Solving Simultaneous Equations (2)

Solving Simultaneous Equations (2)

•Review of techniques •Extended use of the substitution method •One linear / One quadratic

Algebraic Method : Solving by equating Most useful methodcoefficients for “ax + by = c” form of equations E.g.

3x + 4y = 26 7x - y = 9

(1) 3x + 4y = 26 (2) 7x - y = 9

“Solve simultaneously”

Make either coefficient of x or y „same size‟

(3)=(2)x4 (1)

28x - 4y = 36 3x + 4y = 26

(1)+(3) (1)+(3)

31x -4y + 4y = 62 31x = 62 x = 2

3x + 4y = 26 When x = 2 32 + 4y = 26 6 + 4y = 26 4y = 20 y=5

Algebraic Method : Solving by substitution Most useful method for “y = mx + c” graphs

y= 2x + 1 y = -x + 7

“Solve simultaneously” (where would their graphs cross)

They are both equal to y, so:

y = 2x + 1 2x + 1 3x + 1 3x x

= = = = =

-x + 7 -x + 7 (+ x) 7 (-1) 6 ( 3) 2

When x = 2

y= 2x + 1 y= 2  2 + 1 y= 5

Algebraic Method : Solving by substitution extended y= 2x + 1 3x + 2y = 9

“Solve simultaneously by substitution method”

3x + 2y = 9 Since, y= 2x + 1

3x + 2(2x + 1) = 9 3x + 4x + 2 = 9 7x + 2 = 9 7x = 7 x=1

When, x = 1 Put into, y= 2x + 1 y= 2 x 1 + 1

y= 2 + 1 y= 3

Simultaneous Equations and Intersections Example 2

y  x

2

 3

y  4 x  1

Eliminate y:  x

Solving

x

2

x 2

2

      (2)

 3  4 x  1

 4x  4  0

 4x  4  0

 ( x  2 )( x  2 )  0



      (1 )

y x y  4 x  1

x   2 (twice)

x  2 

y  7

The line is a tangent to the curve.

2

3

Algebraic Method : Solving by substitution extended Is the best method when one equation is a quadratic “Solve simultaneously by y= 2x + 1 y = 2x2 + 3x +1 substitution method”

Since, y= 2x + 1 2x2

y= 2x + 1 = + 3x +1 2x + 1 = 2x2 + 3x +1 1 = 2x2 + x +1 0 = 2x2 + x 2x2 + x = 0 x(2x + 1) = 0

x(2x + 1) = 0 Either

or

(0,1) x=0 y= 2x + 1 = 0 + 1 =1

2x + 1 = 0 (-1/2,0) 2x = -1 x = -1/2 y= 2x + 1 = 2(-1/2) + 1 = -1 + 1 = 0

Solving Linear + quadratic Equations Graphically 2

y=2x^2 + 3x + 1 y=2x+1

1.5

1

(0,1)

0.5

0 -2

-1.5

-1

-0.5

0

0.5

1

1.5

-0.5

-1

-1.5

-2

(-1/2,0)

2

Have a go at this one : Solving by substitution extended y= 3x - 1 y = x2 + 2x - 3 Since, y= 3x - 1 x2

y= 3x - 1 = + 2x - 3 3x - 1 = x2 + 2x - 3 -1 = x2 - x - 3 0 = x2 - x -2 x2 - x - 2= 0 (x-2)(x + 1) = 0

“Solve simultaneously by substitution method”

(x-2)(x + 1) = 0 Either

or

(2,5)

x-2=0 x=2 y= 3x - 1 = 3x2 - 1 = 5

x+1=0 (-1,-4) x = -1 y= 3x - 1 = 3(-1) - 1 = -3 - 1 = -4