6 x 9 y 144 5 y 60 y 12
Step 3: Substitute the value of y in first equation
x 12 7 2 3 x 1 3 x 4 7 x 1 3 2 x2 x 3 2 x 6 ANSWER: (x, y) = (-6, 12)
59 .
PROBLEM: x y 3 4 2 4 x y 1 3 6 6 SOLUTION: Step 1: Rewrite the first and second equations in the form
x y 3 4 2 4 x 2y 3 4 4 x 2y 3 x y 1 3 6 6 2x y 1 6 6 2x y 1
Step 2: Multiply second equation by 2 ans then add both the equations to eliminate y
x 2y 3 2 x y 1
2
x 2y 3 4 x 2 y 2
x 2y 3 4x 2 y 2 5x
5
x 1 Step 3: Substitute the value of x in first equation
1 y 3 4 1
2 4 3 y 4 4 2 1 y 2 2 y 1 ANSWER: (x, y) = (1,-1 )
60.
PROBLEM: 2 1 3 x 3 y 3 1 x 1 y 8 3 2 3
SOLUTION: Step 1: Rewrite the first and second equations in the form
1 2 x y 3 3 3 x 2y 3 3 x 2y 9 1 1 8 x y 3 2 3 2x 3y 8 6 3 2 x 3 y 16
Step 2: Multiply first equation by 2 and then subtract second equation from first equation
x 2y 9 2 x 3 y 16
2
2 x 4 y 18 2 x 3 y 16
2 x 4 y 18 2 x 3 y 16 y 2 y 2 Step 3: Substitute the value of y in first equation. 1 2 x 2 3 3 3 1 4 x 3 3 3 1 5 x 3 3 x5 ANSWER: (x, y) = (5, -2)
61.
PROBLEM: 1 4 1 10 x 2 y 5 1 x 1 y 2 7 3 21 SOLUTION: Step 1: Rewrite the first and second equations in the form
1 1 4 x y 10 2 5 1 1 4 x 1 10 2 5 1 4 1 x 10 5 2 1 85 x 10 10 x 3
Step 2: Multiply first equation by 3 and then add both the equations to eliminate x
x 5 y 8 3x 7 y 2
3
3x 15 y 24 3x 7 y 2
3 x 15 y 24 3x 7 y 2 22 y 22 y 1 Step 3: Substitute the value of y in first equation
1 1 4 x y 10 2 5 1 1 4 x 1 10 2 5 1 4 1 x 10 5 2 1 85 x 10 10 x 3
ANSWER: (x, y) = (-3,1 )
62.
PROBLEM: 5 1 y 3 x 2 1 x 1 y 1 3 5 10
SOLUTION: Step 1: Rewrite the first and second equations in the form 5 1 y x 3 2 10 x 3 y 6 6 y 10 x 3
10 x 6 y 3 1 1 1 x y 3 5 10 5x 3 y 1 15 10 3 5x 3 y 2 10 x 6 y 3
Step 2: Subtract second equation from first equation 10 x 6 y 3
10 x 6 y 3 0 0 ANSWER: (x, y) = (x, -5/3 x+1/2)
63.
PROBLEM: 2 1 7 x y 3 1 x 1 y 1 14 2 3
SOLUTION: Step 1: Rewrite the first and second equations in the form 1 2 x y 7 3 2 x 7 y 7 3 3 x 21 y 14 1 1 1 x y 14 2 3 x 7 y 1 14 3 3 x 21 y 14
Step 2: Subtract second equation from first equation 3x 21 y 14 3x 21 y 14
0 28 ANSWER:
Hence there is no solution..
64.
PROBLEM: 1 1 1 15 x 12 y 3 3 x 3 y 3 10 8 2
SOLUTION: Step 1: Rewrite the first and second equations in the form 1 1 1 x y 15 12 3 4x 5 y 1 60 3 4 x 5 y 20 3 3 3 x y 10 8 2 3 12 x 15 y 40 2 12 x 15 y 60
Step 2: Multiply first equation by 2 then add both the equations to eliminate x 4 x 5 y 20 12 x 15 y 60
3
12 x 15 y 60 12 x 15 y 60 0
0
ANSWER: (x, y) = (x, 4/5 x-4)
65.
PROBLEM: x y 4200 0.03x 0.0525 y 193.5 SOLUTION:
12 x 15 y 60 12 x 15 y 60
Step 1:Multiply second equation by 10000
x y 4200 0.03 x 0.0525 y 193.5
10000
x y 4200 300 x 525 y 1935000
Step 2: Multiply first equation by 300 and then add both the equations to eliminate x 300
x y 4200 300 x 525 y 1935000
300 x 300 y 1260000 300 x 525 y 1935000
300 x 300 y 1260000 300 x 525 y 1935000 225 y
675000
y 3000
Step 3: Substitute the value of y in first equation
x 3000 4200 x 1200 ANSWER: (x, y) = (1200,3000)
66.
PROBLEM: x y 350 0.2 x 0.1 y 52.5 SOLUTION: Step 1: Multiply second equation by 10
x y 350 0.2 x 0.1y 52.5
10
x y 350 2 x y 525
Step 2: Multiply first equation by 2 and then subtract second equation from first equation
x y 350 2 x y 525
2
2 x 2 y 700 2 x y 525
2 x 2 y 700 2 x y 525 y
175
Step 3: Substitute the value of y in first equation
x 175 350 x 175 ANSWER: (x, y) = (175, 175)
67.
PROBLEM: 0.2 x 0.05 y 0.43 0.3x 0.1y 0.3
SOLUTION: Step 1:Multiply first equation by 100 and multiply second equation by 10
0.2 x 0.05 y 0.43 0.3x 0.1y 0.3
100
10
20 x 5 y 43 3x y 3
Step 2: Multiply first equation by 300 and then add both the equations to eliminate x 20 x 5 y 43 3 x y 3 5
20 x 5 y 43 15 x 5 y 15
20 x 5 y 43 15 x 5 y 15 35 x
28
4 5 x .8 x
Step 3: Substitute the value of x in first equation
0.2 .8 0.05 y 0.43 .16 .05 y .43 .05 y .27 y 5.4
ANSWER: (x, y) = (0.8, -5.4)
68.
PROBLEM: 0.1x 0.3 y 0.3 0.05 x 0.5 y 0.63 SOLUTION: Step 1: Multiply first equation by 10 and multiply second equation by 100
0.1x 0.3 y 0.3 0.05 x 0.5 y 0.63
10
100
x 3y 3 5 x 50 y 63
Step 2: Multiply first equation by 5 and then subtract second equation from first equation 5
x 3y 3 5 x 50 y 63 5 x 15 y 15 5 x 50 y 63
5 x 15 y 15 5 x 50 y 63
65 y 78 6 y 5 y 1.2 Step 3: Substitute the value of y in first equation 0.1x 0.3 1.2 0.3
.1x .3 .36 .1x .06 x .6 ANSWER: (x, y) = (-0.6, 1.2)
69.
PROBLEM: 0.15 x 0.25 y 0.3 0.75 x 1.25 y 4 SOLUTION: Step 1:Multiply first equation by 100 and multiply second equation by 100 100
0.15 x 0.25 y 0.3 0.75 x 1.25 y 4
15 x 25 y 30 75 x 125 y 40
100
5
15 x 25 y 30 75 x 125 y 40
75 x 125 y 150 75 x 125 y 40
75 x 125 y 150 75 x 125 y 40 0 190 x7 0.15 x 0.25 y 0.3 0.75 x 1.25 y 4 Step 2: Multiply first equation by 5 and then add both the equations 15 x 25 y 30 75 x 125 y 40
5
75 x 125 y 150 75 x 125 y 40 0 190
ANSWER: So, there is no solution.
75 x 125 y 150 75 x 125 y 40
70.
PROBLEM: 0.15 x 1.25 y 0.4 0.03x 0.25 y 0.08 SOLUTION: Step 1: Multiply first equation by 10 and multiply second equation by 100 100
0.15 x 1.25 y 0.4 0.03x 0.25 y 0.08 100
15 x 125 y 40 3x 25 y 8
Step 2: Multiply second equation by 5 and then subtract second equation from first equation 15 x 125 y 40 3 x 25 y 8
5
15 x 125 y 40 15 x 125 y 40
15 x 125 y 40 15 x 125 y 40 0 0
ANSWER: (x, y) = (x, .12x+.32) Part C: Discussion Board Topics
71.
PROBLEM:
How do we choose the best method for solving a linear system? ANSWER:
The best method we choose for solving a linear system depends entirely on the equations that that we find in the questions. Suppose we have the equation like given below----2 x 3 y 9 5 x 8 y 16 In this case, it is easy to use substitution method as in the first equation one variable is there with the coefficient 1. So, it becomes easy to put rest part of the equation in that variable (variable which is with coefficient 1)
Here, first equation can be written as 10 x y 20 y 20 10 x Then this variable can be put in the second equation as 7 x 5 20 10 x 14 After which we can find the value of x 7 x 5 20 10 x 14 7 x 100 50 x 14 43x 86 x2 Keeping this value in the equation y 20 10 x We get the value of y y 20 10 2 y 20 20 y0
Similarly when we find equations where there is no variable is there with coefficient 1, and then we find elimination method very useful there.
2 x 3 y 9 6 x 8 y 7 Here there are no equations where there any variable with coefficient 1, so we will use the elimination method for calculating the value of variables. We will try to eliminate one variable out of the variables given. Here we are eliminating x by multiplying first equation with 3. 2 x 3 y 9 6 x 8 y 7
3
6 x 9 y 27 6 x 8 y 7 17 y 34 y2
6 x 9 y 27 6 x 8 y 7
From the value of y when kept in the first equation, we get the value of x 2 x 3 2 9 2 x 3 3 x 2 72.
PROBLEM:
What does it mean for a system to be dependent? How can we tell if a given system is dependent? SOLUTION: We say that the system is dependent when equations are equal.
We can determine if a given system is dependent or independent based on the equality of the given equations. 6 x 2 y 14 6 x 2 y 14 0 0 In given example, we find that both equations are equal. ANSWER: Hence we can conclude that the lines coincide in this case, and we need y in terms
of x to present the solution in the form of (x, mx+b)
4.4 Applications of Linear Systems Part A: Applications Involving Numbers Set up a linear system and solve.
1.
PROBLEM: The sum of two integers is 54 and the difference is 10. Find the integers. SOLUTION:
Let x be one of the integers Let y be other integer. Given : x + y = 54
x - y = 10 Step 1: Add both the equations
x y 54 x y 10 2 x 64 x 32 Step 2: Keeping the value of x in first equation 32 y 54
y 54 - 32 y 22 ANSWER: The integers are 32 and 22.
2.
PROBLEM: The sum of two integers is 50 and the difference is 24. Find the integers. SOLUTION:
Let x be one of the integers Let y be other integer. Given: x + y = 50 x - y = 24 Step 1: Add both the equations
x y 50 x y 24 74 x 37 Step 2: Keeping the value of x in first equation 37 y 50 2x
y = 50 - 37 y = 13 ANSWER: The integers are 37 and 13.
3.
PROBLEM: The sum of two positive integers is 32. When the smaller integer is subtracted from twice the larger the result is 40. Find the two integers. SOLUTION:
Let x be smaller integers Let y be larger integer. Given : x + y = 32 2y – x = 40 Step 1: Rewrite the second equation in the form 2y x 40
x 2y 40 Step 2: Add both the equations x y 32 x 2y 40 3 y 72 y 24 Step 3: Keeping the value of y in the first equation x 24 32
x 8 ANSWER: The integers are 8 and 24
4.
PROBLEM: The sum of two positive integers is 48. When twice the smaller integer is subtracted from the larger the result is 12. Find the two integers. SOLUTION:
Let x be smaller integer Let y be larger integer. Given: x + y = 48 y - 2x= 12
Step 1: Rewrite the second equation in the form y 2 x 12
2 x y 12 Step 2: Subtract second equation from the first equation x y 48 2 x y 12
3x 36 x 12 Step 3: Keeping the value of y in first equation 12 y 48
y 36 ANSWER: The integers are 12 and 36
5.
PROBLEM: The sum of two integers is 74. The larger is 26 more than twice the smaller. Find the two integers. SOLUTION:
Let x be smaller integers Let y be larger integer. Given : x + y = 74 y - 2x= 26 Step 1: Rewrite the second equation in the form y 2x 26
2x y 26 Step 2: Subtract second equation from the first equation x y 74 2 x y 26
3x 48 x 16 Step 3: Keeping the value of x in first equation
16 y 74 y 58 ANSWER: The integers are 16 and 58
6.
PROBLEM: The sum of two integers is 45. The larger is 3 less than three times the smaller. Find the two integers. SOLUTION:
Let x be smaller integer Let y be larger integer. Given: x + y = 45 3x- y = 3 Step 1: Add both the equations
x y 45 3x y 3 4 x 48 x 12 Step 2: Keeping the value of x in first equation 12 y 45
y 45 - 12 y 33 ANSWER: The integers are 12 and 33
7.
PROBLEM: The sum of two numbers is zero. When 4 times the smaller number is added to 8 times the larger the result is 1. Find the two numbers. SOLUTION:
Let x be smaller number. Let y be larger number.
Given : x+y=0 4x + 8y= 1 Step 1: Multiply the first equation by 4 and then subtract the second equation from first equation 4
x y 0 4 x 8y 1
4x 4 y 0 4 x 8y 1
4x 4 y 0 4 x 8y 1 4 y 1 y
1 4
Step 2: Keeping the value of y in first equation 1 x 0 4 1 x 4 ANSWER:
The integers are -
8.
1 1 and 4 4
PROBLEM: The sum of a larger number and 4 times a smaller number is 5. When 8 times the smaller is subtracted from twice the larger the result is -2. Find the numbers. SOLUTION:
Let x be smaller number. Let y be larger number Given : 4x + y = 5 2y - 8x= -2 Step 1: Rewrite the second equation in the form
2y 8 x -2 8 x 2y -2 Step 2: Multiply first equation by 2 and then add both the equations 2
4x y 5 8 x 2y -2
8 x 2 y 10 8 x 2y -2
8 x 2 y 10 8 x 2y -2 4y 8 y2 Step 3: Keeping the value of y in first equation 4 x 2 5
4 x 5-2 3 x= 4 ANSWER:
The numbers are 9.
3 and 2 4
PROBLEM: The sum of 12 times the larger number and 11 times the smaller is -36. The difference of 12 times the larger and 7 times the smaller is 36. Find the numbers. SOLUTION:
Let x be smaller number. Let y be larger number. Given : 11x + 12y = -36 12y – 7x = 36 Step 1: Rewrite the second equation 12y – 7x 36
–7x 12y 36 Step 2: Subtract second equation from first equation
11x 12y 36 – 7x 12y 36 18 x 72 x 4 Step 3: Keeping the value of x in first equation 11 4 12y 36
12 y 8 2 y 3 ANSWER:
The integers are - 4 and
10.
2 3
PROBLEM: The sum of 4 times the larger number and 3 times the smaller is 7. The difference of 8 times the larger and 6 times the smaller is 10. Find the numbers. SOLUTION:
Let x be smaller number Let y be larger number. Given: 3x 4y 7 8y 6x 10 Step 1: Rewrite the second equation in the form 8y - 6x 10
- 6x 8y 10 Step 2: Multiply the first equation by 2 and then add both the equations
3x 4y 7 -6x 8y 10
2
6 x 8 y 14 6x 8y 10
6 x 8 y 14 -6x 8y 10
16 y 24 3 y 2 Step 2: Keeping the value of y in first equation 3 3x 4 7 2 3x 6 7
3x=1 1 x= 3 ANSWER:
The numbers are
1 3 and 3 2
Part B: Interest and Money Problems Set up a linear system and solve.
11.
PROBLEM: A $7,000 principal was invested in two accounts, one earning 3% and another earning 7% interest. If the total interest for the year was $262, then how much was invested in each account? SOLUTION:
Let x represents the amount invested at 3 %= .03 Let y represents the amount invested at 7% = .07 Given : x y 7000 .03x .07 y 262 Step 1: Multiply the second equation by 100
x y 7000 .03x .07 y 262
100
x y 7000 3x 7 y 26200
Step 2: Multiply first equation by 3 and then subtract second equation from the first equation
x y 7000 3x 7 y 26200
3
3x 3 y 21000 3x 7 y 26200
3x 3 y 21000 3x 7 y 26200 4 y 5200 y 1300 Step 3: Keeping the value of y in first equation x 1300 7000
x 5700 ANSWER: The two amounts are $5700and $1300
12.
PROBLEM: Mary has her total savings of $12,500 in two different CD accounts. One CD earns 4.4% and another earns 3.2% interest. If her total interest for the year was $463, then how much does she have in each CD account? SOLUTION:
Let x represents the amount invested at 4.4 % = 0.044 Let y represents the amount invested at 3.2% = 0.032 Given: x y 12500
.044 x .032 y 463 Step 1: Multiply the second equation by 1000
x y 12500 0.044 x 0.032 y 463
1000
x y 12500 44 x 32 y 463000
Step 2: Multiply the second equation by 44 and then add both the equations
44
x y 12500 44 x 32 y 463000
44 x 44 y 550000 44 x 32 y 463000
44 x 44 y 550000 44 x 32 y 463000 12 y 87000 y 7250 Step 3: Keeping the value of y in first equation x 7250 12500
x 5250 ANSWER: The two amounts are $5250 and $7250
13.
PROBLEM: Sally’s $1800 savings is in two accounts. One account earns 6% annual interest and the another earns 3% . Her total interest for the year was $93. How much does she have in each account? SOLUTION:
Let x represents the amount invested at 6 %= .06 Let y represents the amount invested at 3% = .03 Given: x y 1800 .06 x .03 y 93 Step 1: Multiply the second equation by 100
x y 1800 .06 x .03 y 93
100
x y 1800 6 x 3 y 9300
Step 2: Multiply the first equation by 6 and then add both the equations
x y 1800 .06 x .03 y 93
6
6 x 6 y 10800 6 x 3 y 9300
6 x 6 y 10800 6 x 3 y 9300 3 y 1500 y 500 Step 3: Keeping the value of y in first equation x 500 1800
x 1300 ANSWER: The two amounts are $1300 and $500
14.
PROBLEM: Joe has two savings accounts totaling $4,500. One account earns 3 34 % annual interest and the other earns 2 85 % . If his total interest for the year was $141.75, then how much was in each account? SOLUTION:
15 % .0375 4 21 Let y represents the amount invested at 2 85 % % .02625 8
Let x represents the amount invested at 3 34 %
Given: x y 4500 .0375 x .02625 y 141.75 Step 1: Multiply the second equation by 10000
x y 4500 .0375 x .02625 y 141.75
100000
x y 4500 3750 x 2625 y 14175000
Step 2: Multiply the first equation by 375 and subtract the second equation from the first equation
x y 4500 3750 x 2625 y 14175000
3750
3750 x 3750 y 16875000 3750 x 2625 y 14175000
3750 x 3750 y 16875000 3750 x 2625 y 14175000 1125 y 2700000 y 2400
Step 3: Keeping the value of y in first equation x 2400 4500
x 2100 ANSWER: The two amounts are $2100 and $2400
15.
PROBLEM: Millicent has $10,000 invested in two accounts. For the year she earned $535 more in interest from her 7% Mutual Fund account than she did from her 4% CD. How much does she have in each account? SOLUTION:
Let x represents the amount invested at 7%= .07 Let y represents the amount invested at 4% = .04 Given: x y 10000 .07 x .04 y 535 Step 1: Multiply the second equation by 100
x y 10000 .07 x .04 y 535
100
x y 10000 7 x 4 y 53500
Step 2: Multiply the first equation by 4 and then add both the equations
x y 10000 .07 x .04 y 535
4
4 x 4 y 40000 7 x 4 y 53500
4 x 4 y 40000 7 x 4 y 53500 11x 93500 x 8500 Step 3: Keeping the value of y in first equation 8500 y 10000 y 1500
ANSWER: The two amounts are $8500 and $1500
16.
PROBLEM: A small business has $85,000 invested in two accounts. If the account earning 3% annual interest earned $825 more in interest than the account earning 4.5% annual interest, then how much was invested in each account? SOLUTION:
Let x represents the amount invested at 3% .03 Let y represents the amount invested at 4.5% .045 Given: x y 85000 .03 x .045 y 825 Step 1: Multiply the second equation by 1000
x y 85000 .03x .045 y 825
1000
x y 85000 30 x 45 y 825000
Step 2: Multiply the first equation by 45 and then add both the equations
x y 85000 30 x 45 y 825000
45
45 x 45 y 3825000 30 x 45 y 825000
45 x 45 y 3825000 30 x 45 y 825000 75 x 4650000 62000 x Step 3: Keeping the value of x in first equation 62000 y 85000 y 23000 ANSWER: The two amounts are $6200 and $2300
17.
PROBLEM: Jerry earned a total of $284 in simple interest from two separate accounts. In an account earning 6% interest, Jerry invested $1,000 more than twice the amount he invested in an account earning 4%. How much did he invest in each account? SOLUTION:
Let x represents the amount invested at 4%= .04 Let y represents the amount invested at 6% = .06 Given : y =2x +1000 .04 x .06 y 284 Step 1: Substitute the value of y in second equation .04 x .06 y 284
.04 x .06 2 x 1000 284 .04 x .12 x 60 284 .16 x 224 x 1400 Step 2: Keeping the value of x in first equation
y 2 1400 1000 y 3800 ANSWER: The two amounts are $1400 and $3800
18.
PROBLEM: James earned a total of $68.25 in simple interest from two separate accounts. In an account earning 2.6% interest, James invested one-half as much as he did in the other account that earned 5.2%. How much did he invest in each account? SOLUTION:
Let x represents the amount invested at 5.2% .052 Then y represents the other amount invested at 2.6% .026 Given: y=x/2 0.052 x +0.026 (x/2) = 68.25 Step 1: Multiply equation by 2
.052x .026 x / 2 68.25
2
.104 x .026x 136.5
Step 2: Multiply equation by 1000
.104 x .026x 136.5
1000
104 x 26x 136500
Step 3: 104 x 26x 136500 130 x 136500 x 1050
So, the second amount is x/2 =1050/2=525 ANSWER: The two amounts are $1050 and $525
18.
PROBLEM: James earned a total of $68.25 in simple interest from two separate accounts. In an account earning 2.6% interest, James invested one-half as much as he did in the other account that earned 5.2%. How much did he invest in each account? SOLUTION:
Let x represents the amount invested at 5.2% .052 Then y represents the other amount invested at 2.6% .026 Given:
y=x/2 0.052 x +0.026 (x/2) = 68.25 Step 1: Multiply equation by 2
.052x .026 x / 2 68.25
2
.104 x .026x 136.5
Step 2: Multiply equation by 1000
.104 x .026x 136.5
1000
104 x 26x 136500
Step 3: 104 x 26x 136500 130 x 136500 x 1050
So, the second amount is x/2 =1050/2=525 ANSWER: The two amounts are $1050 and $525
19.
PROBLEM: A cash register contains $10 bills and $20 bills with total value $340. If there are 23 bills total, then how many of each does the register contain. SOLUTION:
Let x represent the number of $10 bills. Let y represent the number of $20 bills. Given: x + y =23 10x + 20y = 340 Step 1: Multiply the first equation by 10 and then subtract the second equation from the first equation
10
x y 23 10 x 20 y 340
10 x 10 y 230 10 x 20 y 340
10 x 10 y 230 10 x 20 y 340 10 y 110 y 11 Step 2: Keeping the value of x in first equation x 11 23
x 12 ANSWER: The number of $10 bills and $20 bills are 12 and 11.
20.
PROBLEM: John was able to purchase a pizza for $10.80 with quarters and dimes. If he used 60 coins to buy the pizza, then how many of each did he have? SOLUTION:
Let x represent the number of quarters. Let y represent the number of dimes. Given: Total number of coins is x + y = 60 A dollar has 4 quarters; hence the number of quarters used is 0.25x. A dollar has 10 dimes; hence the number of dimes used is 0.1y. Total monetary value: 0.25x + 0.1y = $10.80 Step 1: Multiply the second equation by 100
x y 60 x y 60 0.25x 0.1y 10.80 100 25 x 10 y 1080 Step 2: Multiply the first equation by 25 and then subtract second equation from the first equation
25
25 x 25 y 1500 x y 60 25 x 10 y 1080 25 x 10 y 1080 25 x 25 y 1500 25 x 10 y 1080 15 y 420 y 28 Step 3: Keeping the value of y in first equation x 28 60 x 32 ANSWER: John uses 32 quarters and 28 dimes to purchase the pizza.
21.
PROBLEM: Dennis mowed his neighbor’s lawn for a jar of dimes and nickels. Upon completing the job, he counted the coins and found that there were 4 less than twice as many dimes as there were nickels with total value $6.60. How many of each coin did he have? SOLUTION:
Let x represent the number of dimes. Let y represent the number of nickels. Given: Total number of coins: y = 2x – 4 2x – y=4 A dollar has 10 dimes; hence the number of dimes in the jar is 0.1x. A dollar has 20 nickels; hence the numbers of nickels is 0.05y. Total monetary value: 0.1x + 0.05y = $6.60 Step 1: Multiply the second equation by 100
2x – y 4 0.1x 0.05y 6.60 100
2x – y 4 10 x 5 y 660
Step 2: Multiply the first equation by 25 and then subtract the second equation from first equation
2x – y 4 10 x 5 y 660
5
10x – 5 y 20 10 x 5 y 660
10x – 5 y 20 10 x 5 y 660 10 y 640 y 64 Step 3: Keeping the value of y in first equation 2x – 64 4 2 x 68 x 34
ANSWER: The jar contains 34 dimes and 64 nickels.
22.
PROBLEM: Two families bought tickets for the big football game. One family ordered 2 adult tickets and 3 children’s tickets for a total of $26.00. Another family ordered 3 adult tickets and 4 children’s tickets for a total of $37.00. How much was each adult ticket? SOLUTION:
Let x represent price of adult ticket. Let y represent price of children ticket. Given: 2x + 3y = 26 3x + 4y = 37 Step 1: Multiply the first equation by 50 and then subtract the second equation from the first equation
3
2 x 3 y 26 3x 4y 37 2
6 x 9 y 78 6 x 8 y 74
6 x 9 y 78 6 x 8 y 74 y 4 y4 2 x 3 4 26 2 x 14 x7
Step 2: Keeping the value of y in first equation 2 x 3 4 26 2 x 14 x7
ANSWER: So, price of adult ticket is $7.
23. PROBLEM: Two friends found shirts and shorts on sale at the flea market. One bought 5 shirts and 3 shorts for a total of $51.00. The other bought 3 shirts and 7 shorts for a total of $80.00. How much was each shirt and each pair of shorts? SOLUTION: Let x represent the price of shirts. Let y represent the price of shorts.
Given : 5x + 3y = $ 51.00 3x + 7y = $ 80.00 Step 1: Multiply first equation by 3 and multiply second equation by 5 then subtract second equation from the first equation
5x 3y 51 3x 7y 80
3
5
15 x 9 y 153 15 x 35 y 400
15 x 9 y 153 15 x 35 y 400 26 y 247 y 9.50 Step 2: Keeping the value of y in first equation 5x 3 9.50 51.00 5 x 22.50 x 4.50
ANSWER: The prices of shirts and shorts are $9.50 and $4.50.
24.
PROBLEM: On Monday Joe bought 10 cups of coffee and 5 doughnuts for his office at the cost of $16.50. It turns out that the doughnuts were more popular than the coffee. Therefore, on Tuesday he bought 5 cups of coffee and 10 doughnuts for a total of $14.25. How much was each cup of coffee? SOLUTION:
Let x represent price of cup of coffee. Let y represent price of doughnut. Given: 10x + 5y = 16.50 5x + 10y = 14.25 Step 1: Multiply the first equation by 50 and then subtract the second equation from the first equation
25
25 x 25 y 1500 x y 60 25 x 10 y 1080 25 x 10 y 1080 25 x 25 y 1500 25 x 10 y 1080 15 y 420 y 28 0.25x 0.1y 10.80 Step 2: Keeping the value of y in first equation 10 x 5 .8 16.50 10 x 12.50 x 1.25
ANSWER: So, price of cup of coffee is $ 1.25. Part C: Mixture Problems Set up a linear system and solve.
25.
PROBLEM: A 15% acid solution is to be mixed with a 25% acid solution to produce 12 gallons of a 20% acid solution. How much of each is needed? SOLUTION:
Let x represent the amount of 15% acid solution needed. Let y represent the amount of 25% acid solution needed. Given: x + y = 12 (0.15) x + (.25) y = (.20)12 Step 1: Rewrite the second equation in the form 0.15 x .25 y .20 12
0.15 x .25 y 2.4 Step 2: Multiply the second equation by 100
x y 12 .15 x .25 y 2.4
100
x y 12 15 x 25 y 240
Step 3: Multiply the first equation by 15 then subtract the second equation from the first equation
x y 12 15 x 25 y 240
15
15 x 15 y 180 15 x 25 y 240
15 x 15 y 180 15 x 25 y 240 10 y 60 y6 Step 4: Keeping the value of y in first equation x 6 12
x6 ANSWER: The amount of 15% acid and amount of 25% acid solution required is 6 gallons each.
26.
PROBLEM: One alcohol solution contains 12% alcohol and another contains 26% alcohol. How much of each should be mixed together to obtain 5 gallons of a 14.8% alcohol solution? SOLUTION:
Let x represent the amount of 12% acid solution needed. Let y represent the amount of 26% acid solution needed. Given: x+y=5 (0.12)x + (.26)y = (.148)5 Step 1: Rewrite the second equation in the form 0.12 x .26 y .148 5
0.12x .26y .74
Step 2: Multiply second equation by 100
xy 5 0.12x .26y .74
100
xy 5 12 x 26 y 74
Step 3: Multiply the first equation by 15 then subtract second equation from the first equation 12
xy 5 12 x 26 y 74
12 x 12 y 60 12 x 26 y 74
12 x 12 y 60 12 x 26 y 74 14 y 14 y 1 Step 4: Keeping the value of y in first equation x 1 5
x4 ANSWER: The amount of 12% acid and amount of 26% acid solution required are 4 gallons and 1 gallon.
27.
PROBLEM: A nurse wishes to obtain 40 oz of a 1.2% saline solution. How much of a 1% saline solution must she mix with a 2% saline solution to achieve the desired result? SOLUTION:
Let x represent the amount of 1 % saline solution needed. Let y represent the amount of 2 % saline solution needed. Given: x + y = 40 (0.01)x + (.02)y = (.012)40 Step 1: Rewrite the second equation in the form 0.01 x .02 y .012 40
.01x .02 y .48
Step 2: Multiply the second equation by 1000
x y 40 .01x .02 y .48
100
x y 40 x 2 y 48
Step 3: Subtract the second equation from the first equation x y 40 x 2 y 48 y 8 y 8 Step 4: Keeping the value of y in first equation x 8 40
x 32 ANSWER: Hence the amount of 1 % saline solution and 2 % saline solution needed are 32 oz and 8 oz.
28.
PROBLEM: A customer ordered 20 lbs of fertilizer that contains 15% nitrogen. To fill the customer’s order, how much of the stock 30% nitrogen fertilizer must be mixed with the 10% nitrogen fertilizer? SOLUTION:
Let x represent 30% nitrogen fertilizer needed. Let y represent 10% nitrogen fertilizer needed. Given: x + y = 20 (0.30)x + (.10)y = (.15)20 Step 1: Rewrite the second equation in the form 0.30 x .10 y .15 20
0.30x .10y 3 Step 2: Multiply second equation by 100
x y 20 0.30x .10y 3
100
x y 20 30 x 10 y 300
Step 3: Multiply the first equation by 15 then subtract the second equation from the first equation 30
x y 20 30 x 10 y 300
30 x 30 y 600 30 x 10 y 300
30 x 30 y 600 30 x 10 y 300 20 y 300 y 15 Step 4: Keeping the value of y in first equation x 15 20
x5 ANSWER: The amount of 30% nitrogen fertilizer and amount of 10% nitrogen fertilizer required are 5 lbs and 15 lbs.
29.
PROBLEM: A customer ordered 2 lbs of a mixed peanut product containing 15% cashews. The inventory consists of only two mixes containing 10% and 30% cashews. How much of each type must be mixed to fill the order? SOLUTION:
Let x represent the amount of 10 % cashew. Let y represent the amount of 30 % cashew. Given: x+y=2 (0.1)x + (.3)y = (.15)2 Step 1: Rewrite the second equation in the form 0.10 x .30 y .15 2
0.10 x 0.30y .30 Step 2: Multiply second equation by 1000
x y 2 0.10 x 0.30y .30
100
x y 2 x 3y 3
Step 3: Multiply the first equation by 12 then subtract the second equation from the first equation
x y 2 x 3y 3 2 y 1 y 0.5 Step 4: Keeping the value of y in first equation
x .5 2 x .15 ANSWER: The two amounts required are 1.5 lb and .5 lb
30.
PROBLEM: How many pounds of pure peanuts must be combined with a 20% peanut mix to produce 10 lbs of a 32% peanut mix? SOLUTION:
Let x represent pure peanuts needed. Let y represent peanuts mix needed. Given: x + y = 10 (0.30)x + (.10) y = (.15)20 Step 1: Rewrite the second equation in the form x .20 y .32 10
x .20y 3.2 Step 2: Multiply the second equation by 100
x y 10 x .20y 3.2
x y 10 100 x 20 y 320
100
Step 3: Multiply the first equation by 15 then subtract second equation from the first equation x y 10 100 x 20 y 320
20
20 x 20 y 200 100 x 20 y 320
20 x 20 y 200 100 x 20 y 320 80 x 120 x 1.5
ANSWER: The amount pure peanuts needed is 1.5 lbs.
31.
PROBLEM: How much cleaning fluid, with 20% alcohol content, must be mixed with water to obtain a 24 oz mixture with 10% alcohol content? SOLUTION:
Let x represent cleaning fluid, with 20% alcohol content. Let y represent water. Given: x + y = 24 (0.2)x + (0)y = (.10)24 Step 1: Rewrite the second equation 0.2 x 0 y .10 24 .2 x 2.40 x 12
ANSWER: The amount of cleaning fluid needed is 12 oz.
32.
PROBLEM: A chemist wishes to create a 32 oz solution with 12% acid content. He will use two types of stock solutions, one with 30% acid content and another with 10% acid content. How much of each will he need? SOLUTION:
Let x represent the amount of 30% acid solution needed. Let y represent the amount of 10% acid solution needed. Given: x + y = 32 (0.30)x + (.10) y = (.12)32 Step 1: Rewrite the second equation in the form 0.30 x .10 y .12 32
0.30x .10y 3.84 Step 2: Multiply second equation by 100
x y 32 0.30x .10y 3.84 100
x y 32 30 x 10 y 384
Step 3: Multiply first equation by 15 then subtract second equation from the first equation 30
x y 32 30 x 10 y 384
30 x 30 y 960 30 x 10 y 384
30 x 30 y 960 30 x 10 y 384 20 y 576 y 28.8 Step 4: Keeping the value of y in first equation x 28.8 32
x=3.2 ANSWER: The amount of 30% acid and 10% acid solution required are 3.2 oz and 28.8 oz.
33.
PROBLEM: A concentrated cleaning solution that contains 50% ammonia is to be mixed with another solution containing 10% ammonia. How much of each should be mixed to obtain 8 oz of a 32% ammonia cleaning formula. SOLUTION:
Let x represent the amount of 50 % ammonia needed. Let y represent the amount of 10 % acid ammonia needed. Given : x+y=8 (0.5)x + (.1)y = (.32)8 Step 1: Rewrite the second equation in the form 0.5 x .1 y .32 8
.5 x .1 y 2.56 Step 2: Multiply second equation by 100
xy 8 .5 x .1y 2.56 100
xy 8 50 x 10 y 256
Step 3: Multiply the first equation by 15 then subtract the second equation from the first equation 10
x y 8 50 x 10 y 256
10 x 10 y 80 50 x 10 y 256
10 x 10 y 80 50 x 10 y 256 40 x 176 x 4.4 Step 4: Keeping the value of x in first equation 4.4 y 8
y 3.6 ANSWER: The amount of 50 % ammonia and 10 % acid ammonia required is 4.4 oz and 3.6 oz
34. PROBLEM: A 50% fruit juice concentrate can be purchased wholesale. Best taste is achieved when water is mixed with the concentrate in such a way as to obtain a 12% fruit juice mixture. How much water and concentrate is need to make a 50 oz fruit juice drink? SOLUTION:
Let x represent 50% fruit juice concentrate. Let y represent water. Given: x + y = 50 (0.5) x + (0) y = (.12)50 Step 1: Rewrite the second equation 0.5 x 0 y .12 50 0.5 x 6.00 x 12
Step 2: Keeping the value of x in first equation 12 y 50
y 38 ANSWER: The amount of fruit juice concentrate and water needed is 12 oz and 38 oz.
35.
PROBLEM: A 75% antifreeze concentrate is to be mixed with water to obtain 6 gallons of a 25% antifreeze solution. How much water is needed? SOLUTION:
Let x represent 75% antifreeze concentrate. Let y represent water. Given : x+ y=6 (0.75) x + (0) y = (.25)6 Step 1: Rewrite the second equation
0.75 x 0 y .25 6 .75 x 1.50 x2
Step 2: Keeping the value of x in first equation 2 y 6
y4 ANSWER: The amount of water needed is 4 gallons
36.
PROBLEM: Pure sugar is to be mixed with a fruit salad containing 10% sugar to produce 48 ounces of a salad containing 16% sugar. How much pure sugar is required? SOLUTION:
Let x represent the amount of pure sugar needed. Let y represent the amount of fruit salad needed. Given: x + y = 48 x + (.10) y = (.16)48 Step 1: Rewrite the second equation in the form x .10 y .16 48
x .10y 7.68 Step 2: Multiply second equation by 100
x y 48 x .10y 7.68 100
x y 48 100 x 10 y 768
Step 3: Multiply the first equation by 15 then subtract the second equation from the first equation
100
100 x 100 y 4800 100 x 10 y 768 100 x 100 y 4800 100 x 10 y 768
x y 48 100 x 10 y 768
90 y 4032 y 44.8 Step 4: Keeping the value of y in first equation x 44.8 48
x 3.2 ANSWER: The amount of the amount of pure sugar needed is 3.2 ounces. Part D: Uniform Motion Problems Set up a linear system and solve.
37.
PROBLEM: An airplane averaged 460 mph on a trip with the wind and 345 mph on the return trip against the wind. If the total round trip took 7 hours, then how long did the airplane spend on each leg of the trip? SOLUTION: Let x represent the time taken while going with the wind. Let y represent the time taken while going against the wind.
Given: y= 7-x We know that d = r . t Step 1: While going with the wind d = 460 . x Step 2: While going against the wind d= 345 .(7 – x) Step 3: As the distance traveled will be equal, so 460. x 345.(7 – x) 460 x 345x 2415 x3
Step 4: Keeping the value of x we get the value of y
y 73 y4 ANSWER: The time taken while going with the wind and going against the wind are 3 hours and 4 hours
38.
PROBLEM: The two legs of a 330 mile trip took 5 hours. The average speed for the first leg of the trip was 70 mph and the average speed for the second leg of the trip was 60 mph. How long did each leg of the trip take? SOLUTION:
Let x represent the time taken by first leg of the trip. Let y represent the time taken by second leg of the trip. Given: x+y=5 70x + 60y= 330 Step 1: Multiply first equation by 50 and then subtract second equation from the first equation 70
x y 5 70 x 60 y 330
70 x 70 y 350 70 x 60 y 330
70 x 70 y 350 70 x 60 y 330 10 y 20 y2 Step 2: Keeping the value of y in first equation x 2 5
x3 ANSWER: So, time taken by first leg and second legs of the trip are 3 hours and 2 hours
39.
PROBLEM: An executive traveled 1,200 miles part by helicopter and part by private jet. The jet averaged 320 mph while the helicopter averaged 80 mph. If the total trip took 4½ hours, then how long did she spend in the private jet? SOLUTION: Let x represent the time taken by the private jet. Let y represent the time taken by the helicopter.
Given: x + y = 4.5 We know d= r . t 320 x + 80 y = 1200 Step 1: Multiply the first equation by 80 then subtract the second equation from the first equation 80
x y 4.5 320 x 80 y 1200
80 x 80 y 360 320 x 80 y 1200
80 x 80 y 360 320 x 80 y 1200 240 x 840 x 3.5 Step 2: Keeping the value of x in first equation
3.5
y 4.5
y 1
ANSWER: Hence the time taken by the private jet is 3.5 hours.
40.
PROBLEM: Joe took two busses on the 463 mile trip from San Jose to San Diego. The first bus averaged 50 mph and the second bus was able to average 64 mph. If the total trip took 8 hours, then how long was spent in each bus? SOLUTION:
Let x represent the time taken by first bus. Let y represent the time taken by second bus. Given: x+y=8 50x +64y= 463 Step 1: Multiply the first equation by 50 and then subtract the second equation from the first equation 50
x y 8 50 x 64y 463
50 x 50 y 400 50 x 64y 463
50 x 50 y 400 50 x 64y 463 14 y 63 y 4.5 Step 2: Keeping the value of y in first equation x 4.5 8
x 3.5 ANSWER: So, time taken by first bus and second bus are 3.5 hours and 4.5 hours
41.
PROBLEM: Billy canoed downstream to the general store at an average rate of 9 mph. His average rate canoeing back upstream was 4 mph. If the total trip took 6½ hours, then how long did it take Billy to get back on the return trip? SOLUTION: Let x represent the time taken while downstream. Let y represent the time taken while going upstream.
Given: y = 6.5 - x We know that d = r . t Step 1: while downstream d= 9 . x
Step 2: while going upstream. d= 4.(6.5 – x) Step 3: As the distance traveled will be equal, so 9. x 4(6.5 – x) 9. x 26 4 x 13 x 26 x2 Step 4: Keeping the value of x we get the value of y y 6.5 2
y 4.5 ANSWER: The time taken while returning is 4.5 hours
42.
PROBLEM: Two brothers drove the 2,793 miles from Los Angeles to New York. One of the brothers, driving in the day, was able to average 70 mph and the other, driving at night, was able to average 53 mph. If the total trip took 45 hours, then how many hours did each brother drive? SOLUTION:
Let x represent the time taken by one of the brothers. Let y represent the time taken by another brother. Given: x + y = 45 70x +53y= 2793 Step 1: Multiply the first equation by 70 and then subtract the second equation from the first equation
70
x y 45 70 x 70 y 3150 70 x 53y 2793 70 x 53y 2793 70 x 70 y 3150 70 x 53y 2793 17 y 357 y 21 Step 2: Keeping the value of y in first equation x 21 45
x 24 ANSWER: So, time taken by both the brothers is 24 and 21
43.
PROBLEM: A boat traveled 24 miles downstream in 2 hours. The return trip took twice as long. What was the speed of the boat and the current? SOLUTION:
Let x represent the speed of the boat Let y represent the speed of the current Speed of the downstream = x + y Speed of the upstream = x - y Given: In case of downstream 24 x y 2 x y 12 In case of upstream 24 4 x y 6 x y
Step 1: Add both the equations
x y 12 x y 6 2 x 18 x9 Step 2: Keeping the value of x in the first equation
9
y 12
y 3 ANSWER: The speed of the boat and current are as 9 mph and 3 mph
44.
PROBLEM: A helicopter flying with the wind can travel 525 miles in 5 hours. On the return trip, against the wind, it will take 7 hours. What is the speed of the helicopter and the wind? SOLUTION:
Let x represent the speed of the helicopter Let y represent the speed of the wind With the wind = x + y Against the wind = x - y Given: In case of with the wind 525 x y 5 x y 105 In case of against the wind 525 7 x y 75 x y
Step 1: Add both the equations
x y 105 x y 75 2 x 180 x 90 Step 2: Keeping the value of x in the first equation 90 y 105
y 15 ANSWER: The speed of the helicopter and wind are 90 mph and 15 mph
45.
PROBLEM: A boat can travel 42 miles with the current downstream in 3 hours. Returning upstream, against the current, the boat can only travel 33 miles in 3 hours. Find the speed of the current. SOLUTION:
Let x represent the speed of the boat Let y represent the speed of the current Speed of the downstream = x + y Speed of the upstream = x - y Given: In case of downstream 42 x y 3 x y 14 In case of upstream 33 3 x y 11 x y
Step 1: Add both the equations
x y 14 x y 11 2 x 25 x 12.5 Step 2: Keeping the value of x in the first equation 12.5 y 14
y 1.5 ANSWER: The speed of the current is 1.5 mph 46.
PROBLEM: A light aircraft flying with the wind can travel 180 miles in 1½ hours. The aircraft can fly the same distance against the wind in 2 hours. Find the speed of the wind. SOLUTION:
Let x represent the speed of the aircraft Let y represent the speed of the wind With the wind = x + y Against the wind = x - y Given: In case of with the wind 180 x y 1.5 x y 120 In case of against the wind 180 2 x y 90 x y
Step 1: Add both the equations x y 120
x y 90 2 x 210 x 105
Step 2: Keeping the value of x in the first equation 105 y 120
y 15 ANSWER: The speed of the wind is 15 mph. Part E: Discussion Board
47.
PROBLEM:
Compose a number or money problem of your own and share it on the discussion board. SOLUTION: (Students answers may vary) The sum of two integers is 64 and the difference is 10. Find the integers
Let x be one of the integers Let y be other integer. Given : x + y = 64 x - y = 10 Step 1: Add both the equations
x y 64 x y 10 2x
74 x 37
Step 2: Keeping the value of x in first equation 37 y 64
y=64-37 y=27 ANSWER: The integers are 37 and 27.
48.
PROBLEM:
Compose a mixture problem of your own and share it on the discussion board.
SOLUTION: A chemist wishes to create a 32 oz solution with 12% acid content. He will use two types of stock solutions, one with 30% acid content and another with 10% acid content. How much of each will he need?
Let x represent the amount of 30% acid solution needed. Let y represent the amount of 10% acid solution needed. Given: x + y = 32 (0.30) x + (.10) y = (.12)32 Step 1: Rewrite the second equation in the form 0.30 x .10 y .12 32
0.30x .10y 3.84 Step 2: Multiply second equation by 100
x y 32 0.30x .10y 3.84 100
x y 32 30 x 10 y 384
Step 3: Multiply first equation by 15 then subtract second equation from the first equation
x y 32 30 x 10 y 384
30
30 x 30 y 960 30 x 10 y 384
30 x 30 y 960 30 x 10 y 384
20 y 576 y 28.8 Step 4: Keeping the value of y in first equation x 28.8 32
x=3.2 ANSWER: The amount of 30% acid and 10% acid solution required are 3.2 oz and 28.8 oz.
49.
PROBLEM:
Compose a uniform motion problem of your own and share it on the discussion board. SOLUTION: (Students answers may vary)
An airplane averaged 460 mph on a trip with the wind and 345 mph on the return trip against the wind. If the total round trip took 7 hours, then how long did the airplane spend on each leg of the trip? SOLUTION: Let x represent the time taken while going with the wind. Let y represent the time taken while going against the wind.
Given: y= 7-x We know that d = r . t Step 1: While going with the wind d= 460 . x Step 2: While going against the wind d= 345 .(7 – x) Step 3: As the distance traveled will be equal, so 460 . x 345 .(7 – x) 460 x 345x 2415 x=3 Step 4: Keeping the value of x we get the value of y
y 73 y4 ANSWER: The time taken while going with the wind and going against the wind are 3 hours and 4 hours
4.5 Solving Systems of Linear Inequalities (Two Variables) Part A: Solving Systems of Linear Inequalities Determine whether or not the given point is a solution to the given system of linear equations.
1.
PROBLEM: y x3 (3, 2); y x 3 SOLUTION:
Check: (3, 2) Inequality 1: y x 3
2 3 3
Inequality 2: y x 3
2 3 3 2 0
26
ANSWER: Since, it satisfies both the inequalities, so (3, 2) is the solution of the given system.
2.
PROBLEM: y 3x 4 y 2x 1
(3, 2);
SOLUTION:
Check: (3, 2) Inequality 1: y 3 x 4
Inequality 2: y 2 x 1
2 3 3 4
2 2 3 1
2 9 4
2 6 1
2 13
2 7
ANSWER: Since, it satisfies both the inequalities, so (3, 2) is the solution of the given system.
3.
PROBLEM: y x 5 3 y 4 x 2
(5, 0)
SOLUTION:
Check: (5, 0) Inequality 1: y x 5
0 5 5 0 0 Incorrect
4.
Inequality 2: y
3 x2 4
3 4 15 0 2 4 15 8 0 4 7 0 4
0 5 2
ANSWER: Since, it does not satisfy both the inequalities, so (5, 0) is not the solution of the given system PROBLEM: 2 y 3 x 1 (0, 1); y 5 x 2 2 SOLUTION:
Check: (0, 1); Inequality 1: y
2 x 1 3
2 3 1 1 Incorrect
1 0 1
Inequality 2: y
5 x2 2
5 2 1 2
1 0 2
ANSWER: Since, it does not satisfy both the inequalities, so (0, 1) is not the solution of the given system
5.
PROBLEM: 4 x 3 y 12 2x 3y 6
(1, 83 )
SOLUTION:
Check: (1, 83 ) Inequality 1: 4 x 3 y 12
Inequality 2: 2 x 3 y 6
4 1 3 83 12
2 1 3 83 6
4 8 12
2 8 6
12 12
6 6 Incorrect
ANSWER: Since, it does not satisfy both the inequalities, so (1, 83 ) is not the solution of the given system
6.
PROBLEM: x y 0 x y 0 x y 2
(1, 2)
SOLUTION:
Check: (1, 2) Inequality 1: x y 0
Inequality 2: x y 0
Inequality 3: x y 2
1 2 0
1 2 0
1 2 2
1 2 0
3 0
3 2
1 0
ANSWER: Since, it satisfies all three inequalities, so (1, 2) is the solution of the given system.
Part B: Solving Systems of Linear Inequalities Graph the solution set.
7.
PROBLEM: y x3 y x 3 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (0, 3) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (0, 3) Inequality 1: y x 3
Inequality 2: y x 3
3 0 3
3 0 3
33
33
ANSWER: This shows that it satisfies both the inequalities.
8.
PROBLEM: y 3x 4 y 2x 1 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (-1, 1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (-1, 1) Inequality 1: y 3 x 4
Inequality 2: y 2 x 1
1 3 1 4
1 2 1 1
1 7
1 3
ANSWER: This shows that it satisfies both the inequalities.
9.
PROBLEM: yx y 1 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (-3, -2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (-3, -2) Inequality 1: y x
Inequality 2: y 1
2 3
2 1
2 3
2 1
ANSWER: This shows that it satisfies both the inequalities.
10.
PROBLEM: 2 y 3 x 1 y 5 x 2 2 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (-3, -2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (-3, -2) Inequality 1: y 2 3 2 2 1 2 1
2 x 1 3
2 3 1
Inequality 2: y 5 2 15 4 2 2 2 9.5
2 3 2
ANSWER: This shows that it satisfies both the inequalities.
11.
PROBLEM: y x 5 3 y 4 x 2 SOLUTION:
5 x2 2
The above graph shows that shaded region is the intersection area. The graph suggests that (5,1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (5, 1) Inequality 1: y x 5
1 5 5 1 0
Inequality 2: y 3 4 15 8 1 4 7 1 4 1 1.75
3 x2 4
1 5 2
ANSWER: This shows that it satisfies both the inequalities.
12.
PROBLEM: 3 y 5 x 3 y 3 x 3 5 SOLUTION:
Inequality 1: y If x=0, then y=3 If y=0, then x= -5
3 x3 5
Inequality 2: y If x=0, then y= -3 If y=0, then x= 5
3 x 3 5
ANSWER: Since there is no intersection area, so there is no solution.
13.
PROBLEM: x 4 y 12 3x 12 y 12 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (5,1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (5, 1) Inequality 1: x 4 y 12
Inequality 2: 3x 12 y 12
5 4 1 12
3 5 12 1 12
9 12
3 12
ANSWER: This shows that it satisfies both the inequalities.
14.
PROBLEM: x y 6 2x y 1 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (-1, 4) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (-1, 4) Inequality 1: x y 6
Inequality 2: 2 x y 1
1 4 6
2 1 4 1
56
2 1
ANSWER: This shows that it satisfies both the inequalities.
15.
PROBLEM: 2 x 3 y 3 4 x 3 y 15
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (6, 6) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (6, 6) Inequality 1: 2 x 3 y 3
Inequality 2: 4 x 3 y 15
2 6 3 6 3
4 6 3 6 15
12 18 3
24 18 15
63
6 15
ANSWER: This shows that it satisfies both the inequalities.
16.
PROBLEM: 4 x 3 y 12 2x 3y 6
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (2, 0) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (2, 0) Inequality 1: 4 x 3 y 12
Inequality 2: 2 x 3 y 6
4 2 3 0 12
2 2 3 0 6
8 12
46
ANSWER: This shows that it satisfies both the inequalities.
17.
5x y 4 4 x 3 y 6
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (1, -3) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (1, -3) Inequality 1: 5 x y 4
Inequality 2: 4 x 3 y 6
5 1 3 4
4 1 3 3 6
53 4 24
4 9 6 13 6
ANSWER: This shows that it satisfies both the inequalities.
18.
PROBLEM: 3x 5 y 15 x 2 y 0
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (2, 1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (2, 1) Inequality 1: 3x 5 y 15
Inequality 2: x 2 y 0
3 2 5 1 15
2 2 1 0
6 5 15
2 2 0
11 15
00
ANSWER: This shows that it satisfies both the inequalities.
19.
PROBLEM: x0 5 x y 5 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (0, 6) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (0, 6) Inequality 1: x 0
Inequality 2: 5 x y 5
0 0
5 0 6 5
00
65
ANSWER: This shows that it satisfies both the inequalities.
20.
PROBLEM: x 2 y 1
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (-2, 1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (-2, 1) Inequality 1: x 2
Inequality 2: y 1
2 2
1 1
2 2
11
ANSWER: This shows that it satisfies both the inequalities.
21.
PROBLEM: x 3 0 y 2 0
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (2, 2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (2, 2) Inequality 1: x 3 0
Inequality 2: y 2 0
2 3 0
2 2 0
1 0
40
ANSWER: This shows that it satisfies both the inequalities.
22.
PROBLEM: 5 y 2x 5 2 x 5 y 5 SOLUTION:
Inequality 1: 5 y 2 x 5
Inequality 2: 2 x 5 y 5
If x=0, then y=1 If y=0, then 5 x= 2
If x=0, then y= -1 If y=0, then 5 x= 2
ANSWER: Since there is no intersection area, there is no solution.
23.
PROBLEM: x y 0 x y 1 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (1, 1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (1, 1) Inequality 1: x y 0
Inequality 2: x y 1
1 1 0
1 1 1
00
0 1
ANSWER: This shows that it satisfies both the inequalities.
24.
PROBLEM: x y 0 y x 1
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (-2, -2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (-2, -2) Inequality 1: x y 0
Inequality 2: y x 1
2 2 0
2 2 1
22 0 00
2 2 1 0 1
ANSWER: This shows that it satisfies both the inequalities.
25.
PROBLEM: x 2 x 2 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (1, 1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (1, 1) Inequality 1: x 2
Inequality 2: x 2
x 2
x2
1 2
1 2
ANSWER: This shows that it satisfies both the inequalities.
26.
PROBLEM: y 1 y 2
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (0, 1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (0, 1) Inequality 2: y 2
Inequality 1: y 1
1 1
1 2
1 1
1 2
ANSWER: This shows that it satisfies both the inequalities.
27.
PROBLEM: x 2 y 8 3x 6 y 18 SOLUTION:
Inequality 1: x 2 y 8 If x=0, then y=4 If y=0, then x= -8
Inequality 2: 3x 6 y 18 If x=0, then y= -3 If y=0, then x= 6
ANSWER: Since there is no intersection area, so there is no solution.
28.
PROBLEM: 3 x 4 y 4 6 x 8 y 8
SOLUTION:
Inequality 1: 3x 4 y 4
Inequality 2: 6 x 8 y 8
If x=0, then y= 1
If x=0, then y= 1
If y=0, then 4 x= 3
If y=0, then 4 x= 3
ANSWER: Since there is no intersection area, there is no solution.
29.
PROBLEM: 2 x y 3 1 x 2 y SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (0, 2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (0, 2) Inequality 1: 2 x y 3 2 0 2 3 23
Inequality 2: x 0 0 1
ANSWER: This shows that it satisfies both the inequalities.
30.
PROBLEM: 2x 6 y 6 1 3 x y 3 SOLUTION:
1 2 2
1 y 2
The above graph shows that shaded region is the intersection area. The graph suggests that (0, 1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (0, 1) Inequality 1: 2 x 6 y 6 2 0 6 1 6 66
1 Inequality 2: x y 3 3 1 0 1 3 3 1 3
ANSWER: This shows that it satisfies both the inequalities.
31.
PROBLEM: y3 yx x 3 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (2, 2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (2, 2) Inequality 1: y3
Inequality 2: yx
Inequality 3: x 3
2 3
2 2
2 3
23
2 2
2 3
ANSWER: This shows that it satisfies all the inequalities
32.
PROBLEM: y 1 y x 1 y 3 x 3 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (-1, -1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (1, -1) Inequality 1: y 1
Inequality 2: y x 1
Inequality 3: y 3x 3
1 1
1 1 1
1 3 1 3
1 1
1 2
1 6
ANSWER: This shows that it satisfies all the inequalities.
33.
PROBLEM: 4 x 3 y 12 y2 2x 3y 6 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (5, 3) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (5, 3) Inequality 1: 4 x 3 y 12
Inequality 2: y2
Inequality 3: 2x 3y 6
4 5 3 3 12
3 2
2 5 3 3 6
20 9 12 11 12
3 2
10 9 6 19 6
ANSWER: This shows that it satisfies all the inequalities
34.
PROBLEMS: x y 0 x y 0 x y 2 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (1, -2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (1, -2) Inequality 1: x y 0
Inequality 2: x y 0
Inequality 3: x y 2
1 2 0
1 2 0
1 2 2
3 0
1 0
1 2
ANSWER: This shows that it satisfies all the inequalities.
35.
x y 2 x3 x y 2
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (2, -2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (2, -2) Inequality 1: x y 2
Inequality 2: x3
Inequality 3: x y 2
2 2 2
2 3
2 2 2
4 2
2 3
22 2 02
ANSWER: This shows that it satisfies all the inequalities.
36.
PROBLEM: y40 1 1 x y 1 3 2 1 1 2 x 3 y 1
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (0, 3) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (0, 3) Inequality 1: y40
3 4 0 70
Inequality 2: 1 1 x y 1 2 3 1 1 0 3 1 2 3 11
Inequality 3: 1 1 x y 1 2 3 1 1 0 3 1 2 3 11
ANSWER: This shows that it satisfies all the inequalities.
37.
PROBLEM: Construct a system of linear inequalities that describes all points in the first quadrant. SOLUTION:
In the first quadrant, all the values for x and y are positive. It means that x and y will be greater than 0. So, a system of linear inequalities that describes all points in the first quadrant is x>0 y>0
38.
PROBLEM: Construct a system of linear inequalities that describes all points in the second quadrant. SOLUTION:
In the second quadrant, all the values for x are negative but the values of y are positive. It means that x will be less than 0 but the values of y will be greater than 0. ANSWER: So, a system of linear inequalities that describes all points in the second quadrant is
x0 39.
PROBLEM: Construct a system of linear inequalities that describes all points in the third quadrant. SOLUTION:
In the third quadrant, all the values for x and y are negative. It means that x and y will be less than 0. ANSWER: So, a system of linear inequalities that describes all points in the third quadrant is x
PROBLEM: x y 1 (3, 2 ); 2 x 2 y 2 SOLUTION: Check: (3, 2) Equation 1: x y 1
Equation 2: 2 x 2 y 2
x y 1
2 x 2 y 2
3 2 1
2 3 2 2 2
3 2 1 1 1
6 4 2 6 4 2 2 2
ANSWER: (3, -2) does not satisfy both equations, hence, it is not a solution.
2.
PROBLEM: x y 1 (5, 0); 2 x 2 y 2 SOLUTION:
Check: (5, 0) Equation 1: x y 1
Equation 2: 2 x 2 y 2
x y 1
2 x 2 y 2
5 0 1 5 1
2 5 2 0 2 10 0 2 10 2
ANSWER: Since (5, 0) does not satisfy both equations, hence, it is not a solution
3.
PROBLEM: x y 4 (2, 6); 3x y 12 SOLUTION:
Check (2, 6) Equation 1: x y 4
Equation 2: 3x y 12 3x y 12 3 2 6 12
x y 4 2 6 4
6 6 12 0 12
2 6 4 4 4
ANSWER: Since (3, 4) does not satisfy both equations, hence, it is not a solution
4.
PROBLEM: 3 x 2 y 8 (2, 7); 5 x 3 y 11 SOLUTION:
Check: (2, 7) Equation 1: 3x 2 y 8
Equation 2: 5 x 3 y 11
3x 2 y 8
5 x 3 y 11
3 2 2 7 8
5 2 3 7 11
6 14 8
10 (21) 11 10 21 11 11 11
8 8
ANSWER: Since (2, 7) satisfy both equations, hence, it is a solution
5.
PROBLEM: 5 x 5 y 15 (0, 3); 13x 2 y 6 SOLUTION:
Check (0, 3) Equation 1: 5 x 5 y 15
Equation 2: 13 x 2 y 6
5 x 5 y 15
13x 2 y 6
5 0 5 3 15
13 0 2 3 6
0 (15) 15 15 15
0 6 6 6 6
ANSWER: Since (0, 3) satisfy both equations, hence, it is a solution
6.
PROBLEM: 1 x y 1 1 ( 2 , 4 ) ; 4 2 x 4 y 0 SOLUTION:
Check: ( 12 , 14 ) Equation 1: x y x y
1 4
1 1 1 2 4 4 1 1 4 4
1 4
Equation 2: 2 x 4 y 0 2 x 4 y 0 2 12 4 14 0 1 1 0 00
ANSWER: Since ( 12 , 14 ) satisfy both equations, hence, it is a solution
7.
PROBLEM: ( 34 , 14 ) ; x y 1 4 x 8 y 5 SOLUTION:
Check: ( 34 , 14 ) Equation 1: x y 1
Equation 2: 4 x 8 y 5
x y 1
4 34 8 14 5
34 14 1
3 2 5 5 5
1 1
ANSWER: Since ( 34 , 14 ) does not satisfy both equations, hence, it is not a solution.
8.
PROBLEM: 1 1 3 x 2 y 1 (3, 4); 2 x 3 y 8 3 2 SOLUTION:
Check: (3, 4)
1 1 x y 1 3 2 1 1 3 4 1 3 2 1 2 1 11
Equation 1:
2 3 x y 8 3 2 2 3 x y 8 3 2 2 3 3 4 8 3 2 2 6 8 8 8
Equation 2:
ANSWER: Since (3, 4) satisfy both equations, hence, it is a solution
9.
PROBLEM: y 3 (5, 3); 5 x 10 y 5 SOLUTION:
Check: (5, 3) Equation 1: y 3
Equation 2: 5 x 10 y 5
y 3
5 x 10 y 5
3 3
5 5 10 3 5 25 30 5 25 30 5 55
ANSWER: Since (5, 3) satisfy both equations, hence, it is a solution
10. .
PROBLEM: x4 (4, 2); 7 x 4 y 8 SOLUTION:
Check: (4, 2) Equation 1: x 4 x4 44
Equation 2: 7 x 4 y 8 7x 4 y 8 7 4 4 2 8 28 8 8 20 8
ANSWER: Since (4, 2) does not satisfy both equations, hence, it is not a solution
11.
PROBLEM:
SOLUTION: In the above graph, the point of intersection appears to be (5, 0). ANSWER: Hence, solution is (5, 0).
12.
PROBLEM:
SOLUTION: In the above graph, the point of intersection appears to be ( 2, -2). ANSWER: So, solution is (2, -2).
13.
PROBLEM:
SOLUTION: In the above graph, the point of intersection appears to be (2, 1). ANSWER: Hence, solution is (2, 1).
14.
PROBLEM:
SOLUTION:
In the above graph, the point of intersection appears to be ( 5, -2). ANSWER: So, solution is ( 5, -2).
15.
PROBLEM:
SOLUTION: In the above graph, the point of intersection appears to be (0, 0). ANSWER: Hence, solution is (0, 0).
16.
PROBLEM:
SOLUTION: In the above graph, the point of intersection is not there. ANSWER: So, there is no solution.
17.
PROBLEM:
ANSWER: In the above graph, y depends on x and so we express all the ordered pair solutions ( x, y ) in the form ( x, 2 x 2) where x is any real number.
18.
PROBLEM:
SOLUTION: In the above graph, the point of intersection appears to be ( -1, -1). ANSWER: So, solution is (-1, -1).
19.
PROBLEM:
SOLUTION: In the above graph, the point of intersection is not there. ANSWER: Hence there is no solution.. Part B: Solving Linear Systems
20.
PROBLEM:
ANSWER: In the above graph, y depends on x and so we may express all the ordered pair solutions ( x, y ) in the form ( x, 3x 6) where x is any real number
21.
PROBLEM: 3 y x 6 2 y x 1 SOLUTION:
Equation 1: y
3 x6 2
Equation 2: y x 1
If x = 0, then y=0+6 y=6
If x = 0, then y=0+1 y=1
If y = 0, then 0=3/2x + 6 - 6 = 3/2x x = -4
If y = 0, then -x = -1 x=1
ANSWER: The solution is (-2, 3)
22.
PROBLEM: 3 y 4 x 2 y 1 x 2 4 SOLUTION:
Equation 1: y
3 x2 4
1 Equation 2: y x 2 4
If x=0, then y=0+2 y=2
If x=0, then y=0-2 y= -2
If y=0, then 3 y x2 4 3 0 x2 4 3 x2 4 3x 2* 4 x 8 / 3
If y=0, then 1 0 x2 4 1 x 2 4 1 x 2 4 x 8
23.
PROBLEM: y x 4 y x 2 SOLUTION:
Equation 1: y x 4
Equation 2: y x 2
If x = 0, then y=0-4 y = -4
If x = 0, then y=0+2 y=2
If y =0, then 0=x-4
If y = 0, then 0= -x +2
x=4
ANSWER: The solution is (3,-1)
24.
PROBLEM: y 5 x 4 y 4x 5 SOLUTION:
x=2
Equation 1: y 5 x 4
Equation 2: y 4 x 5
If x=0, then y 5 0 4
If x=0, then y 4 0 5
y4 If y=0, then 0 5 x 4
y 5 If y=0, then 0 4x 5
5x 4
4x 5
x .8
x 1.25
ANSWER:
25.
PROBLEM: 2 y 5 x 1 y 3 x 5
SOLUTION:
Equation 1: y If x=0, then 2 y 0 1 5 y 0 1
y 1 If y=0, then 2 y x 1 5 2 0 x 1 5 2 x 1 5 5 x 2
ANSWER:
2 x 1 5
Equation 2: y If x=0, then 3 y x 5 3 y 0 5 y0 If y=0, then 3 y x 5 3 0 x 5 x0
3 x 5
26.
PROBLEM: 2 y 5 x 6 y 2 x 10 5
SOLUTION:
2 Equation 1: y x 6 5
Equation 2: y
2 x 10 5
If x=0, then y 06 y6 If y=0, then 2 y x6 5 2 0 x6 5 2 x6 5 x 15 ANSWER:
If x=0, then y=0+10 y= 10 If y=0, then 2 0 x 10 5 2 10 x 5 x 25
27.
PROBLEM: y 2 y x 1 SOLUTION:
Equation 1: y 2 If every real value of x, y= -2
Equation 2: y x 1 If x=0, then y=0+1 y=1 If y=0, then y x 1 0 x 1 x 1
ANSWER:
28.
PROBLEM: y 3 x 3 SOLUTION:
Equation 1: y 3 For x = any real number, y=3
ANSWER:
Equation 2: x 3 For y = any real number x= -3
29.
PROBLEM: y 0 2 y 5 x 4 SOLUTION: Equation 1: y 0
For every real value of x, y=0
Equation 2: y If x=0, then y=0-4 y=-4 If y=0, then 2 0 x4 5 2 4 x 5 x 10
ANSWER:
2 x4 5
30.
PROBLEM: x 2 y 3x SOLUTION:
Equation 1: x 2 For y = any real number x=2
Equation 2: y 3x If x=0, then y= 0 If y= 0, then x=0
ANSWER:
31.
PROBLEM: 3 y 5 x 6 y 3 x 3 5 SOLUTION:
Equation 1: y If x=0, then y= 0-6 y= -6 If y=0, then 3 0 x 6 5 3 x6 5 x 10 ANSWER:
3 x6 5
Equation 2: y If x=0, then y=0-3 y=-3 If y=0, then 3 0 x 3 5 3 3 x 5 x5
3 x3 5
32.
PROBLEM: 1 y 2 x 1 y 1 x 1 2
SOLUTION:
1 Equation 1: y x 1 2 If x=0, then 1 y 0 1 2 y 1 If y= 0, then 1 0 x 1 2 1 x 1 2 x2 ANSWER:
1 Equation 2: y x 1 2 If x=0, then 1 y 0 1 2 y 1 If y= 0, then 1 0 x 1 2 1 x 1 2 x2
33.
PROBLEM: 2 x 3 y 18 6 x 3 y 6 SOLUTION:
Equation 1: 2 x 3 y 18
Equation 2: 6 x 3 y 6
If x = 0, then 2 0 3 y 18
If x = 0, then 3 y 6 6 x
0 3 y 18 y6 If y = 0, then 2 x 3 0 18
3 y 6 0 y 2 If y = 0, then 6 x 3 0 6
2 x 0 18 x9
ANSWER:
6 x 0 6 6 x 6 x 1
34.
PROBLEM: 3x 4 y 20 2x 8 y 8 SOLUTION:
Equation 1: 3x 4 y 20
Equation 2: 2 x 8 y 8
If x=0, then 3 0 4 y 20
If x=0, then 2 0 8 y 8
0 4 y 20 y5
0 8y 8 y 1
If y= 0, then 3 x 4 0 20
If y= 0, then 2 x 8 0 8
3 x 0 20
2x 0 8
x 6.67
x4
ANSWER:
35.
PROBLEM: 2 x y 1 2x 3y 9 SOLUTION:
Equation 1: 2 x y 1
Equation 2: 2 x 3 y 9
If x = 0, then 2 0 y 1
If x = 0, then 2 0 3 y 9
0 y 1 y 1 If y = 0, then 2 x 0 1 2 x 1 x 0.5
0 3y 9 y 3 If y = 0, then 2x 3 0 9
ANSWER:
2x 0 9 x 4.5
36.
PROBLEM: x 2 y 8 5 x 4 y 4 SOLUTION:
Equation 1: x 2 y 8
Equation 2: 5 x 4 y 4
If x=0, then 0 2 y 8 2 y 8
If x=0, then 5 0 4 y 4
y 4
y 1
0 4 y 4
If y= 0, then x 2 0 8
If y= 0, then 5 x 4 0 4
x 0 8
5 x 0 4
x 8
x .8
ANSWER:
37.
PROBLEM: 4 x 6 y 36 2x 3 y 6 SOLUTION:
Equation 1: 4 x 6 y 36
Equation 2: 2 x 3 y 6
If x = 0, then 4 0 6 y 36
If x = 0, then 2 0 3 y 6
0 6 y 36
0 3y 6 y 2 If y = 0, then 2x 3 0 6
6 y 36 y6 If y = 0, then 4 x 6 0 36 4 x 0 36 x9
ANSWER:
2x 0 6 x3
38.
PROBLEM: 2 x 3 y 18 6 x 3 y 6 SOLUTION:
Equation 1: 2 x 3 y 18
Equation 2: 6 x 3 y 6
If x=0, then 2 0 3 y 18
If x=0, then 6 0 3 y 6
0 3 y 18 3 y 18
0 3 y 6 3y 6
y 6 If y= 0, then
y2 If y= 0, then
2 x 3 0 18
6 x 3 0 6
2 x 0 18
6 x 0 6 6 x 6
x9
x 1 ANSWER:
39.
PROBLEM: 3 x 5 y 30 6 x 10 y 10 SOLUTION:
Equation 1: 3 x 5 y 30
Equation 2: 6 x 10 y 10
If x = 0, then 3 0 5 y 30
If x = 0, then 6 0 10 y 10
0 5 y 30
0 10 y 10
y6 If y = 0, then 3 x 5 0 30
10 y 10
3 x 0 30 x 10
y 1 If y = 0, then 6 x 10 0 10
6 x 0 10 6 x 10 x 1.67
ANSWER:
40.
PROBLEM: x 3y 3 5 x 15 y 15 SOLUTION:
Equation 1: x 3 y 3
Equation 2: 5 x 15 y 15
If x=0, then 0 3 y 3
If x=0, then 5 0 15 y 15
0 3y 3 y 1 If y= 0, then
0 15 y 15 15 y 15 y 1
x 3 0 3 x 0 3 x 3
If y= 0, then 5 x 15 0 15 5 x 0 15 5 x 15 x 3
ANSWER:
41.
PROBLEM: x y 0 x y 0 SOLUTION:
Equation 1: x y 0
Equation 2: x y 0
If x = 0, then 0 y 0
If x = 0, then 0 y 0
y 0 y0 If y = 0, then x 0 0
y0 If y = 0, then x 0 0
x0 ANSWER:
x 0 x0
42.
PROBLEM: yx y x 1 SOLUTION:
Equation 1: y x X= Y
Equation 2: y x 1 If x=0, then y 0 1 y 1 If y= 0, then
0 x 1 x 1 ANSWER:
43.
PROBLEM: 3x 2 y 0 x2 SOLUTION:
Equation 1: 3x 2 y 0 If x=0, then 3 0 2 y 0 0 2y 0 2y 0 y0 If y=0, then 3x 2 0 0 3x 0 0 x0
ANSWER:
Equation 2: x 2 For every value of y, x will remain 2
44.
PROBLEM: 1 2 2 x 3 y 3 3 x 1 y 2 2 SOLUTION:
Equation 1: 2 x If x=0, then 1 2 2 0 y 3 3 1 2 0 y 3 3 1 2 y 3 3 y2 If y= 0, then 1 2 0 3 3 2 2x 3 1 x 3 x .33
2x
ANSWER:
1 2 y 3 3
Equation 2: 3x If x=0, then 1 3 0 y 2 2 1 y 2 2 y 4 If y= 0, then 1 3 x 0 2 2 3 x 0 2 3x 2 x .67
1 y 2 2
45.
PROBLEM: 1 1 10 x 5 y 2 1 x 1 y 1 5 5
SOLUTION:
Equation 1 :
1 1 x y 2 10 5
If x = 0, then 1 1 0 y 2 10 5 1 0 y 2 5 y 10 If y = 0, then 1 1 x 0 2 10 5 1 x0 2 10 1 x2 10 x 20
ANSWER:
1 1 Equation 2: x y 1 5 5 If x = 0, then 1 1 0 y 1 5 5 1 y 1 5 y 5 If y = 0, then 1 1 x 0 1 5 5 1 x 1 5 x5
46.
PROBLEM: 1 1 3 x 2 y 1 1 x 1 y 1 3 5 SOLUTION:
Equation 1: If x=0, then
1 1 x y 1 3 2
Equation 2: If x=0, then
1 1 x y 1 3 5
1 1 0 y 1 3 2 1 y 1 2 y 2
1 1 0 y 1 3 5 1 y 1 5 y5
If y= 0, then 1 1 x 0 1 3 2 1 x 1 3 x3
If y= 0, then 1 1 x 0 1 3 5 1 x 1 3 x3
ANSWER:
47.
PROBLEM: 1 1 9 x 6 y 0 1 x 1 y 1 9 4 2 SOLUTION:
1 1 Equation 1 : x y 0 9 6 If x = 0, then 1 1 0 y 0 9 6 1 y0 6 y0 If y = 0, then 1 1 x 0 0 9 6 1 x0 9 x0 ANSWER:
Equation 2: If x = 0, then 1 1 1 0 y 9 4 2 1 1 y 4 2 y2 If y = 0, then 1 1 1 x 0 9 4 2 1 1 x 9 2 x 4.5
1 1 1 x y 9 4 2
48.
PROBLEM: 1 5 x y 5 16 2 5 x 1 y 5 16 2 2 SOLUTION:
Equation 1: If x=0, then
5 1 x y 5 16 2
Equation 2: If x=0, then
5 1 5 x y 16 2 2
5 1 0 y 5 16 2 1 y5 2 y 10 If y= 0, then 5 1 x 0 5 16 2 5 x5 16 x 16 ANSWER:
5 1 5 0 y 16 2 2 1 5 y 2 2 y5
If y= 0, then 5 1 5 x 0 16 2 2 5 5 x 16 2 x 8
49.
PROBLEM: 1 9 1 6 x 2 y 2 1 x 1 y 3 18 6 2 SOLUTION:
1 1 9 Equation 1 : x y 6 2 2
Equation 2:
If x = 0, then 1 1 9 0 y 6 2 2 1 9 y 2 2 y 9 If y = 0, then 1 1 9 x 0 6 2 2 1 9 x 6 2 x 27
If x = 0, then 1 1 3 0 y 18 6 2 1 3 y 6 2 y 9 If y = 0, then 1 1 3 x 0 18 6 2 1 3 x 18 2 x 27
ANSWER:
1 1 3 x y 18 6 2
50.
PROBLEM: 1 1 1 2 x 4 y 2 1 x 1 y 3 3 2 SOLUTION:
Equation 1:
1 1 1 x y 2 4 2
Equation 2:
If x=0, then 1 1 1 0 y 2 4 2 1 1 y 4 2 y2
If x=0, then 1 1 0 y 3 3 2 1 y3 2 y 6
If y= 0, then 1 1 1 x 0 2 4 2 1 1 x 2 2 x 1
If y= 0, then 1 1 x 0 3 3 2 1 x3 3 x9
ANSWER:
1 1 x y 3 3 2
51.
PROBLEM: y 4 x 5 SOLUTION:
Equation 1 : y 4 For every value of x, y will remain 4
Equation 2: x 5
For every value of y, x will remain -5
ANSWER:
52.
PROBLEM: y 3 x 2 SOLUTION:
Equation 1 : y 3 For every value of x, y will remain -3
Equation 2: x 2
For every value of y, x will remain 2
ANSWER:
53.
PROBLEM: y 0 x 0 SOLUTION:
Equation 1 : y 0 For every value of x, y will remain 0
Equation 2: x 0 For every value of y, x will remain 0
ANSWER:
54.
PROBLEM: y 2 y 3 SOLUTION:
Equation 1 : y 2 For every value of x, y will remain -2
Equation 2: y 3 For every value of x, y will remain 3
ANSWER:
55.
PROBLEM: y 5 y 5 SOLUTION:
Equation 1 : y 5 For every value of x, y will remain 5
Equation 2: y 5 For every value of x, y will remain -5
ANSWER:
56.
PROBLEM: y2 y 2 0 SOLUTION:
Equation 1 : y 2 For every value of x, y will remain 2
Equation 2: y 2 0 For every value of x, y will remain 2
ANSWER:
57.
PROBLEM: x 5 x 1 SOLUTION:
Equation 1 : x 5 For every value of y, x will remain -5
Equation 2: x 1 For every value of y, x will remain 1
ANSWER:
58.
PROBLEM: y x x 0 SOLUTION:
Equation 1 : y x For every value of x, y will remain 2
Equation 2: x 0 For every value of x, y will remain 2
ANSWER:
59.
PROBLEM: 4 x 6 y 3 x y 2 SOLUTION:
Equation 1 : 4 x 6 y 3 If x=0, then
Equation 2: x y 2 If x=0, then
4 0 6 y 3
0 y 2
6y 3 y .5 If y=0, then 4x 6 0 3
y 2 If y=0, then x 0 2
4x 3 x .75
ANSWER:
60.
PROBLEM: 2 x 20 y 20 3x 10 y 10
x 2 x2
SOLUTION:
Equation 1: 2 x 20 y 20
Equation 2: 3x 10 y 10
If x=0, then 2 0 20 y 20
If x=0, then 3 0 10 y 10
20 y 20 y 1
10 y 10 y 1
If y=0, then 2 x 20 0 20
If y=0, then 3x 10 0 10
2 x 20
3x 10 x 3.33
x 10
ANSWER:
61.
PROBLEM: The sum of two numbers is 20. The larger number is 10 less than five times the smaller. SOLUTION:
Let the two numbers be x and y Then two equations are x+ y = 20 5x- y = 10 Equation 1 : 4 x 6 y 3
Equation 2: x y 2
If x = 0, then 0 y 20
If x = 0, then 5 0 y 10
y 20
y 10
If y = 0, then x 0 20
y 10 If y = 0, then 5x 0 10
x 20
ANSWER:
5x 10 x2
62.
PROBLEM: The difference between two numbers is 12 and the sum is 4. SOLUTION: Let the two numbers be x and y Then two equations are x - y = 12 x+y=4
Equation 1: x y 12 If x=0, then
Equation 2: x y 4 If x=0, then 0 y 4 y4
0 y
12
y 12 y 12 If y=0, then x 0 12 x 12 ANSWER:
63.
PROBLEM: 3x 2 y 6 SOLUTION:
If y=0, then x 0 4 x4
x 1 2 3 4 5 6
ANSWER:
y -3/2 0 3/2 3 9/2 6
3x 6 2
64.
PROBLEM: 5 x 2 y 30 SOLUTION:
x 0 -2 -4 -6 -8 -10
y 15 10 5 0 -5 -10 ANSWER:
30 5 x 2
65.
PROBLEM: Determine the price at which the quantity demanded is equal to the quantity supplied. SOLUTION:
The price at which the quantity demanded is equal to the quantity supplied is = $1.25. ANSWER:
66.
PROBLEM: If production of bottled water slips to 20 tons, then what price does the demand curve predict for a bottle of water? SOLUTION:
Production of bottled water slips to 20 tons, the price demand curve predict for a bottle of water is $2.00 ANSWER:
67.
PROBLEM: If production of bottled water increases to 40 tons, then what price does the demand curve predict for a bottle of water? SOLUTION: Production of bottled water increases to 40 tons, the predicted price by the demand curve predict for a bottle of water is $1.00. ANSWER:
68.
PROBLEM: If the price of bottled water is set at $2.50 dollars per bottle, what quantity does the demand curve predict? SOLUTION:
The price of bottled water is set at $2.50 dollars per bottle. The quantity predicted by the demand curve is 10 tons. ANSWER:
4.2 Solving Linear Systems by Substitution Part A: Substitution Method 1.
PROBLEM: y 4x 1 3 x y 1 SOLUTION: Step 1: Put the value of y of first equation in second equation 3 x 4 x 1 1 3 x 4 x 1 1 x2
Step 2: Put the value of x in first equation
y 4 2 1 y 8 1 y7 ANSWER: (x, y) = (2,7)
2.
PROBLEM: y 3x 8 4 x y 2 SOLUTION: Step 1: Put the value of y of first equation in second equation 4 x 3x 8 2 4 x 3x 8 2 x 6 Step 2: Put the value of x in first equation
y 3 6 8 y 18 8 y 26 ANSWER: (x, y) = (-6,-26)
3.
PROBLEM: x 2y 3 x 3 y 8 SOLUTION: Step 1: Put the value of x of first equation in second equation
2 y 3 3 y 8 5 y 3 8 5 y 5 y 1
Step 2: Put the value of y in first equation
x 2 1 3 x 5 ANSWER: (x, y) = (-5,-1)
4.
PROBLEM: x 4 y 1 2 x 3 y 12 SOLUTION: Step 1: Put the value of x of first equation in second equation 2 4 y 1 3 y 12 8 y 2 3 y 12 5 y 10 y 2 Step 2: Put the value of y in first equation x 4 2 1 x 8 1 x9 ANSWER: (x, y) = (9,-2)
5.
PROBLEM: y 3x 5 x 2 y 2 SOLUTION: Step 1: Put the value of y of first equation in second equation 5 x 2 3 x 2 5 x 6 x 2 x2
Step 2: Put the value of x in first equation y 3 2 y6
ANSWER: (x, y) = (2,6)
6.
PROBLEM: yx 2 x 3 y 10 SOLUTION: Step 1: Put the value of y of first equation in second equation 2 x 3 x 10 5 x 10 x2 Step 2: Put the value of x in first equation y 2 y2
ANSWER: (x, y) = (2,2)
7.
PROBLEM: y 4x 1 4 x y 2 SOLUTION: Step 1: Put the value of y of first equation in second equation
4 x 4 x 1 2 1 2 ANSWER: Hence there is no solution.
8.
PROBLEM: y 3x 5 3x y 5 SOLUTION: Step 1: Put the value of y of first equation in second equation 3 x 3 x 5 5
55 ANSWER: (x, y) = (x,-3x +5)
9.
PROBLEM: y 2x 3 2 x y 3 SOLUTION: Step 1: Put the value of y of first equation in second equation
2 x 2 x 3 3 3 3 ANSWER: (x, y) = (x, 2x+3)
10.
PROBLEM: y 5x 1 x 2 y 5 SOLUTION: Step 1: Put the value of y of first equation in second equation x 2 5 x 1 5
x 10 x 2 5 9 x 3 1 x 3 Step 2: Put the value of x in first equation
1 y 5 1 3 8 y 3 ANSWER: (x, y) = (-1/3,-8/3)
11.
PROBLEM: y 7 x 1 3x y 4 SOLUTION: Step 1: Put the value of y of first equation in second equation 3x 7 x 1 4
10 x 1 4 10 x 5 x
1 2
Step 2: Put the value of x in first equation
1 y 7 1 2 7 y 1 2 5 y 2 ANSWER: (x, y) = (1/2, - 5/2)
12.
PROBLEM: x 6 y 2 5 x 2 y 0
SOLUTION: Step 1: Put the value of x of first equation in second equation 5 6 y 2 2 y 0
30 y 10 2 y 0 28 y 10 10 y 28 5 y 14 Step 2: Put the value of y in first equation 5 x 6 2 14 15 x 2 7 1 x 7 ANSWER: (x, y) = (-1/7, -5/14)
13.
PROBLEM: y 2 2 x y 6 SOLUTION: Step 1: Put the value of y = -2 of first equation in second equation
2 x 2 6 2 x 2 6 2 x 8 x4 ANSWER: (x, y) = (4, -2)
14.
PROBLEM: x 3 x 4 y 3 SOLUTION: Step 1: Put the value of x =-3 of first equation in second equation
3 4 y 3 4 y 0 y0 ANSWER: (x, y) = (-3, 0)
15.
PROBLEM: 1 y x 3 5 7 x 5 y 9 SOLUTION: Step 1: Put the value of y of first equation in second equation 1 7x 5 x 3 9 5 7 x x 15 9 7 x x 15 9 8 x 24 x3
Step 2: Put the value of x in first equation
1 3 3 5 3 y 3 5 12 y 5 y
ANSWER: (x, y) = (3, 12/5)
16.
PROBLEM: 2 y x 1 3 6 x 9 y 0 SOLUTION: Step 1: Put the value of y of first equation in second equation 2 6 x 9 x 1 0 3 6x 6x 9 0 90
ANSWER: So, there is no solution
17.
PROBLEM: 1 1 y x 2 3 x 6 y 4 SOLUTION: Step 1: Put the value of y of first equation in second equation 1 1 x 6 x 4 3 2 x 3x 2 4 2 x 6 x 3
Step 2: Put the value of x in first equation 1 1 y 3 2 3 3 1 y 2 3 7 y 6 ANSWER: (x, y) = (-3, -7/6)
18.
PROBLEM: 3 1 y x 8 2 2 x 4 y 1 SOLUTION: Step 1: Put the value of y of first equation in second equation
1 3 2x 4 x 1 2 8 3 2x x 2 1 2 1 x 1 2 x 2 Step 2: Put the value of x in first equation 3 1 2 8 2 3 1 y 4 2 5 y 4 ANSWER: (x, y) = (-2, 5/4) y
19.
PROBLEM: x y 6 2 x 3 y 16 SOLUTION: Step 1: Write the first equation in the form of y
x y 6 y 6 x Step 2: Put the value of y of first equation in second equation 2 x 3 6 x 16
2 x 18 3x 16 x 2 x2 Step 3: Put the value of x in first equation 2 y 6
y 62 y4 ANSWER: (x, y) = (2, 4)
20.
PROBLEM: x y 3 2 x 3 y 2 SOLUTION: Step 1: Write the first equation in the form of y x y 3
y x3
Step 2: Put the value of y of first equation in second equation 2 x 3 x 3 2 2 x 3 x 9 2 x7
Step 3: Put the value of x in first equation 7 3 y 4 y
ANSWER: (x, y) = (7, 4)
21.
PROBLEM: 2x y 2 3x 2 y 17 SOLUTION: Step 1: Write the first equation in the form of y 2x y 2
2x 2 y y 2 2x Step 2: Put the value of y of first equation in second equation 3x 2 2 2 x 17
3x 4 4 x 17 7 x 21 x3 Step 3: Put the value of x in first equation
2 3 y 2 6 2 y y 4 ANSWER: (x, y) = (3, -4)
22.
PROBLEM: x 3 y 11 3x 5 y 5 SOLUTION: Step 1: Write the first equation in the form of x x 3 y 11 x 11 3 y Step 2: Put the value of x of first equation in second equation 3 11 3 y 5 y 5 33 9 y 5 y 5 14 y 28 y2
Step 3: Put the value of y in first equation x 3 2 11
x 5 ANSWER: (x, y) = (-5, 2)
23.
PROBLEM: x 2 y 3 3x 4 y 2 SOLUTION: Step 1: Write the first equation in the form of x
x 2 y 3 x 3 2 y Step 2: Put the value of x of first equation in second equation 3 3 2 y 4 y 2
9 6 y 4 y 2 10 y 7 7 y 10 Step 3: Put the value of y in first equation 7 x 2 3 10 7 x 3 5 8 x 5 ANSWER: (x, y) = (-8/5, -7/10)
24.
PROBLEM: 5 x y 12 9 x y 10
SOLUTION: Step 1: Write the first equation in the form of y 5 x y 12
5 x 12 y Step 2: Put the value of y of first equation in second equation
9 x 5 x 12 10 9 x 5 x 12 10 4 x 2 1 x 2 Step 3: Put the value of x in first equation 1 5 y 12 2 5 12 y 2 29 y 2 ANSWER: (x, y) = (-1/2, -29/2)
25.
PROBLEM: x 2 y 6 4 x 8 y 24
SOLUTION: Step 1: Write the first equation in the form of x
x 2 y 6 x 6 2 y Step 2: Put the value of x of first equation in second equation 4 6 2 y 8 y 24
24 8 y 8 y 24 24 24 ANSWER: (x, y) = (x, -x/2-3)
26.
PROBLEM: x 3 y 6 2 x 6 y 12
SOLUTION: Step 1: Write the first equation in the form of x x 3 y 6 x 6 3 y Step 2: Put the value of x of first equation in second equation 2 6 3 y 6 y 12
12 6 y 6 y 12 12 12 ANSWER: So, there is no solution.
27.
PROBLEM: 3x y 4 6 x 2 y 2 SOLUTION: Step 1: Write the first equation in the form of y 3x y 4
y 4 3 x Step 2: Put the value of y of first equation in second equation 6 x 2 4 3 x 2 6 x 8 6 x 2 8 2
ANSWER: Hence there is no solution.
28.
PROBLEM: x 5 y 10 2 x 10 y 20 SOLUTION: Step 1: Write the first equation in the form of x
x 5 y 10 x 5 y 10 Step 1: Put the value of x of first equation in second equation 2 5 y 10 10 y 20 10 y 20 10 y 20 20 20
ANSWER: (x, y) = (x, x/5+2)
29.
PROBLEM: 3x y 9 4 x 3 y 1 SOLUTION: Step 1: Write the first equation in the form of y 3x y 9
3x 9 y Step 2: Put the value of y of first equation in second equation 4 x 3 3x 9 1
4 x 9 x 27 1 13 x 26 x2 Step 2: Put the value of x in first equation 3 2 y 9
69 y y 3 ANSWER: (x, y) = (2, -3)
30.
PROBLEM: 2x y 5 4 x 2 y 2 SOLUTION: Step 1: Write the first equation in the form of y 2x y 5
2x 5 y Step 2: Put the value of y of first equation in second equation 4 x 2 2 x 5 2
4 x 4 x 10 2 8x 8 x 1 Step 3: Put the value of x in first equation
2 1 y 5 25 y y 3 ANSWER: (x, y) = (1, -3)
31.
PROBLEM: x 4 y 0 2 x 5 y 6 SOLUTION: Step 1: Write the first equation in the form of x x 4 y 0
x 4 y x 4y Step 2: Put the value of x of first equation in second equation
2 4 y 5 y 6 8 y 5 y 6 3 y 6 y 2
Step 3: Put the value of y in first equation x 4 2 0 x 8 x 8
ANSWER: (x, y) = (-8, -2)
32.
PROBLEM: 3y x 5 5 x 2 y 8 SOLUTION: Step 1: Write the first equation in the form of x 3y x 5 3y 5 x Step 2: Put the value of x of first equation in second equation 5 3 y 5 2 y 8 15 y 25 2 y 8 17 y 17 y 1
Step 3: Put the value of y in first equation 3 1 x 5 35 x x 2
ANSWER: (x, y) = (-2, 1)
33.
PROBLEM: 2x 5 y 1 4 x 10 y 2
SOLUTION: Step 1: Write the first equation in the form of x 2x 5 y 1 2x 1 5 y 1 5y x 2 Step 2: Put the value of x of first equation in second equation
1 5y 4 10 y 2 2 2 10 y 10 y 2 20 y 0 y0 Step 3: Put the value of y in first equation
2x 5 0 1 2x 1 0 1 x 2 ANSWER: (x, y) = (1/2, 0)
34.
PROBLEM: 3x 7 y 3 6 x 14 y 0 SOLUTION: Step 1: Write the first equation in the form of y 3x 7 y 3 7 y 3 3x 7 y 3 3x 3 3 y x 7 7 Step 2: Put the value of y of first equation in second equation 3 3 6 x 14 x 0 7 7 6x 6 6x 0 12 x 6 1 x 2 Step 3: Put the value of x in first equation 1 3 7 y 3 2 3 7 y 3 2 3 7 y 2 3 y 14 ANSWER: (x, y) = (-1/2, 3/14)
35.
PROBLEM: 10 x y 3 1 5 x 2 y 1 SOLUTION: Step 1: Write the first equation in the form of y 10 x y 3
10 x 3 y Step 2: Put the value of y of first equation in second equation 1 5 x 10 x 3 1 2 3 5 x 5 x 1 2 3 1 2 ANSWER: Hence there is no solution.
36.
PROBLEM: 1 2 1 3 x 6 y 3 1 x1 y 3 2 3 2
SOLUTION: Step 1: Write the first equation in the form of y
1 1 2 x y 3 6 3 1 1 2 y x 6 3 3 x2 1 y 6 3 y 2x 4
Step 2: Put the value of y of first equation in second equation 1 1 3 x 2x 4 2 3 2 1 2 4 3 x x 2 3 3 2 3x 4 x 4 3 6 3 2 x 8 9 6 6 x 1 6 6 x 1 Step 3: Put the value of x in first equation 1 1 2 1 y 3 6 3 1 1 2 y 6 3 3 1 y 1 6 y6 ANSWER: (x, y) = (1, 6)
37.
PROBLEM: 2 1 3 x 3 y 1 1 x 1 y 1 4 3 12 SOLUTION: Step 1: Write the first equation in the form of y
1 2 x y 1 3 3 1 2 x 1 y 3 3 2 x3 y 3 3 x 3 y 2 3 1 x y 2 2 Step 2: Put the value of y of first equation in second equation 1 1 3 1 1 x x 4 3 2 2 12 1 1 1 1 x x 4 2 6 12 6x 4x 1 1 24 2 12 10 x 6 1 24 12 x 1
Step 3: Put the value of x in first equation 1 2 1 y 1 3 3 2 1 y 1 3 3 2 2 y 3 3 y 1 ANSWER: (x, y) = (1, 1)
38.
PROBLEM: 1 1 7 x y 2 1 x 1 y 2 4 2 SOLUTION: Step 1: Write the first equation in the form of y
1 1 x y 7 2 1 1 x y 7 2 Step 2: Put the value of y of first equation in second equation 1 11 1 x x 2 4 27 2 1 1 1 x x 2 4 14 4 7x 2x 1 2 28 4 9x 1 2 28 4 9x 9 28 4 x7
Step 3: Put the value of x in first equation
1 1 7 y 7 2 1 1 y 2 1 y 2 ANSWER: (x, y) = (7, 1/2)
39.
PROBLEM: 2 1 3 5 x 5 y 2 1 x 1 y 1 3 12 3 SOLUTION: Step 1: Write the first equation in the form of y 3 2 1 x y 5 5 2 2 1 3 y x 5 2 5 2 5 6x y 5 10 5 6x y 4 Step 2: Put the value of y of first equation in second equation 1 1 5 6x 1 x 3 12 4 3 1 5 1 1 x x 3 48 8 3 8x 3x 1 5 24 3 48 5 x 5 16 24 48 11 x 10 Step 3: Put the value of x in first equation 3 11 2 1 y 5 10 5 2 2 1 33 y 5 2 50 2 25 33 y 5 50 2 y 5 ANSWER: (x, y) = (-11/10, -2/5)
40.
PROBLEM: 1 2 x y 2 3 x 2 y 2 3 SOLUTION: Step 1: Write the first equation in the form of y
1 2 x y 2 3 3 x y 4 Step 2: Put the value of y of first equation in second equation 23 x x 2 34 1 x x 2 2 1 x2 2 x4 Step 3: 1 4 2 3 4 4 y3
Put the value of x in first equation 2 y 3
y
ANSWER: (x, y) = (4, 3)
41.
PROBLEM: 1 5 1 2 x 2 y 8 1 x 1 y 1 4 2 4 SOLUTION: Step 1: Write the first equation in the form of y
1 1 5 x y 2 2 8 1 5 1 y x 2 8 2 5 y x 4 Step 2: Put the value of y of first equation in second equation 1 15 1 x x 4 24 4 1 5 1 1 x x 4 8 2 4 3 1 5 x 4 4 8 3 25 x 4 8 3 x 6 1 x 2 Step 3: Put the value of x in first equation 1 1 1 5 y 2 2 2 8 1 5 1 y x 2 8 4 52 y 4 3 y 4 ANSWER: (x, y) = (-1/2, 3/4)
42.
PROBLEM: x y 0 x 2 y 3 SOLUTION: Step 1: Write the first equation in the form of y
x y 0 x y Step 2: Put the value of y of first equation in second equation x 2 x 3
x3 Step 3: Put the value of x in first equation x 3 0
x3 ANSWER: (x, y) = (3, 3)
43.
PROBLEM: y 3x 2 x 3 y 0
SOLUTION: Step 1: Put the value of y = 3x of first equation in second equation 2x 3y 0
2 x 3 3x 0 2x 9x 0 7 x 0 x0 Step 2: Put the value of x in first equation y 3 0 y0
ANSWER: (x, y) = (0,0)
44.
PROBLEM: 2 x 3 y 18 6 x 3 y 6
SOLUTION: Step 1: Write the first equation in the form of y
2 x 3 y 18 3 y 18 2 x 2 y 6 x 3 Step 2: Put the value of y of first equation in second equation 2 6 x 3 6 x 6 3 6 x 18 2 x 6 8 x 24 x3 Step 3: Put the value of x in first equation 2 3 3 y 18 3 y 18 6 3 y 12 y4
ANSWER: (x, y) = (3, 4)
45.
PROBLEM: 3 x 4 y 20 2x 8 y 8 SOLUTION: Step 1: Write the first equation in the form of y 3 x 4 y 20 4 y 20 3x
y 5
3 x 4
Step 2: Put the value of y of first equation in second equation 3 2x 8 5 x 8 4 2 x 40 6 x 8 8 x 8 40 8 x 32 x 4
Step 3: Put the value of x in first equation 3 4 4 y 20
4 y 20 12 8 y 4 y2 ANSWER: (x, y) = (-4, 2)
46.
PROBLEM: 5 x 3 y 1 3x 2 y 7
SOLUTION: Step 1: Write the first equation in the form of y 5 x 3 y 1
5x 1 3 y 5 1 y x 3 3 Step 2: Put the value of y of first equation in second equation
1 5 3x 2 x 7 3 3 10 2 3x x 7 3 3 19 2 x 7 3 3 19 19 x 3 3 x 1 Step 3: Put the value of x in first equation 5 1 3 y 1
5 1 3y y2 ANSWER: (x, y) = (1, 2)
47.
PROBLEM: 3 x 7 y 2 2x 7 y 1 SOLUTION: Step 1: Write the first equation in the form of y 3 x 7 y 2 7 y 2 3x
y
2 3 x 7 7
Step 2: Put the value of y of first equation in second equation 2 3 2x 7 x 1 7 7 2 x 2 3x 1
5 x 1 1 x 5 Step 3: Put the value of x in first equation
1 3 7 y 2 5 3 7y 2 5 2 3 y 7 35 10 3 y 35 1 y 5
ANSWER: (x, y) = (-1/5, 1/5)
48.
PROBLEM: y3 y 3 ANSWER: As the values of y in both the equations are not equal, the solution for the two equations is not defined.
49.
PROBLEM: x5 x 2 ANSWER: Since the values of x in both the equations are not equal, the solution for the two equations is not defined
50.
PROBLEM: y 4 y 4 ANSWER: As the values of y in both the equations are equal, the solution is (x, y) = (x, 4)
51.
PROBLEM: The sum of two numbers is 19. The larger number is 1 less than three times the smaller. SOLUTION:
Let the two numbers be x and y x + y = 19 3x - y = 1 Step 1: Write the first equation in the form of y x y 19
y 19 x Step 2: Put the value of y of first equation in second equation 3x 19 x 1
3x 19 x 1 4x=20 x=5 Step 3: Put the value of x in first equation
5
y 19
y 14 ANSWER: (x, y) = (5, 14)
52.
PROBLEM: The sum of two numbers is 15. The larger is 3 more than twice the smaller. SOLUTION:
Let the two numbers be x and y x + y = 15 y - 2x = 3 Step 1: Write the first equation in the form of y x y 15
y 15-x Step 2: Put the value of y of first equation in second equation
15-x 2x
3
15- x- 2x=3 3x=12 x=4 Step 3: Put the value of x in first equation y 15- 4
y 11 ANSWER: (x, y) = (4, 11)
53.
PROBLEM: The difference of two numbers is 7 and the sum is 1. SOLUTION:
Let the two numbers be x and y x-y=7 x+y=1 Step 1: Write the first equation in the form of y x y 7
x 7 y Step 2: Put the value of y of first equation in second equation x x 7 1
2x = 8 x=4 Step 3: Put the value of x in first equation 4 y 7
y 47 y 3 ANSWER: (x, y) = (4, -3)
54.
PROBLEM: The difference of two numbers is 3 and the sum is -7.
Let the two numbers be x and y x-y=3 x + y = -7 SOLUTION: Step 1: Write the first equation in the form of y x y 3
y x3 Step 2: Put the value of y of first equation in second equation
x x 3 7 2x-3= -7 2x = -4 x=-2 Step 3: Put the value of x in first equation 2 y 3
2 3 y y 5 ANSWER: (x, y) = (-2, -5)
55.
PROBLEM: 5 x 3 y 30 SOLUTION: x
0 -3 -6 -9 -12 -15
5 y 10 x 3 10 5 0 -5 -10 -15
ANSWER: So the solution is (-15, -15)
56.
PROBLEM: 1 1 x y 1 2 3 SOLUTION:
X 0 1 2 3 4 5 6
y
3x 6 2
-3 -3/2 0 3/2 3 9/2 6
ANSWER: So the solution is (6, 6)
58.
PROBLEM: Merits and drawbacks of the substitution method ANSWER: Merits: The substitution method is the simplest method when compared to the other methods available. It is most helpful when the question deals with just two unknown variables.
Drawbacks: Substitution is a time consuming process, since we have to move the equation around. First, we need to take out equation in the form of one variable then substitute this variable in the next equation. This method becomes quite difficult when it comes to dealing with square roots. It is long and complicated when any question consists of more than two equations. There is always an element of confusion involved as to which would be the best way to go ahead when we are working with three or four variables.
Part B: Discussion Board Topics
1.
PROBLEM:
Describe what drives the choice of variable to solve for when beginning the process of solving by substitution. ANSWER: The most important drive that makes any student to go for the choice of any variable in the equations is that variable which is there in the equation with coefficient 1. This variable is expressed in the form of rest part of the equation. When all the variables are available with the coefficient 1 then out of which the variable which is positive is generally chosen by students and is expressed in the rest part of the equation. These perceptions can include the tendency to convert negative signs to positive by shifting the value to the other side of the equation. In the substitution method, the student will try to make the substitution as simple as possible, which will make it easier to solve the value of the other variable.
4. 3 Solving Linear Systems by Elimination Part A : Elimination Method 1.
PROBLEM: x y 3 2 x y 9 SOLUTION: Step 1: Add both the equations to eliminate y, we get x y 3 2x y 9
3x
12
x4 Step 2 Substitute the value of x in first equation 4 y 3 y 3 4 y 1 ANSWER: (x, y) = (4, -1)
2.
PROBLEM: x y 6 5 x y 18 SOLUTION: Step 1: Add both the equations to eliminate y, we get x y 6 5 x y 18
6 x 24 x 4 Step 2 Substitute the value of x in first equation
4 y 6 4 6 y y2 ANSWER: (x, y) = (-4, 2)
3.
PROBLEM: x 3y 5 x 2 y 0 SOLUTION: Step 1: Add both the equations to eliminate x, we get x 3y 5 x 2y 0
y
5
y5 Step 2 Substitute the value of y in first equation x 3 5 5
x 15 5 x 5 15 x 10 ANSWER: (x, y) = (-10, 5)
4.
PROBLEM: x 4 y 4 x y 7
SOLUTION: Step 1: Add both the equations to eliminate x , we get x 4y 4 x y 7
3 y 3 y 1 Step 2 Substitute the value of y in first equation x 4 1 4 x 8 x 8
ANSWER: (x, y) = (-8, -1)
5.
PROBLEM: x y 2 x y 3 SOLUTION: Step 1: Add both the equations , we get
x y 2 x y 3 0
1
ANSWER: Hence there is no solution..
6.
PROBLEM: 3x y 2 6 x 4 y 2 SOLUTION: Step 1: Multiply first equation by 4 and then add both the equations to eliminate y 4
3 x y 2 6 x 4 y 2
12 x 4 y 8 6x 4 y 2
12 x 4 y 8 6x 4 y 2 18 x 1 x 3
6
Step 2: Substitute the value of x in first equation 1 3 y 2 3 1 y 2
y 2 1 y 1 y 1 ANSWER: (x, y) = (-1/3, 1)
7.
PROBLEM: 5 x 2 y 3 10 x y 4 SOLUTION: Step 1: Multiply second equation by 2 then add both the equations to eliminate y
5 x 2 y 3 2 10 x y 4
5 x 2 y 3 20 x 2 y 8
5 x 2 y 3 20 x 2 y 8 25 x x
5
1 5
Step 2 Substitute the value of x in first equation 1 5 2 y 3 5 1 2 y 3
2 y 4 y 2
ANSWER: (x, y) = (1/5, -2)
8.
PROBLEM: 2 x 14 y 28 x 7 y 21
SOLUTION: Step 1: Multiply second equation by 2 and then add both the equations
2 x 14 y 28 x 7 y 21
2
2 x 14 y 28 2 x 14 y 42 0
70
ANSWER: So, there is no solution
2 x 14 y 28 2 x 14 y 42
9.
PROBLEM: 2 x y 4 12 x 6 y 24 SOLUTION: Step 1: Multiply first equation by 6 then add both the equations 6
2 x y 4 12 x 6 y 24
12 x 6 y 24 12 x 6 y 24
12 x 6 y 24 12 x 6 y 24 0
0
ANSWER: (x, y) = (x, 2x+4)
10.
PROBLEM: x 8y 3 3x 12 y 6 SOLUTION: Step 1: Multiply first equation by 3 and then subtract second equation from first equation. x 8y 3 3 x 12 y 6
3
3 x 24 y 9 3 x 12 y 6
3 x 24 y 9 3 x 12 y 6 12 y
3
1 4 Step 2 Substitute the value of y in first equation y
1 x 8 3 4 x23 x 1
ANSWER: (x, y) = (1, 1/4)
11.
PROBLEM: 2 x 3 y 15 4 x 10 y 14
SOLUTION: Step 1: Multiply first equation by 2 then subtract second equation from first equation
2 x 3 y 15 4 x 10 y 14
2
4 x 6 y 30 4 x 10 y 14
4 x 6 y 30 4 x 10 y 14 16 y y 1
16
Step 2 Substitute the value of x in first equation 2 x 3 1 15 2 x 15 3 x6
ANSWER: (x, y) = (6, -1)
12.
PROBLEM: 4 x 3 y 10 3x 9 y 15 SOLUTION:
Step 1: Multiply first equation by 3 and then add both the equations to eliminate y
4 x 3 y 10 3 x 9 y 15
3
12 x 9 y 30 3 x 9 y 15
12 x 9 y 30 3 x 9 y 15 15 x 15 x 1 Step 2 Substitute the value of x in first equation 4 1 3 y 10 3 y 10 4 3 y 6 y 2
ANSWER: (x, y) = (-1, -2)
13.
PROBLEM: 4 x 5 y 3 8 x 3 y 15 SOLUTION: Step 1: Multiply first equation by 2 then add both the equations to eliminate x
4 x 5 y 3 8 x 3 y 15
2
8 x 10 y 6 8 x 3 y 15
8 x 10 y 6 8 x 3 y 15 7 y 21 y3 Step 2 Substitute the value of x in first equation 4 x 5 3 3 4 x 12 x 3 ANSWER: (x, y) = (-3, 3)
14.
PROBLEM: 2 x 7 y 56 4 x 2 y 112 SOLUTION: Step 1: Multiply first equation by 2 then add both the equations to eliminate
x
2 x 7 y 56 4 x 2 y 112
2
4 x 14 y 112 4 x 2 y 112
4 x 14 y 112 4 x 2 y 112 12 y
0
y0 Step 2 Substitute the value of y in first equation 2 x 7 0 56 2 x 56 x 28
ANSWER: (x, y) = (-28, 0)
15.
PROBLEM: 9 x 15 y 15 3 x 5 y 10 SOLUTION: Step 1: Multiply second equation by 3 then add both the equations to eliminate y
9 x 15 y 15 3 3x 5 y 10
9 x 15 y 15 9 x 15 y 30
9 x 15 y 15 9 x 15 y 30 0 45 ANSWER: Hence there is no solution..
16.
PROBLEM: 6 x 7 y 4 2 x 6 y 7 SOLUTION: Step 1: Multiply second equation by 3 and then subtract second equation from first equation
6 x 7 y 4 3 2 x 6 y 7
6x 7 y 4 6 x 18 y 21
6x 7 y 4 6 x 18 y 21 25 y y 1
25
Step 2 Substitute the value of y in first equation 6 x 7 1 4
6x 4 7 3 x 6 1 x 2 ANSWER: (x, y) = (-1/2, -1)
17.
PROBLEM: 4x 2 y 4 5 x 3 y 7
SOLUTION: Step 1: Multiply second equation by 2 and multiply first equation by 3 then add both the equations to eliminate y
3
4x 2 y 4 5 x 3 y 7 2
12 x 6 y 12 10 x 6 y 14
12 x 6 y 12 10 x 6 y 14 2
2x x 1
Step 2 Substitute the value of x in first equation 4 1 2 y 4
2y 8 y4 ANSWER: (x, y) = ( -1, 4)
18.
PROBLEM: 5 x 3 y 1 3x 2 y 7 SOLUTION: Step 1: Multiply first equation by 2 and multiply second equation by 3 then add both the equations to eliminate y 2
5 x 3 y 1 3 3x 2 y 7
10 x 6 y 2 9 x 6 y 21
10 x 6 y 2 9 x 6 y 21 19 x
19
x 1 Step 2: Substitute the value of x in first equation
5 1 3 y 1 3 y 5 1 3 y 6 y2
ANSWER: (x, y) = (1, 2)
19.
PROBLEM: 7 x 3 y 9 2 x 5 y 14 SOLUTION: Step 1: Multiply second equation by 2 then add both the equations to eliminate y 5
7 x 3 y 9 2 x 5 y 14 3
35 x 15 y 45 6 x 15 y 42
35 x 15 y 45 6 x 15 y 42 29 x
87
x3
Step 2 Substitute the value of x in first equation 7 3 3 y 9
3 y 12 y 4
ANSWER: (x, y) = (3, -4)
20.
PROBLEM: 9x 3y 3 7 x 2 y 15 SOLUTION: Step 1: Multiply first equation by 2 and multiply second equation by 3 then add both the equations to eliminate y 2
9x 3y 3 7 x 2 y 15 3
18 x 6 y 6 21x 6 y 45
18 x 6 y 6 21x 6 y 45 x
39 x 1
39
Step 2: Substitute the value of y in first equation 9 1 3 y 3 3 y 3 9 3 y 12 y 4
ANSWER: (x, y) = (-1, -4) 21.
PROBLEM: 5 x 3 y 7 7 x 6 y 11 SOLUTION: Step 1: Multiply first equation by 2 then add both the equations to eliminate y
2
5 x 3 y 7 7 x 6 y 11
10 x 6 y 14 7 x 6 y 11
10 x 6 y 14 7 x 6 y 11 3 x 3 x 1 Step 2 Substitute the value of x in first equation 5 1 3 y 7
3 y 2 2 y 3 ANSWER: (x, y) = ( -1, 2/3)
22.
PROBLEM: 2 x 9 y 8 3x 7 y 1 SOLUTION: Step 1: Multiply first equation by 3 and multiply second equation by 2 then subtract second equation from first equation 3
2 x 9 y 8 3 x 7 y 1 2
6 x 27 y 24 6 x 14 y 2
6 x 27 y 24 6 x 14 y 2 13 y y
26
2
Step 2: Substitute the value of y in first equation
2x 9 2 8 2 x 8 18 2 x 10 x 5 ANSWER: (x, y) = (-5, 2)
23.
PROBLEM: 2 x 2 y 5 3x 3 y 5
SOLUTION: Step 1: Multiply first equation by 3 and multiply second equation by 2 then subtract second equation from first equation 3
2 x 2 y 5 3 x 3 y 5 2
6 x 6 y 15 6 x 6 y 10
6 x 6 y 15 6 x 6 y 10 0 25
ANSWER: Hence there is no solution..
24.
PROBLEM: 3x 6 y 12 2x 4 y 8 SOLUTION: Step 1: Multiply first equation by 2 and multiply second equation by 3 then add both the equations 2
3 x 6 y 12 2x 4 y 8
3
6 x 12 y 24 6 x 12 y 24
6 x 12 y 24 6 x 12 y 24 0 0
ANSWER: (x, y) = (x, 1/2 x-2)
25.
PROBLEM: 25 x 15 y 1 15 x 10 y 1
SOLUTION: Step 1: Multiply first equation by 6 and multiply second equation by 10 then subtract second equation from first equation 6
25 x 15 y 1 15 x 10 y 1 10 150 x 90 y 6 150 x 100 y 10 10 y 4 2 y 5
150 x 90 y 6 150 x 100 y 10
Step 2 Substitute the value of y in first equation
25 x 15 y 1 2 25 x 15 1 5 25 x 6 1 25 x 5 1 x 5 ANSWER: (x, y) = (1/5, -2/5)
26.
PROBLEM: 2x 3y 2 18 x 12 y 5 SOLUTION: Step 1: Multiply first equation by 9 and then subtract second equation from first equation
2x 3y 2 18 x 12 y 5
9
18 x 27 y 18 18 x 12 y 5
18 x 27 y 18 18 x 12 y 5 15 y 13 13 y 15 Step 2 Substitute the value of y in first equation
13 2x 3 2 15 13 2x 2 5 10 13 2x 5 3 2x 5 3 x 10
ANSWER: (x, y) = (-3/10, -13/15)
27.
PROBLEM: y 2 x 3 3 x 2 y 4
SOLUTION: Step 1: Rewrite the first equation as y 2 x 3
2 x y 3 Step 2: Multiply first equation by 2 then add both the equations to eliminate y
2 x y 3 3 x 2 y 4
2
4 x 2 y 6 3 x 2 y 4
4 x 2 y 6 3x 2 y 4 x 2 Step 3: Substitute the value of x in first equation y 2 2 3 y 1 ANSWER: (x, y) = (-2, 1)
28.
PROBLEM: 28 x 6 y 9 6 y 4 x 15 SOLUTION: Step 1: Rewrite the second equation as 6 y 4 x 15
4 x 6 y 15 Step 2: Subtract second equation from the first equation 28 x 6 y 9 4 x 6 y 15 32 x 24 x
3 4
Step 3: Substitute the value of x in first equation
3 28 6 y 9 4 21 6 y 9 6 y 12 y 2 ANSWER: (x, y) = (3/4, -2) 29.
PROBLEM: y 5 x 15 y 5 x 5
SOLUTION: Step 1: Add both the equations to eliminate x
y 5 x 15 y 5 x 5 2 y 20 y 10
Step 2 Substitute the value of y in first equation
10 5 x 15 5 x 5 x 1
ANSWER: (x, y) = ( -1,10)
30.
PROBLEM: 2 x 3 y 9 5 x 8 y 16 SOLUTION: Step 1: Multiply first equation by 5 and multiply second equation by 2 then subtract second equation from first equation 5
2 x 3 y 9 5 x 8 y 16 2
10 x 15 y 45 10 x 16 y 32
10 x 15 y 45 10 x 16 y 32 y 77
Step 2 Substitute the value of y in first equation 2 x 3 77 9
2 x 231 9 2 x 240 x 120 ANSWER: (x, y) = (120, 77)
31.
PROBLEM: 1 1 1 2 x 3 y 6 5 x y 7 2 2
SOLUTION: Step 1: Multiply first and second equations by 6 1 1 6 1 2 x 3 y 6 3x 2 y 1 6 15 x 6 y 21 5 x y 7 2 2
3x 2 y 1 15 x 6 y 21 Step 2: Multiply first equation by 5 then subtract second equation from first equation 5
3x 2 y 1 15 x 6 y 21
15 x 10 y 5 15 x 6 y 21
15 x 10 y 5 15 x 6 y 21 16 y 16 y 1
Step 3: Substitute the value of y in first equation 1 1 1 x 1 2 3 6 1 1 1 x 2 6 3 1 1 x 2 2 x 1 ANSWER: (x, y) = (1, 1)
32.
PROBLEM: 1 1 4 x 9 y 1 3 x y 4 SOLUTION: Step 1: Rewrite the first equation as 1 1 x y 1 4 9 9x 4 y 1 36 9 x 4 y 36 Step 2: Multiply second equation by 4 then add both the equations to eliminate y 9 x 4 y 36 9 x 4 y 36 3 4 4 x 4 y 3 x y 4
9 x 4 y 36 4x 4 y 3 13 x 39 x3 Step 3: Substitute the value of x in first equation 1 1 3 y 1 4 9 3 1 1 y 4 9 1 1 y 4 9 9 y 4 ANSWER: (x, y) = (3, -9/4)
33.
PROBLEM: 1 1 1 2 x 4 y 3 1 x 1 y 19 4 2 6
SOLUTION: Step 1: Multiply first and second equations by 12 1 1 1 12 6 x 3 y 4 2 x 4 y 3 12 3x 6 y 38 1 x 1 y 19 6 2 4
6x 3y 4 3x 6 y 38 Step 2: Multiply first equation by 2 then subtract second equation from first equation
6 x 3 y 4 3 x 6 y 38
2
12 x 6 y 8 3 x 6 y 38
12 x 6 y 8 3 x 6 y 38 15 x 30 x 2
Step 3: Substitute the value of x in first equation 1 1 1 2 y 2 4 3 1 4 y 4 3 16 y 3 ANSWER: (x, y) = (-2, -16/3)
34.
PROBLEM: 14 3 x 2 y 4 1 x 1 y 4 3 7 21 SOLUTION: Step 1: Rewrite the first and second equations as 14 x 2y 4 3 14 x 6 y 4 3 14 x 6 y 12 1 1 4 x y 3 7 21 7 x 3 y 4 21 21 7 x 3 y 4
Step 2: Multiply second equation by 2 then subtract second equation from first equation
14 x 6 y 12 7 x 3 y 4
2
14 x 6 y 12 14 x 6 y 8 0
12
ANSWER: So, there is no solution.
14 x 6 y 12 14 x 6 y 8
35.
PROBLEM: 0.025 x 0.1 y 0.5 0.11x 0.04 y 0.2 SOLUTION: Step 1: Multiply first equation by 1000 and multiply second equation by 100 to get equations in the form 1000
0.025 x 0.1 y 0.5 0.11x 0.04 y 0.2
100
25 x 100 y 500 11x 4 y 20
25 x 100 y 500 11x 4 y 20
Step 2: Multiply second equation by 25 then subtract second equation from first equation
25 x 100 y 500 11x 4 y 20
25
25 x 100 y 500 275 x 100 y 500
25 x 100 y 500 275 x 100 y 500 250 x x 4
1000
Step 3: Substitute the value of x in first equation 0.025 4 0.1 y 0.5 .1 .1 y .5 .1 y .6 y6
ANSWER: (x, y) = (-4, 6)
36.
PROBLEM: 1.3x 0.1y 0.35 0.5 x y 2.75
SOLUTION: Step 1: Multiply first equation by 100 and multiply second equation by 100 to rewrite equations in the form 100
1.3 x 0.1 y 0.35 0.5 x y 2.75 100
130 x 10 y 35 50 x 100 y 275
130 x 10 y 35 50 x 100 y 275
Step 2: Multiply first equation by 10 then subtract second equation from first equation
130 x 10 y 35 50 x 100 y 275
10
1300 x 100 y 350 50 x 100 y 275
1300 x 100 y 350 50 x 100 y 275 1250 x 625 1 2 x .5 x
Step 3: Substitute the value of x in first equation 1.3 .5 0.1 y 0.35 .65 0.1 y 0.35 0.1 y .30 y 3 ANSWER: (x, y) = ( .5, -3)
37.
PROBLEM: x y 5 0.02 x 0.03 y 0.125 SOLUTION: Step 1: Multiply second equation by 1000 to get equations in the form as
x y 5 1000 0.02 x 0.03 y 0.125 x y 5
x y 5 20 x 30 y 125
20 x 30 y 125 Step 2: Multiply first equation by 20 then subtract second equation from first equation x y 5 20 x 30 y 125
20
20 x 20 y 100 20 x 30 y 125
20 x 20 y 100 20 x 30 y 125 10 y 25 5 2 y 2.5 y
Step 3: Substitute the value of y in first equation x 2.5 5
x 2.5 ANSWER: (x, y) = (2.5, 2.5)
38.
PROBLEM: x y 30 0.05 x 0.1y 2.4
SOLUTION:
Step 1: Multiply second equation by 100 to rewrite equations in the form
x y 30 0.05 x 0.1y 2.4
100
x y 30 5 x 10 y 240
x y 30 5 x 10 y 240 Step 2: Multiply first equation by 5 then subtract second equation from first equation 5
x y 30 5 x 10 y 240
5 x 5 y 150 5 x 10 y 240
5 x 5 y 150 5 x 10 y 240 5y y 18
90
Step 3: Substitute the value of y in first equation
x 18 3 x 12 ANSWER: (x, y) = (12, 18)
39.
PROBLEM: The sum of two numbers is 14. The larger number is 1 less than two times the smaller. SOLUTION:
Let the two numbers be x and y Given: x + y =14 2x - y =1
Step 1: Multiply first equation by 2 then subtract second equation from first equation
x y 14 2 x y 1
2
2 x 2 y 28 2x y 1
2 x 2 y 28 2x y 1 3 y 27 y9 Step 2 Substitute the value of y in first equation x 9 14
x 5 ANSWER: (x, y) = (5, 9)
40.
PROBLEM: The sum of two numbers is 30. The larger is 2 more than three times the smaller. SOLUTION: Given: x + y = 30 - 3x + y = 2 Step 1: Subtract second equation from first equation x y 30 3 x y 2
4 x 28 x7 Step 2: Substitute the value of x in first equation 7 y 30 y 23
41.
ANSWER: (x, y) = (7, 23) PROBLEM: The difference of two numbers is 13 and the sum is 11.
SOLUTION: Given: x - y = 13 x + y =11 Step 1: Add both the equations to eliminate y
x y 13 x y 11 2 x 24 x 12 Step 2 Substitute the value of y in first equation 12 y 13 y = -1 ANSWER: (x, y) = (12, -1)
42.
PROBLEM: The difference of two numbers is 2 and the sum is -12. SOLUTION: Given: x + y = -12 x-y=2 Step 1: Subtract second equation from first equation x y 12 x y 2
2 x 10 x 5 Step 2: Substitute the value of x in first equation 5 y 12 y 7
ANSWER: (x, y) = (-5,-7) Part B: Mixed Exercises
43.
PROBLEM: y 2x 3 3x y 12 SOLUTION: Step 1: Rewrite the first equation in the form y 2x 3
2 x y 3 Step 2: Subtract second equation from first equation 2 x y 3 3x y 12
5 x 15 x3 Step 3: Substitute the value of x in first equation 2 3 y 3 y3
ANSWER: (x, y) = (3, 3)
44.
PROBLEM: x 3 y 5 1 y 3 x 5
SOLUTION: Step 1: Rewrite the second equation as 1 y x5 3 3 y x 15 x 3 y 15 Step 2: Add both the equations to eliminate x
x 3 y 5 x 3 y 15 6 y 10 y
5 3
Step 3: Substitute the value of y in first equation 5 x 3 5 3 x 10 ANSWER: (x, y) = (-10, 5/3)
45.
PROBLEM: x 1 y3 ANSWER: From above equations it is clear that the solution is (x, y) = (-1, 3)
46.
PROBLEM: 1 y 2 x 9 0 SOLUTION: Step 1: Rewrite the second equation as x9 0 x 9 ANSWER: (x, y) = (-9, 1/2)
47.
PROBLEM: yx x y 1 SOLUTION: Step 1: Rewrite the second equation as
x y 1 y 1 x Step 2: yx y 1 x ANSWER: Hence there is no solution..
48.
PROBLEM: y 5x y 10
SOLUTION: Step 1: Put the value of y of second equation in the first equation, we get
10 5 x 5 x 10 x 2
ANSWER: (x, y) = (-2, -10)
49.
PROBLEM: 3 y 2 x 24 3x 4 y 2
SOLUTION: Step 1: Rewrite the first equation in the form 3 y 2 x 24
2 x 3 y 24 Step 2: Multiply first equation by 3 and multiply second equation by 2 then add both the equations 2 x 3 y 24 3x 4 y 2
3
2
6 x 9 y 72 6x 8 y 4
6 x 9 y 72 6x 8 y 4 17 y 68 y 4
Step 2 Substitute the value of y in first equation 3 4 2 x 24 3 4 2 x 24 2 x 12 x6
ANSWER: (x, y) = (6, -4)
50.
PROBLEM: 3 y x 1 2 2 y 2 3 x
SOLUTION: Step 1: Rewrite the first and second equations in the form 3 y x 1 2 3 x y 1 2 3x 2 y 2
2 y 2 3 x 3x 2 y 2 Step 2: Subtract second equation from first equation 3x 2 y 2 3x 2 y 2
0 0 ANSWER: (x, y) = (x, -3/2 x +1)
51.
PROBLEM: 7 y 2 x 1 7 x 2 y 23 SOLUTION: Step 1: Rewrite the first and second equations in the form
7 y 2 x 1 2 x 7 y 1 7 x 2 y 23 7 x 2 y 23 2 x 7 y 1 7 x 2 y 23 Step 2: Multiply first equation by 3 and multiply second equation by 2 then add both the equations 2
2 x 7 y 1 7 x 2 y 23 7
4 x 14 y 2 49 x 14 y 161
4 x 14 y 2 49 x 14 y 161 53x 159 x3
Step 3: Substitute the value of x in first equation
7 y 2 3 1 7 y 7 y 1 ANSWER: (x, y) = (3, -1)
52.
PROBLEM: 5 x 9 y 14 0 3x 2 y 5 0 SOLUTION: Step 1: Rewrite the first and second equations in the form 5 x 9 y 14 0 5 x 9 y 14
3x 2 y 5 0 3x 2 y 5 Step 2: Multiply first equation by 3 and multiply second equation by 5 and then subtract second equation from first equation 5 x 9 y 14 3 x 2 y 5
3
5
15 x 27 y 42 15 x 10 y 25
15 x 27 y 42 15 x 10 y 25 17 y 17 y 1
Step 3: Substitute the value of y in first equation 5 x 9 1 14
5 x 14 9 5x 5 x 1 ANSWER: (x, y) = (1, 1)
53 .
PROBLEM: 5 y 16 x 10 y 5 x 10 16
Step 1: Rewrite the first and second equations in the form. 5 y x 10 16 5 x y 10 16 5 x 16 y 10 16 5 x 16 y 160
5 x 10 16 5 x 160 y 16 16 y 5 x 160 5 x 16 y 160 y
5 x 16 y 160 5 x 16 y 160 Step 2: Add both the equations 5 x 16 y 160 5 x 16 y 160
10 x 320 x 32 Step 3: Substitute the value of x in first equation 5 y 32 10 16 y0 ANSWER: (x, y) = (32, 0)
54.
PROBLEM: 6 y x 12 5 x6 SOLUTION:
Given x=6 Step 1: Substitute the value of x in first equation 6 y 6 12 5 36 60 y 5 24 y 5 ANSWER: (x, y) = (6, 24/5)
55.
PROBLEM: 2 x 3 y 0 3 2 x y 1 15 SOLUTION: Step 1: Rewrite the first and second equations in the form 2 x 3 y 0
2x 6 y 0 2x y 6 3 2 x y 1 15 2x y 1 5 2x y 6 Step 2: Subtract second equation from first equation 2x y 6
2x y 6 0 0 ANSWER: (x, y) = (x, -2x+6)
56.
PROBLEM: 3 2 x y 3 4 x 3 y 1 8 SOLUTION: Step 1: Rewrite the first and second equations in the form 3 2 x y 3
3 2 x 2 y 3 2 x 2 y 6 2x 2 y 6 x y 3 4 x 3 y 1 8 4x 3y 3 8 4 x 3 y 11 Step 2: Multiply first equation by 4 and then subtract second equation from first equation
x y 3 4 x 3 y 11
4
4 x 4 y 12 4 x 3 y 11
4 x 4 y 12 4 x 3 y 11 y 1 y 1 Step 3: Substitute the value of y in first equation x 1 3 x 1 3 x2
ANSWER: (x, y) = (2, -1)
57.
PROBLEM: 2 x 1 3 2 y 1 21 3 x 2 1 (3 y 2)
SOLUTION: Step 1: Rewrite the first and second equations in the form
2 x 1 3 2 y 1 21 2 x 2 6 y 3 21 2 x 6 y 26 x 3 y 13 3 x 2 1 (3 y 2) 3x 6 1 3 y 2 3x 3 y 3 x y 1 Step 2: Subtract second equation from first equation x 3 y 13 x y 1
4 y 12 y 3 Step 3: Substitute the value of y in first equation
2 x 2 6 3 3 21 2 x 6 2 2 x 8 x 4 ANSWER: (x, y) = (-4, 3)
58.
PROBLEM: x y 2 3 7 x y 8 3 2
SOLUTION: Step 1: Rewrite the first and second equations in the form 100
0.15 x 1.25 y 0.4 0.03x 0.25 y 0.08 100
15 x 125 y 40 3 x 25 y 80
15 x 27 y 42 15 x 10 y 25 17 y 17 y 1 0.15 x 1.25 y 0.4 0.03x 0.25 y 0.08 Step 2: Multiply first equation by 2 and multiply second equation by 3 and then subtract second equation from first equation 3 x 2 y 42 2 x 3 y 48
2
3
6 x 4 y 84 6 x 9 y 144
6 x 4 y 84 6 x 9 y 144 5 y 60 y 12
Step 3: Substitute the value of y in first equation
x 12 7 2 3 x 1 3 x 4 7 x 1 3 2 x2 x 3 2 x 6 ANSWER: (x, y) = (-6, 12)
59 .
PROBLEM: x y 3 4 2 4 x y 1 3 6 6 SOLUTION: Step 1: Rewrite the first and second equations in the form
x y 3 4 2 4 x 2y 3 4 4 x 2y 3 x y 1 3 6 6 2x y 1 6 6 2x y 1
Step 2: Multiply second equation by 2 ans then add both the equations to eliminate y
x 2y 3 2 x y 1
2
x 2y 3 4 x 2 y 2
x 2y 3 4x 2 y 2 5x
5
x 1 Step 3: Substitute the value of x in first equation
1 y 3 4 1
2 4 3 y 4 4 2 1 y 2 2 y 1 ANSWER: (x, y) = (1,-1 )
60.
PROBLEM: 2 1 3 x 3 y 3 1 x 1 y 8 3 2 3
SOLUTION: Step 1: Rewrite the first and second equations in the form
1 2 x y 3 3 3 x 2y 3 3 x 2y 9 1 1 8 x y 3 2 3 2x 3y 8 6 3 2 x 3 y 16
Step 2: Multiply first equation by 2 and then subtract second equation from first equation
x 2y 9 2 x 3 y 16
2
2 x 4 y 18 2 x 3 y 16
2 x 4 y 18 2 x 3 y 16 y 2 y 2 Step 3: Substitute the value of y in first equation. 1 2 x 2 3 3 3 1 4 x 3 3 3 1 5 x 3 3 x5 ANSWER: (x, y) = (5, -2)
61.
PROBLEM: 1 4 1 10 x 2 y 5 1 x 1 y 2 7 3 21 SOLUTION: Step 1: Rewrite the first and second equations in the form
1 1 4 x y 10 2 5 1 1 4 x 1 10 2 5 1 4 1 x 10 5 2 1 85 x 10 10 x 3
Step 2: Multiply first equation by 3 and then add both the equations to eliminate x
x 5 y 8 3x 7 y 2
3
3x 15 y 24 3x 7 y 2
3 x 15 y 24 3x 7 y 2 22 y 22 y 1 Step 3: Substitute the value of y in first equation
1 1 4 x y 10 2 5 1 1 4 x 1 10 2 5 1 4 1 x 10 5 2 1 85 x 10 10 x 3
ANSWER: (x, y) = (-3,1 )
62.
PROBLEM: 5 1 y 3 x 2 1 x 1 y 1 3 5 10
SOLUTION: Step 1: Rewrite the first and second equations in the form 5 1 y x 3 2 10 x 3 y 6 6 y 10 x 3
10 x 6 y 3 1 1 1 x y 3 5 10 5x 3 y 1 15 10 3 5x 3 y 2 10 x 6 y 3
Step 2: Subtract second equation from first equation 10 x 6 y 3
10 x 6 y 3 0 0 ANSWER: (x, y) = (x, -5/3 x+1/2)
63.
PROBLEM: 2 1 7 x y 3 1 x 1 y 1 14 2 3
SOLUTION: Step 1: Rewrite the first and second equations in the form 1 2 x y 7 3 2 x 7 y 7 3 3 x 21 y 14 1 1 1 x y 14 2 3 x 7 y 1 14 3 3 x 21 y 14
Step 2: Subtract second equation from first equation 3x 21 y 14 3x 21 y 14
0 28 ANSWER:
Hence there is no solution..
64.
PROBLEM: 1 1 1 15 x 12 y 3 3 x 3 y 3 10 8 2
SOLUTION: Step 1: Rewrite the first and second equations in the form 1 1 1 x y 15 12 3 4x 5 y 1 60 3 4 x 5 y 20 3 3 3 x y 10 8 2 3 12 x 15 y 40 2 12 x 15 y 60
Step 2: Multiply first equation by 2 then add both the equations to eliminate x 4 x 5 y 20 12 x 15 y 60
3
12 x 15 y 60 12 x 15 y 60 0
0
ANSWER: (x, y) = (x, 4/5 x-4)
65.
PROBLEM: x y 4200 0.03x 0.0525 y 193.5 SOLUTION:
12 x 15 y 60 12 x 15 y 60
Step 1:Multiply second equation by 10000
x y 4200 0.03 x 0.0525 y 193.5
10000
x y 4200 300 x 525 y 1935000
Step 2: Multiply first equation by 300 and then add both the equations to eliminate x 300
x y 4200 300 x 525 y 1935000
300 x 300 y 1260000 300 x 525 y 1935000
300 x 300 y 1260000 300 x 525 y 1935000 225 y
675000
y 3000
Step 3: Substitute the value of y in first equation
x 3000 4200 x 1200 ANSWER: (x, y) = (1200,3000)
66.
PROBLEM: x y 350 0.2 x 0.1 y 52.5 SOLUTION: Step 1: Multiply second equation by 10
x y 350 0.2 x 0.1y 52.5
10
x y 350 2 x y 525
Step 2: Multiply first equation by 2 and then subtract second equation from first equation
x y 350 2 x y 525
2
2 x 2 y 700 2 x y 525
2 x 2 y 700 2 x y 525 y
175
Step 3: Substitute the value of y in first equation
x 175 350 x 175 ANSWER: (x, y) = (175, 175)
67.
PROBLEM: 0.2 x 0.05 y 0.43 0.3x 0.1y 0.3
SOLUTION: Step 1:Multiply first equation by 100 and multiply second equation by 10
0.2 x 0.05 y 0.43 0.3x 0.1y 0.3
100
10
20 x 5 y 43 3x y 3
Step 2: Multiply first equation by 300 and then add both the equations to eliminate x 20 x 5 y 43 3 x y 3 5
20 x 5 y 43 15 x 5 y 15
20 x 5 y 43 15 x 5 y 15 35 x
28
4 5 x .8 x
Step 3: Substitute the value of x in first equation
0.2 .8 0.05 y 0.43 .16 .05 y .43 .05 y .27 y 5.4
ANSWER: (x, y) = (0.8, -5.4)
68.
PROBLEM: 0.1x 0.3 y 0.3 0.05 x 0.5 y 0.63 SOLUTION: Step 1: Multiply first equation by 10 and multiply second equation by 100
0.1x 0.3 y 0.3 0.05 x 0.5 y 0.63
10
100
x 3y 3 5 x 50 y 63
Step 2: Multiply first equation by 5 and then subtract second equation from first equation 5
x 3y 3 5 x 50 y 63 5 x 15 y 15 5 x 50 y 63
5 x 15 y 15 5 x 50 y 63
65 y 78 6 y 5 y 1.2 Step 3: Substitute the value of y in first equation 0.1x 0.3 1.2 0.3
.1x .3 .36 .1x .06 x .6 ANSWER: (x, y) = (-0.6, 1.2)
69.
PROBLEM: 0.15 x 0.25 y 0.3 0.75 x 1.25 y 4 SOLUTION: Step 1:Multiply first equation by 100 and multiply second equation by 100 100
0.15 x 0.25 y 0.3 0.75 x 1.25 y 4
15 x 25 y 30 75 x 125 y 40
100
5
15 x 25 y 30 75 x 125 y 40
75 x 125 y 150 75 x 125 y 40
75 x 125 y 150 75 x 125 y 40 0 190 x7 0.15 x 0.25 y 0.3 0.75 x 1.25 y 4 Step 2: Multiply first equation by 5 and then add both the equations 15 x 25 y 30 75 x 125 y 40
5
75 x 125 y 150 75 x 125 y 40 0 190
ANSWER: So, there is no solution.
75 x 125 y 150 75 x 125 y 40
70.
PROBLEM: 0.15 x 1.25 y 0.4 0.03x 0.25 y 0.08 SOLUTION: Step 1: Multiply first equation by 10 and multiply second equation by 100 100
0.15 x 1.25 y 0.4 0.03x 0.25 y 0.08 100
15 x 125 y 40 3x 25 y 8
Step 2: Multiply second equation by 5 and then subtract second equation from first equation 15 x 125 y 40 3 x 25 y 8
5
15 x 125 y 40 15 x 125 y 40
15 x 125 y 40 15 x 125 y 40 0 0
ANSWER: (x, y) = (x, .12x+.32) Part C: Discussion Board Topics
71.
PROBLEM:
How do we choose the best method for solving a linear system? ANSWER:
The best method we choose for solving a linear system depends entirely on the equations that that we find in the questions. Suppose we have the equation like given below----2 x 3 y 9 5 x 8 y 16 In this case, it is easy to use substitution method as in the first equation one variable is there with the coefficient 1. So, it becomes easy to put rest part of the equation in that variable (variable which is with coefficient 1)
Here, first equation can be written as 10 x y 20 y 20 10 x Then this variable can be put in the second equation as 7 x 5 20 10 x 14 After which we can find the value of x 7 x 5 20 10 x 14 7 x 100 50 x 14 43x 86 x2 Keeping this value in the equation y 20 10 x We get the value of y y 20 10 2 y 20 20 y0
Similarly when we find equations where there is no variable is there with coefficient 1, and then we find elimination method very useful there.
2 x 3 y 9 6 x 8 y 7 Here there are no equations where there any variable with coefficient 1, so we will use the elimination method for calculating the value of variables. We will try to eliminate one variable out of the variables given. Here we are eliminating x by multiplying first equation with 3. 2 x 3 y 9 6 x 8 y 7
3
6 x 9 y 27 6 x 8 y 7 17 y 34 y2
6 x 9 y 27 6 x 8 y 7
From the value of y when kept in the first equation, we get the value of x 2 x 3 2 9 2 x 3 3 x 2 72.
PROBLEM:
What does it mean for a system to be dependent? How can we tell if a given system is dependent? SOLUTION: We say that the system is dependent when equations are equal.
We can determine if a given system is dependent or independent based on the equality of the given equations. 6 x 2 y 14 6 x 2 y 14 0 0 In given example, we find that both equations are equal. ANSWER: Hence we can conclude that the lines coincide in this case, and we need y in terms
of x to present the solution in the form of (x, mx+b)
4.4 Applications of Linear Systems Part A: Applications Involving Numbers Set up a linear system and solve.
1.
PROBLEM: The sum of two integers is 54 and the difference is 10. Find the integers. SOLUTION:
Let x be one of the integers Let y be other integer. Given : x + y = 54
x - y = 10 Step 1: Add both the equations
x y 54 x y 10 2 x 64 x 32 Step 2: Keeping the value of x in first equation 32 y 54
y 54 - 32 y 22 ANSWER: The integers are 32 and 22.
2.
PROBLEM: The sum of two integers is 50 and the difference is 24. Find the integers. SOLUTION:
Let x be one of the integers Let y be other integer. Given: x + y = 50 x - y = 24 Step 1: Add both the equations
x y 50 x y 24 74 x 37 Step 2: Keeping the value of x in first equation 37 y 50 2x
y = 50 - 37 y = 13 ANSWER: The integers are 37 and 13.
3.
PROBLEM: The sum of two positive integers is 32. When the smaller integer is subtracted from twice the larger the result is 40. Find the two integers. SOLUTION:
Let x be smaller integers Let y be larger integer. Given : x + y = 32 2y – x = 40 Step 1: Rewrite the second equation in the form 2y x 40
x 2y 40 Step 2: Add both the equations x y 32 x 2y 40 3 y 72 y 24 Step 3: Keeping the value of y in the first equation x 24 32
x 8 ANSWER: The integers are 8 and 24
4.
PROBLEM: The sum of two positive integers is 48. When twice the smaller integer is subtracted from the larger the result is 12. Find the two integers. SOLUTION:
Let x be smaller integer Let y be larger integer. Given: x + y = 48 y - 2x= 12
Step 1: Rewrite the second equation in the form y 2 x 12
2 x y 12 Step 2: Subtract second equation from the first equation x y 48 2 x y 12
3x 36 x 12 Step 3: Keeping the value of y in first equation 12 y 48
y 36 ANSWER: The integers are 12 and 36
5.
PROBLEM: The sum of two integers is 74. The larger is 26 more than twice the smaller. Find the two integers. SOLUTION:
Let x be smaller integers Let y be larger integer. Given : x + y = 74 y - 2x= 26 Step 1: Rewrite the second equation in the form y 2x 26
2x y 26 Step 2: Subtract second equation from the first equation x y 74 2 x y 26
3x 48 x 16 Step 3: Keeping the value of x in first equation
16 y 74 y 58 ANSWER: The integers are 16 and 58
6.
PROBLEM: The sum of two integers is 45. The larger is 3 less than three times the smaller. Find the two integers. SOLUTION:
Let x be smaller integer Let y be larger integer. Given: x + y = 45 3x- y = 3 Step 1: Add both the equations
x y 45 3x y 3 4 x 48 x 12 Step 2: Keeping the value of x in first equation 12 y 45
y 45 - 12 y 33 ANSWER: The integers are 12 and 33
7.
PROBLEM: The sum of two numbers is zero. When 4 times the smaller number is added to 8 times the larger the result is 1. Find the two numbers. SOLUTION:
Let x be smaller number. Let y be larger number.
Given : x+y=0 4x + 8y= 1 Step 1: Multiply the first equation by 4 and then subtract the second equation from first equation 4
x y 0 4 x 8y 1
4x 4 y 0 4 x 8y 1
4x 4 y 0 4 x 8y 1 4 y 1 y
1 4
Step 2: Keeping the value of y in first equation 1 x 0 4 1 x 4 ANSWER:
The integers are -
8.
1 1 and 4 4
PROBLEM: The sum of a larger number and 4 times a smaller number is 5. When 8 times the smaller is subtracted from twice the larger the result is -2. Find the numbers. SOLUTION:
Let x be smaller number. Let y be larger number Given : 4x + y = 5 2y - 8x= -2 Step 1: Rewrite the second equation in the form
2y 8 x -2 8 x 2y -2 Step 2: Multiply first equation by 2 and then add both the equations 2
4x y 5 8 x 2y -2
8 x 2 y 10 8 x 2y -2
8 x 2 y 10 8 x 2y -2 4y 8 y2 Step 3: Keeping the value of y in first equation 4 x 2 5
4 x 5-2 3 x= 4 ANSWER:
The numbers are 9.
3 and 2 4
PROBLEM: The sum of 12 times the larger number and 11 times the smaller is -36. The difference of 12 times the larger and 7 times the smaller is 36. Find the numbers. SOLUTION:
Let x be smaller number. Let y be larger number. Given : 11x + 12y = -36 12y – 7x = 36 Step 1: Rewrite the second equation 12y – 7x 36
–7x 12y 36 Step 2: Subtract second equation from first equation
11x 12y 36 – 7x 12y 36 18 x 72 x 4 Step 3: Keeping the value of x in first equation 11 4 12y 36
12 y 8 2 y 3 ANSWER:
The integers are - 4 and
10.
2 3
PROBLEM: The sum of 4 times the larger number and 3 times the smaller is 7. The difference of 8 times the larger and 6 times the smaller is 10. Find the numbers. SOLUTION:
Let x be smaller number Let y be larger number. Given: 3x 4y 7 8y 6x 10 Step 1: Rewrite the second equation in the form 8y - 6x 10
- 6x 8y 10 Step 2: Multiply the first equation by 2 and then add both the equations
3x 4y 7 -6x 8y 10
2
6 x 8 y 14 6x 8y 10
6 x 8 y 14 -6x 8y 10
16 y 24 3 y 2 Step 2: Keeping the value of y in first equation 3 3x 4 7 2 3x 6 7
3x=1 1 x= 3 ANSWER:
The numbers are
1 3 and 3 2
Part B: Interest and Money Problems Set up a linear system and solve.
11.
PROBLEM: A $7,000 principal was invested in two accounts, one earning 3% and another earning 7% interest. If the total interest for the year was $262, then how much was invested in each account? SOLUTION:
Let x represents the amount invested at 3 %= .03 Let y represents the amount invested at 7% = .07 Given : x y 7000 .03x .07 y 262 Step 1: Multiply the second equation by 100
x y 7000 .03x .07 y 262
100
x y 7000 3x 7 y 26200
Step 2: Multiply first equation by 3 and then subtract second equation from the first equation
x y 7000 3x 7 y 26200
3
3x 3 y 21000 3x 7 y 26200
3x 3 y 21000 3x 7 y 26200 4 y 5200 y 1300 Step 3: Keeping the value of y in first equation x 1300 7000
x 5700 ANSWER: The two amounts are $5700and $1300
12.
PROBLEM: Mary has her total savings of $12,500 in two different CD accounts. One CD earns 4.4% and another earns 3.2% interest. If her total interest for the year was $463, then how much does she have in each CD account? SOLUTION:
Let x represents the amount invested at 4.4 % = 0.044 Let y represents the amount invested at 3.2% = 0.032 Given: x y 12500
.044 x .032 y 463 Step 1: Multiply the second equation by 1000
x y 12500 0.044 x 0.032 y 463
1000
x y 12500 44 x 32 y 463000
Step 2: Multiply the second equation by 44 and then add both the equations
44
x y 12500 44 x 32 y 463000
44 x 44 y 550000 44 x 32 y 463000
44 x 44 y 550000 44 x 32 y 463000 12 y 87000 y 7250 Step 3: Keeping the value of y in first equation x 7250 12500
x 5250 ANSWER: The two amounts are $5250 and $7250
13.
PROBLEM: Sally’s $1800 savings is in two accounts. One account earns 6% annual interest and the another earns 3% . Her total interest for the year was $93. How much does she have in each account? SOLUTION:
Let x represents the amount invested at 6 %= .06 Let y represents the amount invested at 3% = .03 Given: x y 1800 .06 x .03 y 93 Step 1: Multiply the second equation by 100
x y 1800 .06 x .03 y 93
100
x y 1800 6 x 3 y 9300
Step 2: Multiply the first equation by 6 and then add both the equations
x y 1800 .06 x .03 y 93
6
6 x 6 y 10800 6 x 3 y 9300
6 x 6 y 10800 6 x 3 y 9300 3 y 1500 y 500 Step 3: Keeping the value of y in first equation x 500 1800
x 1300 ANSWER: The two amounts are $1300 and $500
14.
PROBLEM: Joe has two savings accounts totaling $4,500. One account earns 3 34 % annual interest and the other earns 2 85 % . If his total interest for the year was $141.75, then how much was in each account? SOLUTION:
15 % .0375 4 21 Let y represents the amount invested at 2 85 % % .02625 8
Let x represents the amount invested at 3 34 %
Given: x y 4500 .0375 x .02625 y 141.75 Step 1: Multiply the second equation by 10000
x y 4500 .0375 x .02625 y 141.75
100000
x y 4500 3750 x 2625 y 14175000
Step 2: Multiply the first equation by 375 and subtract the second equation from the first equation
x y 4500 3750 x 2625 y 14175000
3750
3750 x 3750 y 16875000 3750 x 2625 y 14175000
3750 x 3750 y 16875000 3750 x 2625 y 14175000 1125 y 2700000 y 2400
Step 3: Keeping the value of y in first equation x 2400 4500
x 2100 ANSWER: The two amounts are $2100 and $2400
15.
PROBLEM: Millicent has $10,000 invested in two accounts. For the year she earned $535 more in interest from her 7% Mutual Fund account than she did from her 4% CD. How much does she have in each account? SOLUTION:
Let x represents the amount invested at 7%= .07 Let y represents the amount invested at 4% = .04 Given: x y 10000 .07 x .04 y 535 Step 1: Multiply the second equation by 100
x y 10000 .07 x .04 y 535
100
x y 10000 7 x 4 y 53500
Step 2: Multiply the first equation by 4 and then add both the equations
x y 10000 .07 x .04 y 535
4
4 x 4 y 40000 7 x 4 y 53500
4 x 4 y 40000 7 x 4 y 53500 11x 93500 x 8500 Step 3: Keeping the value of y in first equation 8500 y 10000 y 1500
ANSWER: The two amounts are $8500 and $1500
16.
PROBLEM: A small business has $85,000 invested in two accounts. If the account earning 3% annual interest earned $825 more in interest than the account earning 4.5% annual interest, then how much was invested in each account? SOLUTION:
Let x represents the amount invested at 3% .03 Let y represents the amount invested at 4.5% .045 Given: x y 85000 .03 x .045 y 825 Step 1: Multiply the second equation by 1000
x y 85000 .03x .045 y 825
1000
x y 85000 30 x 45 y 825000
Step 2: Multiply the first equation by 45 and then add both the equations
x y 85000 30 x 45 y 825000
45
45 x 45 y 3825000 30 x 45 y 825000
45 x 45 y 3825000 30 x 45 y 825000 75 x 4650000 62000 x Step 3: Keeping the value of x in first equation 62000 y 85000 y 23000 ANSWER: The two amounts are $6200 and $2300
17.
PROBLEM: Jerry earned a total of $284 in simple interest from two separate accounts. In an account earning 6% interest, Jerry invested $1,000 more than twice the amount he invested in an account earning 4%. How much did he invest in each account? SOLUTION:
Let x represents the amount invested at 4%= .04 Let y represents the amount invested at 6% = .06 Given : y =2x +1000 .04 x .06 y 284 Step 1: Substitute the value of y in second equation .04 x .06 y 284
.04 x .06 2 x 1000 284 .04 x .12 x 60 284 .16 x 224 x 1400 Step 2: Keeping the value of x in first equation
y 2 1400 1000 y 3800 ANSWER: The two amounts are $1400 and $3800
18.
PROBLEM: James earned a total of $68.25 in simple interest from two separate accounts. In an account earning 2.6% interest, James invested one-half as much as he did in the other account that earned 5.2%. How much did he invest in each account? SOLUTION:
Let x represents the amount invested at 5.2% .052 Then y represents the other amount invested at 2.6% .026 Given: y=x/2 0.052 x +0.026 (x/2) = 68.25 Step 1: Multiply equation by 2
.052x .026 x / 2 68.25
2
.104 x .026x 136.5
Step 2: Multiply equation by 1000
.104 x .026x 136.5
1000
104 x 26x 136500
Step 3: 104 x 26x 136500 130 x 136500 x 1050
So, the second amount is x/2 =1050/2=525 ANSWER: The two amounts are $1050 and $525
18.
PROBLEM: James earned a total of $68.25 in simple interest from two separate accounts. In an account earning 2.6% interest, James invested one-half as much as he did in the other account that earned 5.2%. How much did he invest in each account? SOLUTION:
Let x represents the amount invested at 5.2% .052 Then y represents the other amount invested at 2.6% .026 Given:
y=x/2 0.052 x +0.026 (x/2) = 68.25 Step 1: Multiply equation by 2
.052x .026 x / 2 68.25
2
.104 x .026x 136.5
Step 2: Multiply equation by 1000
.104 x .026x 136.5
1000
104 x 26x 136500
Step 3: 104 x 26x 136500 130 x 136500 x 1050
So, the second amount is x/2 =1050/2=525 ANSWER: The two amounts are $1050 and $525
19.
PROBLEM: A cash register contains $10 bills and $20 bills with total value $340. If there are 23 bills total, then how many of each does the register contain. SOLUTION:
Let x represent the number of $10 bills. Let y represent the number of $20 bills. Given: x + y =23 10x + 20y = 340 Step 1: Multiply the first equation by 10 and then subtract the second equation from the first equation
10
x y 23 10 x 20 y 340
10 x 10 y 230 10 x 20 y 340
10 x 10 y 230 10 x 20 y 340 10 y 110 y 11 Step 2: Keeping the value of x in first equation x 11 23
x 12 ANSWER: The number of $10 bills and $20 bills are 12 and 11.
20.
PROBLEM: John was able to purchase a pizza for $10.80 with quarters and dimes. If he used 60 coins to buy the pizza, then how many of each did he have? SOLUTION:
Let x represent the number of quarters. Let y represent the number of dimes. Given: Total number of coins is x + y = 60 A dollar has 4 quarters; hence the number of quarters used is 0.25x. A dollar has 10 dimes; hence the number of dimes used is 0.1y. Total monetary value: 0.25x + 0.1y = $10.80 Step 1: Multiply the second equation by 100
x y 60 x y 60 0.25x 0.1y 10.80 100 25 x 10 y 1080 Step 2: Multiply the first equation by 25 and then subtract second equation from the first equation
25
25 x 25 y 1500 x y 60 25 x 10 y 1080 25 x 10 y 1080 25 x 25 y 1500 25 x 10 y 1080 15 y 420 y 28 Step 3: Keeping the value of y in first equation x 28 60 x 32 ANSWER: John uses 32 quarters and 28 dimes to purchase the pizza.
21.
PROBLEM: Dennis mowed his neighbor’s lawn for a jar of dimes and nickels. Upon completing the job, he counted the coins and found that there were 4 less than twice as many dimes as there were nickels with total value $6.60. How many of each coin did he have? SOLUTION:
Let x represent the number of dimes. Let y represent the number of nickels. Given: Total number of coins: y = 2x – 4 2x – y=4 A dollar has 10 dimes; hence the number of dimes in the jar is 0.1x. A dollar has 20 nickels; hence the numbers of nickels is 0.05y. Total monetary value: 0.1x + 0.05y = $6.60 Step 1: Multiply the second equation by 100
2x – y 4 0.1x 0.05y 6.60 100
2x – y 4 10 x 5 y 660
Step 2: Multiply the first equation by 25 and then subtract the second equation from first equation
2x – y 4 10 x 5 y 660
5
10x – 5 y 20 10 x 5 y 660
10x – 5 y 20 10 x 5 y 660 10 y 640 y 64 Step 3: Keeping the value of y in first equation 2x – 64 4 2 x 68 x 34
ANSWER: The jar contains 34 dimes and 64 nickels.
22.
PROBLEM: Two families bought tickets for the big football game. One family ordered 2 adult tickets and 3 children’s tickets for a total of $26.00. Another family ordered 3 adult tickets and 4 children’s tickets for a total of $37.00. How much was each adult ticket? SOLUTION:
Let x represent price of adult ticket. Let y represent price of children ticket. Given: 2x + 3y = 26 3x + 4y = 37 Step 1: Multiply the first equation by 50 and then subtract the second equation from the first equation
3
2 x 3 y 26 3x 4y 37 2
6 x 9 y 78 6 x 8 y 74
6 x 9 y 78 6 x 8 y 74 y 4 y4 2 x 3 4 26 2 x 14 x7
Step 2: Keeping the value of y in first equation 2 x 3 4 26 2 x 14 x7
ANSWER: So, price of adult ticket is $7.
23. PROBLEM: Two friends found shirts and shorts on sale at the flea market. One bought 5 shirts and 3 shorts for a total of $51.00. The other bought 3 shirts and 7 shorts for a total of $80.00. How much was each shirt and each pair of shorts? SOLUTION: Let x represent the price of shirts. Let y represent the price of shorts.
Given : 5x + 3y = $ 51.00 3x + 7y = $ 80.00 Step 1: Multiply first equation by 3 and multiply second equation by 5 then subtract second equation from the first equation
5x 3y 51 3x 7y 80
3
5
15 x 9 y 153 15 x 35 y 400
15 x 9 y 153 15 x 35 y 400 26 y 247 y 9.50 Step 2: Keeping the value of y in first equation 5x 3 9.50 51.00 5 x 22.50 x 4.50
ANSWER: The prices of shirts and shorts are $9.50 and $4.50.
24.
PROBLEM: On Monday Joe bought 10 cups of coffee and 5 doughnuts for his office at the cost of $16.50. It turns out that the doughnuts were more popular than the coffee. Therefore, on Tuesday he bought 5 cups of coffee and 10 doughnuts for a total of $14.25. How much was each cup of coffee? SOLUTION:
Let x represent price of cup of coffee. Let y represent price of doughnut. Given: 10x + 5y = 16.50 5x + 10y = 14.25 Step 1: Multiply the first equation by 50 and then subtract the second equation from the first equation
25
25 x 25 y 1500 x y 60 25 x 10 y 1080 25 x 10 y 1080 25 x 25 y 1500 25 x 10 y 1080 15 y 420 y 28 0.25x 0.1y 10.80 Step 2: Keeping the value of y in first equation 10 x 5 .8 16.50 10 x 12.50 x 1.25
ANSWER: So, price of cup of coffee is $ 1.25. Part C: Mixture Problems Set up a linear system and solve.
25.
PROBLEM: A 15% acid solution is to be mixed with a 25% acid solution to produce 12 gallons of a 20% acid solution. How much of each is needed? SOLUTION:
Let x represent the amount of 15% acid solution needed. Let y represent the amount of 25% acid solution needed. Given: x + y = 12 (0.15) x + (.25) y = (.20)12 Step 1: Rewrite the second equation in the form 0.15 x .25 y .20 12
0.15 x .25 y 2.4 Step 2: Multiply the second equation by 100
x y 12 .15 x .25 y 2.4
100
x y 12 15 x 25 y 240
Step 3: Multiply the first equation by 15 then subtract the second equation from the first equation
x y 12 15 x 25 y 240
15
15 x 15 y 180 15 x 25 y 240
15 x 15 y 180 15 x 25 y 240 10 y 60 y6 Step 4: Keeping the value of y in first equation x 6 12
x6 ANSWER: The amount of 15% acid and amount of 25% acid solution required is 6 gallons each.
26.
PROBLEM: One alcohol solution contains 12% alcohol and another contains 26% alcohol. How much of each should be mixed together to obtain 5 gallons of a 14.8% alcohol solution? SOLUTION:
Let x represent the amount of 12% acid solution needed. Let y represent the amount of 26% acid solution needed. Given: x+y=5 (0.12)x + (.26)y = (.148)5 Step 1: Rewrite the second equation in the form 0.12 x .26 y .148 5
0.12x .26y .74
Step 2: Multiply second equation by 100
xy 5 0.12x .26y .74
100
xy 5 12 x 26 y 74
Step 3: Multiply the first equation by 15 then subtract second equation from the first equation 12
xy 5 12 x 26 y 74
12 x 12 y 60 12 x 26 y 74
12 x 12 y 60 12 x 26 y 74 14 y 14 y 1 Step 4: Keeping the value of y in first equation x 1 5
x4 ANSWER: The amount of 12% acid and amount of 26% acid solution required are 4 gallons and 1 gallon.
27.
PROBLEM: A nurse wishes to obtain 40 oz of a 1.2% saline solution. How much of a 1% saline solution must she mix with a 2% saline solution to achieve the desired result? SOLUTION:
Let x represent the amount of 1 % saline solution needed. Let y represent the amount of 2 % saline solution needed. Given: x + y = 40 (0.01)x + (.02)y = (.012)40 Step 1: Rewrite the second equation in the form 0.01 x .02 y .012 40
.01x .02 y .48
Step 2: Multiply the second equation by 1000
x y 40 .01x .02 y .48
100
x y 40 x 2 y 48
Step 3: Subtract the second equation from the first equation x y 40 x 2 y 48 y 8 y 8 Step 4: Keeping the value of y in first equation x 8 40
x 32 ANSWER: Hence the amount of 1 % saline solution and 2 % saline solution needed are 32 oz and 8 oz.
28.
PROBLEM: A customer ordered 20 lbs of fertilizer that contains 15% nitrogen. To fill the customer’s order, how much of the stock 30% nitrogen fertilizer must be mixed with the 10% nitrogen fertilizer? SOLUTION:
Let x represent 30% nitrogen fertilizer needed. Let y represent 10% nitrogen fertilizer needed. Given: x + y = 20 (0.30)x + (.10)y = (.15)20 Step 1: Rewrite the second equation in the form 0.30 x .10 y .15 20
0.30x .10y 3 Step 2: Multiply second equation by 100
x y 20 0.30x .10y 3
100
x y 20 30 x 10 y 300
Step 3: Multiply the first equation by 15 then subtract the second equation from the first equation 30
x y 20 30 x 10 y 300
30 x 30 y 600 30 x 10 y 300
30 x 30 y 600 30 x 10 y 300 20 y 300 y 15 Step 4: Keeping the value of y in first equation x 15 20
x5 ANSWER: The amount of 30% nitrogen fertilizer and amount of 10% nitrogen fertilizer required are 5 lbs and 15 lbs.
29.
PROBLEM: A customer ordered 2 lbs of a mixed peanut product containing 15% cashews. The inventory consists of only two mixes containing 10% and 30% cashews. How much of each type must be mixed to fill the order? SOLUTION:
Let x represent the amount of 10 % cashew. Let y represent the amount of 30 % cashew. Given: x+y=2 (0.1)x + (.3)y = (.15)2 Step 1: Rewrite the second equation in the form 0.10 x .30 y .15 2
0.10 x 0.30y .30 Step 2: Multiply second equation by 1000
x y 2 0.10 x 0.30y .30
100
x y 2 x 3y 3
Step 3: Multiply the first equation by 12 then subtract the second equation from the first equation
x y 2 x 3y 3 2 y 1 y 0.5 Step 4: Keeping the value of y in first equation
x .5 2 x .15 ANSWER: The two amounts required are 1.5 lb and .5 lb
30.
PROBLEM: How many pounds of pure peanuts must be combined with a 20% peanut mix to produce 10 lbs of a 32% peanut mix? SOLUTION:
Let x represent pure peanuts needed. Let y represent peanuts mix needed. Given: x + y = 10 (0.30)x + (.10) y = (.15)20 Step 1: Rewrite the second equation in the form x .20 y .32 10
x .20y 3.2 Step 2: Multiply the second equation by 100
x y 10 x .20y 3.2
x y 10 100 x 20 y 320
100
Step 3: Multiply the first equation by 15 then subtract second equation from the first equation x y 10 100 x 20 y 320
20
20 x 20 y 200 100 x 20 y 320
20 x 20 y 200 100 x 20 y 320 80 x 120 x 1.5
ANSWER: The amount pure peanuts needed is 1.5 lbs.
31.
PROBLEM: How much cleaning fluid, with 20% alcohol content, must be mixed with water to obtain a 24 oz mixture with 10% alcohol content? SOLUTION:
Let x represent cleaning fluid, with 20% alcohol content. Let y represent water. Given: x + y = 24 (0.2)x + (0)y = (.10)24 Step 1: Rewrite the second equation 0.2 x 0 y .10 24 .2 x 2.40 x 12
ANSWER: The amount of cleaning fluid needed is 12 oz.
32.
PROBLEM: A chemist wishes to create a 32 oz solution with 12% acid content. He will use two types of stock solutions, one with 30% acid content and another with 10% acid content. How much of each will he need? SOLUTION:
Let x represent the amount of 30% acid solution needed. Let y represent the amount of 10% acid solution needed. Given: x + y = 32 (0.30)x + (.10) y = (.12)32 Step 1: Rewrite the second equation in the form 0.30 x .10 y .12 32
0.30x .10y 3.84 Step 2: Multiply second equation by 100
x y 32 0.30x .10y 3.84 100
x y 32 30 x 10 y 384
Step 3: Multiply first equation by 15 then subtract second equation from the first equation 30
x y 32 30 x 10 y 384
30 x 30 y 960 30 x 10 y 384
30 x 30 y 960 30 x 10 y 384 20 y 576 y 28.8 Step 4: Keeping the value of y in first equation x 28.8 32
x=3.2 ANSWER: The amount of 30% acid and 10% acid solution required are 3.2 oz and 28.8 oz.
33.
PROBLEM: A concentrated cleaning solution that contains 50% ammonia is to be mixed with another solution containing 10% ammonia. How much of each should be mixed to obtain 8 oz of a 32% ammonia cleaning formula. SOLUTION:
Let x represent the amount of 50 % ammonia needed. Let y represent the amount of 10 % acid ammonia needed. Given : x+y=8 (0.5)x + (.1)y = (.32)8 Step 1: Rewrite the second equation in the form 0.5 x .1 y .32 8
.5 x .1 y 2.56 Step 2: Multiply second equation by 100
xy 8 .5 x .1y 2.56 100
xy 8 50 x 10 y 256
Step 3: Multiply the first equation by 15 then subtract the second equation from the first equation 10
x y 8 50 x 10 y 256
10 x 10 y 80 50 x 10 y 256
10 x 10 y 80 50 x 10 y 256 40 x 176 x 4.4 Step 4: Keeping the value of x in first equation 4.4 y 8
y 3.6 ANSWER: The amount of 50 % ammonia and 10 % acid ammonia required is 4.4 oz and 3.6 oz
34. PROBLEM: A 50% fruit juice concentrate can be purchased wholesale. Best taste is achieved when water is mixed with the concentrate in such a way as to obtain a 12% fruit juice mixture. How much water and concentrate is need to make a 50 oz fruit juice drink? SOLUTION:
Let x represent 50% fruit juice concentrate. Let y represent water. Given: x + y = 50 (0.5) x + (0) y = (.12)50 Step 1: Rewrite the second equation 0.5 x 0 y .12 50 0.5 x 6.00 x 12
Step 2: Keeping the value of x in first equation 12 y 50
y 38 ANSWER: The amount of fruit juice concentrate and water needed is 12 oz and 38 oz.
35.
PROBLEM: A 75% antifreeze concentrate is to be mixed with water to obtain 6 gallons of a 25% antifreeze solution. How much water is needed? SOLUTION:
Let x represent 75% antifreeze concentrate. Let y represent water. Given : x+ y=6 (0.75) x + (0) y = (.25)6 Step 1: Rewrite the second equation
0.75 x 0 y .25 6 .75 x 1.50 x2
Step 2: Keeping the value of x in first equation 2 y 6
y4 ANSWER: The amount of water needed is 4 gallons
36.
PROBLEM: Pure sugar is to be mixed with a fruit salad containing 10% sugar to produce 48 ounces of a salad containing 16% sugar. How much pure sugar is required? SOLUTION:
Let x represent the amount of pure sugar needed. Let y represent the amount of fruit salad needed. Given: x + y = 48 x + (.10) y = (.16)48 Step 1: Rewrite the second equation in the form x .10 y .16 48
x .10y 7.68 Step 2: Multiply second equation by 100
x y 48 x .10y 7.68 100
x y 48 100 x 10 y 768
Step 3: Multiply the first equation by 15 then subtract the second equation from the first equation
100
100 x 100 y 4800 100 x 10 y 768 100 x 100 y 4800 100 x 10 y 768
x y 48 100 x 10 y 768
90 y 4032 y 44.8 Step 4: Keeping the value of y in first equation x 44.8 48
x 3.2 ANSWER: The amount of the amount of pure sugar needed is 3.2 ounces. Part D: Uniform Motion Problems Set up a linear system and solve.
37.
PROBLEM: An airplane averaged 460 mph on a trip with the wind and 345 mph on the return trip against the wind. If the total round trip took 7 hours, then how long did the airplane spend on each leg of the trip? SOLUTION: Let x represent the time taken while going with the wind. Let y represent the time taken while going against the wind.
Given: y= 7-x We know that d = r . t Step 1: While going with the wind d = 460 . x Step 2: While going against the wind d= 345 .(7 – x) Step 3: As the distance traveled will be equal, so 460. x 345.(7 – x) 460 x 345x 2415 x3
Step 4: Keeping the value of x we get the value of y
y 73 y4 ANSWER: The time taken while going with the wind and going against the wind are 3 hours and 4 hours
38.
PROBLEM: The two legs of a 330 mile trip took 5 hours. The average speed for the first leg of the trip was 70 mph and the average speed for the second leg of the trip was 60 mph. How long did each leg of the trip take? SOLUTION:
Let x represent the time taken by first leg of the trip. Let y represent the time taken by second leg of the trip. Given: x+y=5 70x + 60y= 330 Step 1: Multiply first equation by 50 and then subtract second equation from the first equation 70
x y 5 70 x 60 y 330
70 x 70 y 350 70 x 60 y 330
70 x 70 y 350 70 x 60 y 330 10 y 20 y2 Step 2: Keeping the value of y in first equation x 2 5
x3 ANSWER: So, time taken by first leg and second legs of the trip are 3 hours and 2 hours
39.
PROBLEM: An executive traveled 1,200 miles part by helicopter and part by private jet. The jet averaged 320 mph while the helicopter averaged 80 mph. If the total trip took 4½ hours, then how long did she spend in the private jet? SOLUTION: Let x represent the time taken by the private jet. Let y represent the time taken by the helicopter.
Given: x + y = 4.5 We know d= r . t 320 x + 80 y = 1200 Step 1: Multiply the first equation by 80 then subtract the second equation from the first equation 80
x y 4.5 320 x 80 y 1200
80 x 80 y 360 320 x 80 y 1200
80 x 80 y 360 320 x 80 y 1200 240 x 840 x 3.5 Step 2: Keeping the value of x in first equation
3.5
y 4.5
y 1
ANSWER: Hence the time taken by the private jet is 3.5 hours.
40.
PROBLEM: Joe took two busses on the 463 mile trip from San Jose to San Diego. The first bus averaged 50 mph and the second bus was able to average 64 mph. If the total trip took 8 hours, then how long was spent in each bus? SOLUTION:
Let x represent the time taken by first bus. Let y represent the time taken by second bus. Given: x+y=8 50x +64y= 463 Step 1: Multiply the first equation by 50 and then subtract the second equation from the first equation 50
x y 8 50 x 64y 463
50 x 50 y 400 50 x 64y 463
50 x 50 y 400 50 x 64y 463 14 y 63 y 4.5 Step 2: Keeping the value of y in first equation x 4.5 8
x 3.5 ANSWER: So, time taken by first bus and second bus are 3.5 hours and 4.5 hours
41.
PROBLEM: Billy canoed downstream to the general store at an average rate of 9 mph. His average rate canoeing back upstream was 4 mph. If the total trip took 6½ hours, then how long did it take Billy to get back on the return trip? SOLUTION: Let x represent the time taken while downstream. Let y represent the time taken while going upstream.
Given: y = 6.5 - x We know that d = r . t Step 1: while downstream d= 9 . x
Step 2: while going upstream. d= 4.(6.5 – x) Step 3: As the distance traveled will be equal, so 9. x 4(6.5 – x) 9. x 26 4 x 13 x 26 x2 Step 4: Keeping the value of x we get the value of y y 6.5 2
y 4.5 ANSWER: The time taken while returning is 4.5 hours
42.
PROBLEM: Two brothers drove the 2,793 miles from Los Angeles to New York. One of the brothers, driving in the day, was able to average 70 mph and the other, driving at night, was able to average 53 mph. If the total trip took 45 hours, then how many hours did each brother drive? SOLUTION:
Let x represent the time taken by one of the brothers. Let y represent the time taken by another brother. Given: x + y = 45 70x +53y= 2793 Step 1: Multiply the first equation by 70 and then subtract the second equation from the first equation
70
x y 45 70 x 70 y 3150 70 x 53y 2793 70 x 53y 2793 70 x 70 y 3150 70 x 53y 2793 17 y 357 y 21 Step 2: Keeping the value of y in first equation x 21 45
x 24 ANSWER: So, time taken by both the brothers is 24 and 21
43.
PROBLEM: A boat traveled 24 miles downstream in 2 hours. The return trip took twice as long. What was the speed of the boat and the current? SOLUTION:
Let x represent the speed of the boat Let y represent the speed of the current Speed of the downstream = x + y Speed of the upstream = x - y Given: In case of downstream 24 x y 2 x y 12 In case of upstream 24 4 x y 6 x y
Step 1: Add both the equations
x y 12 x y 6 2 x 18 x9 Step 2: Keeping the value of x in the first equation
9
y 12
y 3 ANSWER: The speed of the boat and current are as 9 mph and 3 mph
44.
PROBLEM: A helicopter flying with the wind can travel 525 miles in 5 hours. On the return trip, against the wind, it will take 7 hours. What is the speed of the helicopter and the wind? SOLUTION:
Let x represent the speed of the helicopter Let y represent the speed of the wind With the wind = x + y Against the wind = x - y Given: In case of with the wind 525 x y 5 x y 105 In case of against the wind 525 7 x y 75 x y
Step 1: Add both the equations
x y 105 x y 75 2 x 180 x 90 Step 2: Keeping the value of x in the first equation 90 y 105
y 15 ANSWER: The speed of the helicopter and wind are 90 mph and 15 mph
45.
PROBLEM: A boat can travel 42 miles with the current downstream in 3 hours. Returning upstream, against the current, the boat can only travel 33 miles in 3 hours. Find the speed of the current. SOLUTION:
Let x represent the speed of the boat Let y represent the speed of the current Speed of the downstream = x + y Speed of the upstream = x - y Given: In case of downstream 42 x y 3 x y 14 In case of upstream 33 3 x y 11 x y
Step 1: Add both the equations
x y 14 x y 11 2 x 25 x 12.5 Step 2: Keeping the value of x in the first equation 12.5 y 14
y 1.5 ANSWER: The speed of the current is 1.5 mph 46.
PROBLEM: A light aircraft flying with the wind can travel 180 miles in 1½ hours. The aircraft can fly the same distance against the wind in 2 hours. Find the speed of the wind. SOLUTION:
Let x represent the speed of the aircraft Let y represent the speed of the wind With the wind = x + y Against the wind = x - y Given: In case of with the wind 180 x y 1.5 x y 120 In case of against the wind 180 2 x y 90 x y
Step 1: Add both the equations x y 120
x y 90 2 x 210 x 105
Step 2: Keeping the value of x in the first equation 105 y 120
y 15 ANSWER: The speed of the wind is 15 mph. Part E: Discussion Board
47.
PROBLEM:
Compose a number or money problem of your own and share it on the discussion board. SOLUTION: (Students answers may vary) The sum of two integers is 64 and the difference is 10. Find the integers
Let x be one of the integers Let y be other integer. Given : x + y = 64 x - y = 10 Step 1: Add both the equations
x y 64 x y 10 2x
74 x 37
Step 2: Keeping the value of x in first equation 37 y 64
y=64-37 y=27 ANSWER: The integers are 37 and 27.
48.
PROBLEM:
Compose a mixture problem of your own and share it on the discussion board.
SOLUTION: A chemist wishes to create a 32 oz solution with 12% acid content. He will use two types of stock solutions, one with 30% acid content and another with 10% acid content. How much of each will he need?
Let x represent the amount of 30% acid solution needed. Let y represent the amount of 10% acid solution needed. Given: x + y = 32 (0.30) x + (.10) y = (.12)32 Step 1: Rewrite the second equation in the form 0.30 x .10 y .12 32
0.30x .10y 3.84 Step 2: Multiply second equation by 100
x y 32 0.30x .10y 3.84 100
x y 32 30 x 10 y 384
Step 3: Multiply first equation by 15 then subtract second equation from the first equation
x y 32 30 x 10 y 384
30
30 x 30 y 960 30 x 10 y 384
30 x 30 y 960 30 x 10 y 384
20 y 576 y 28.8 Step 4: Keeping the value of y in first equation x 28.8 32
x=3.2 ANSWER: The amount of 30% acid and 10% acid solution required are 3.2 oz and 28.8 oz.
49.
PROBLEM:
Compose a uniform motion problem of your own and share it on the discussion board. SOLUTION: (Students answers may vary)
An airplane averaged 460 mph on a trip with the wind and 345 mph on the return trip against the wind. If the total round trip took 7 hours, then how long did the airplane spend on each leg of the trip? SOLUTION: Let x represent the time taken while going with the wind. Let y represent the time taken while going against the wind.
Given: y= 7-x We know that d = r . t Step 1: While going with the wind d= 460 . x Step 2: While going against the wind d= 345 .(7 – x) Step 3: As the distance traveled will be equal, so 460 . x 345 .(7 – x) 460 x 345x 2415 x=3 Step 4: Keeping the value of x we get the value of y
y 73 y4 ANSWER: The time taken while going with the wind and going against the wind are 3 hours and 4 hours
4.5 Solving Systems of Linear Inequalities (Two Variables) Part A: Solving Systems of Linear Inequalities Determine whether or not the given point is a solution to the given system of linear equations.
1.
PROBLEM: y x3 (3, 2); y x 3 SOLUTION:
Check: (3, 2) Inequality 1: y x 3
2 3 3
Inequality 2: y x 3
2 3 3 2 0
26
ANSWER: Since, it satisfies both the inequalities, so (3, 2) is the solution of the given system.
2.
PROBLEM: y 3x 4 y 2x 1
(3, 2);
SOLUTION:
Check: (3, 2) Inequality 1: y 3 x 4
Inequality 2: y 2 x 1
2 3 3 4
2 2 3 1
2 9 4
2 6 1
2 13
2 7
ANSWER: Since, it satisfies both the inequalities, so (3, 2) is the solution of the given system.
3.
PROBLEM: y x 5 3 y 4 x 2
(5, 0)
SOLUTION:
Check: (5, 0) Inequality 1: y x 5
0 5 5 0 0 Incorrect
4.
Inequality 2: y
3 x2 4
3 4 15 0 2 4 15 8 0 4 7 0 4
0 5 2
ANSWER: Since, it does not satisfy both the inequalities, so (5, 0) is not the solution of the given system PROBLEM: 2 y 3 x 1 (0, 1); y 5 x 2 2 SOLUTION:
Check: (0, 1); Inequality 1: y
2 x 1 3
2 3 1 1 Incorrect
1 0 1
Inequality 2: y
5 x2 2
5 2 1 2
1 0 2
ANSWER: Since, it does not satisfy both the inequalities, so (0, 1) is not the solution of the given system
5.
PROBLEM: 4 x 3 y 12 2x 3y 6
(1, 83 )
SOLUTION:
Check: (1, 83 ) Inequality 1: 4 x 3 y 12
Inequality 2: 2 x 3 y 6
4 1 3 83 12
2 1 3 83 6
4 8 12
2 8 6
12 12
6 6 Incorrect
ANSWER: Since, it does not satisfy both the inequalities, so (1, 83 ) is not the solution of the given system
6.
PROBLEM: x y 0 x y 0 x y 2
(1, 2)
SOLUTION:
Check: (1, 2) Inequality 1: x y 0
Inequality 2: x y 0
Inequality 3: x y 2
1 2 0
1 2 0
1 2 2
1 2 0
3 0
3 2
1 0
ANSWER: Since, it satisfies all three inequalities, so (1, 2) is the solution of the given system.
Part B: Solving Systems of Linear Inequalities Graph the solution set.
7.
PROBLEM: y x3 y x 3 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (0, 3) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (0, 3) Inequality 1: y x 3
Inequality 2: y x 3
3 0 3
3 0 3
33
33
ANSWER: This shows that it satisfies both the inequalities.
8.
PROBLEM: y 3x 4 y 2x 1 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (-1, 1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (-1, 1) Inequality 1: y 3 x 4
Inequality 2: y 2 x 1
1 3 1 4
1 2 1 1
1 7
1 3
ANSWER: This shows that it satisfies both the inequalities.
9.
PROBLEM: yx y 1 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (-3, -2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (-3, -2) Inequality 1: y x
Inequality 2: y 1
2 3
2 1
2 3
2 1
ANSWER: This shows that it satisfies both the inequalities.
10.
PROBLEM: 2 y 3 x 1 y 5 x 2 2 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (-3, -2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (-3, -2) Inequality 1: y 2 3 2 2 1 2 1
2 x 1 3
2 3 1
Inequality 2: y 5 2 15 4 2 2 2 9.5
2 3 2
ANSWER: This shows that it satisfies both the inequalities.
11.
PROBLEM: y x 5 3 y 4 x 2 SOLUTION:
5 x2 2
The above graph shows that shaded region is the intersection area. The graph suggests that (5,1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (5, 1) Inequality 1: y x 5
1 5 5 1 0
Inequality 2: y 3 4 15 8 1 4 7 1 4 1 1.75
3 x2 4
1 5 2
ANSWER: This shows that it satisfies both the inequalities.
12.
PROBLEM: 3 y 5 x 3 y 3 x 3 5 SOLUTION:
Inequality 1: y If x=0, then y=3 If y=0, then x= -5
3 x3 5
Inequality 2: y If x=0, then y= -3 If y=0, then x= 5
3 x 3 5
ANSWER: Since there is no intersection area, so there is no solution.
13.
PROBLEM: x 4 y 12 3x 12 y 12 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (5,1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (5, 1) Inequality 1: x 4 y 12
Inequality 2: 3x 12 y 12
5 4 1 12
3 5 12 1 12
9 12
3 12
ANSWER: This shows that it satisfies both the inequalities.
14.
PROBLEM: x y 6 2x y 1 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (-1, 4) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (-1, 4) Inequality 1: x y 6
Inequality 2: 2 x y 1
1 4 6
2 1 4 1
56
2 1
ANSWER: This shows that it satisfies both the inequalities.
15.
PROBLEM: 2 x 3 y 3 4 x 3 y 15
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (6, 6) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (6, 6) Inequality 1: 2 x 3 y 3
Inequality 2: 4 x 3 y 15
2 6 3 6 3
4 6 3 6 15
12 18 3
24 18 15
63
6 15
ANSWER: This shows that it satisfies both the inequalities.
16.
PROBLEM: 4 x 3 y 12 2x 3y 6
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (2, 0) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (2, 0) Inequality 1: 4 x 3 y 12
Inequality 2: 2 x 3 y 6
4 2 3 0 12
2 2 3 0 6
8 12
46
ANSWER: This shows that it satisfies both the inequalities.
17.
5x y 4 4 x 3 y 6
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (1, -3) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (1, -3) Inequality 1: 5 x y 4
Inequality 2: 4 x 3 y 6
5 1 3 4
4 1 3 3 6
53 4 24
4 9 6 13 6
ANSWER: This shows that it satisfies both the inequalities.
18.
PROBLEM: 3x 5 y 15 x 2 y 0
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (2, 1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (2, 1) Inequality 1: 3x 5 y 15
Inequality 2: x 2 y 0
3 2 5 1 15
2 2 1 0
6 5 15
2 2 0
11 15
00
ANSWER: This shows that it satisfies both the inequalities.
19.
PROBLEM: x0 5 x y 5 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (0, 6) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (0, 6) Inequality 1: x 0
Inequality 2: 5 x y 5
0 0
5 0 6 5
00
65
ANSWER: This shows that it satisfies both the inequalities.
20.
PROBLEM: x 2 y 1
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (-2, 1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (-2, 1) Inequality 1: x 2
Inequality 2: y 1
2 2
1 1
2 2
11
ANSWER: This shows that it satisfies both the inequalities.
21.
PROBLEM: x 3 0 y 2 0
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (2, 2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (2, 2) Inequality 1: x 3 0
Inequality 2: y 2 0
2 3 0
2 2 0
1 0
40
ANSWER: This shows that it satisfies both the inequalities.
22.
PROBLEM: 5 y 2x 5 2 x 5 y 5 SOLUTION:
Inequality 1: 5 y 2 x 5
Inequality 2: 2 x 5 y 5
If x=0, then y=1 If y=0, then 5 x= 2
If x=0, then y= -1 If y=0, then 5 x= 2
ANSWER: Since there is no intersection area, there is no solution.
23.
PROBLEM: x y 0 x y 1 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (1, 1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (1, 1) Inequality 1: x y 0
Inequality 2: x y 1
1 1 0
1 1 1
00
0 1
ANSWER: This shows that it satisfies both the inequalities.
24.
PROBLEM: x y 0 y x 1
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (-2, -2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (-2, -2) Inequality 1: x y 0
Inequality 2: y x 1
2 2 0
2 2 1
22 0 00
2 2 1 0 1
ANSWER: This shows that it satisfies both the inequalities.
25.
PROBLEM: x 2 x 2 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (1, 1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (1, 1) Inequality 1: x 2
Inequality 2: x 2
x 2
x2
1 2
1 2
ANSWER: This shows that it satisfies both the inequalities.
26.
PROBLEM: y 1 y 2
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (0, 1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (0, 1) Inequality 2: y 2
Inequality 1: y 1
1 1
1 2
1 1
1 2
ANSWER: This shows that it satisfies both the inequalities.
27.
PROBLEM: x 2 y 8 3x 6 y 18 SOLUTION:
Inequality 1: x 2 y 8 If x=0, then y=4 If y=0, then x= -8
Inequality 2: 3x 6 y 18 If x=0, then y= -3 If y=0, then x= 6
ANSWER: Since there is no intersection area, so there is no solution.
28.
PROBLEM: 3 x 4 y 4 6 x 8 y 8
SOLUTION:
Inequality 1: 3x 4 y 4
Inequality 2: 6 x 8 y 8
If x=0, then y= 1
If x=0, then y= 1
If y=0, then 4 x= 3
If y=0, then 4 x= 3
ANSWER: Since there is no intersection area, there is no solution.
29.
PROBLEM: 2 x y 3 1 x 2 y SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (0, 2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (0, 2) Inequality 1: 2 x y 3 2 0 2 3 23
Inequality 2: x 0 0 1
ANSWER: This shows that it satisfies both the inequalities.
30.
PROBLEM: 2x 6 y 6 1 3 x y 3 SOLUTION:
1 2 2
1 y 2
The above graph shows that shaded region is the intersection area. The graph suggests that (0, 1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (0, 1) Inequality 1: 2 x 6 y 6 2 0 6 1 6 66
1 Inequality 2: x y 3 3 1 0 1 3 3 1 3
ANSWER: This shows that it satisfies both the inequalities.
31.
PROBLEM: y3 yx x 3 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (2, 2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (2, 2) Inequality 1: y3
Inequality 2: yx
Inequality 3: x 3
2 3
2 2
2 3
23
2 2
2 3
ANSWER: This shows that it satisfies all the inequalities
32.
PROBLEM: y 1 y x 1 y 3 x 3 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (-1, -1) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (1, -1) Inequality 1: y 1
Inequality 2: y x 1
Inequality 3: y 3x 3
1 1
1 1 1
1 3 1 3
1 1
1 2
1 6
ANSWER: This shows that it satisfies all the inequalities.
33.
PROBLEM: 4 x 3 y 12 y2 2x 3y 6 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (5, 3) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (5, 3) Inequality 1: 4 x 3 y 12
Inequality 2: y2
Inequality 3: 2x 3y 6
4 5 3 3 12
3 2
2 5 3 3 6
20 9 12 11 12
3 2
10 9 6 19 6
ANSWER: This shows that it satisfies all the inequalities
34.
PROBLEMS: x y 0 x y 0 x y 2 SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (1, -2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (1, -2) Inequality 1: x y 0
Inequality 2: x y 0
Inequality 3: x y 2
1 2 0
1 2 0
1 2 2
3 0
1 0
1 2
ANSWER: This shows that it satisfies all the inequalities.
35.
x y 2 x3 x y 2
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (2, -2) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (2, -2) Inequality 1: x y 2
Inequality 2: x3
Inequality 3: x y 2
2 2 2
2 3
2 2 2
4 2
2 3
22 2 02
ANSWER: This shows that it satisfies all the inequalities.
36.
PROBLEM: y40 1 1 x y 1 3 2 1 1 2 x 3 y 1
SOLUTION:
The above graph shows that shaded region is the intersection area. The graph suggests that (0, 3) is a common point. As a check, we could substitute that point into the inequalities and verify that it solves both conditions. Check: (0, 3) Inequality 1: y40
3 4 0 70
Inequality 2: 1 1 x y 1 2 3 1 1 0 3 1 2 3 11
Inequality 3: 1 1 x y 1 2 3 1 1 0 3 1 2 3 11
ANSWER: This shows that it satisfies all the inequalities.
37.
PROBLEM: Construct a system of linear inequalities that describes all points in the first quadrant. SOLUTION:
In the first quadrant, all the values for x and y are positive. It means that x and y will be greater than 0. So, a system of linear inequalities that describes all points in the first quadrant is x>0 y>0
38.
PROBLEM: Construct a system of linear inequalities that describes all points in the second quadrant. SOLUTION:
In the second quadrant, all the values for x are negative but the values of y are positive. It means that x will be less than 0 but the values of y will be greater than 0. ANSWER: So, a system of linear inequalities that describes all points in the second quadrant is
x0 39.
PROBLEM: Construct a system of linear inequalities that describes all points in the third quadrant. SOLUTION:
In the third quadrant, all the values for x and y are negative. It means that x and y will be less than 0. ANSWER: So, a system of linear inequalities that describes all points in the third quadrant is x
