K S Sandhu

Simultaneous equationsSolve y = 2x2 – 3x + 4 and y = 4x + 1

2x2 – 3x + 4 = 4x + 1

2x2 – 7x + 3 = 0

(2x – 1)(x – 3) = 0

x = ½ or x = 3

y( ½ ) = 4 ½ + 1 = 3

y( 3 ) = 4 3+ 1 = 13

Complete solution ( ½ , 3) and (3, 13)

K S Sandhu

Simultaneous equationsSolve y = x2 + 3x - 18 and y = 2x + 2

x2 + 3x - 18 = 2x + 2

x2 + x - 20 = 0

(x + 5)(x - 4) = 0

x = -5 or x = 4

y( -5 ) = 2 (-5) + 2 = -8

y( 4 ) = 2 4+ 2 = 10

Complete solution ( -5 , -8) and (4, 10)

K S Sandhu

Simultaneous equationsShow that the simultaneous equations y = x2 + 3x - 18 and y = 2x – 20 have no real solution.

x2 + 3x - 18 = 2x - 20

x2 + x + 2 = 0

a =1, b = 1 and c = 2

b2 – 4ac = 1 – 4 1 2 = - 7 < 0

The line and the curve do not intersect i.e. no real solutions

K S Sandhu

Simultaneous equationsThe graph of y = 2x + k meets the graph of y = x2 + 3x – 18 at only one point. Find the value of k.

x2 + 3x - 18 = 2x + k

x2 + x - k - 18 = 0

a =1, b = 1 and c = -k - 18

b2 – 4ac = 1 – 4 1 (- k – 18) = 0

1 + 4k + 72 = 0

k = -73/4

K S Sandhu

Simultaneous equationsIt is given that x and y satisfy the simultaneous equations:

y – x = 4 and 2x2 + xy + y2 = 8

y – x = 4 y = x + 4

2x2 + xy + y2 = 8

4x2 + 12x + 8 = 0

(i) Show that x2 + 3x + 2 = 0 (ii) Solve the equations.

2x2 + x(x + 4) + (x + 4)2 = 8

2x2 + x2 + 4x + x2 + 8x + 16 = 8

x2 + 3x + 2 = 0

(i)

(x + 2)(x + 1)= 0(ii) x = - 2 or x = - 1

Complete solution (-2, 2) or (-1, 3)

K S Sandhu

Simultaneous equationsSolve the simultaneous equations.

x + y = 1 and x2 - xy + y2 = 7

x + y = 1 y = 1 - x

x2 – x(1 – x) + (1 – x)2 = 7

3x2 - 3x - 6 = 0

x2 – x + x2 + 1 – 2x + x2 = 7

x2 - x - 2 = 0

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

x = - 1 or x = 2

Complete solution (-1, 2) or (2, -1)

K S Sandhu

Substitute for x from (1) into (2)

Factorise and solve for y

Substitute for y in (1)

Solve the simultaneous equationsx + y = 2 (1) x2 + 2y2 = 11 (2)

(2 – y)2 + 2y2 = 11

4 – 4y + y2 + 2y2 = 11

(3y – 7)(y + 1) = 0

3y2 – 4y – 7 = 0

73 or 1y y

7 7 13 3 3When , 2y x x

When 1, ( 1) 2 3y x x 71

3 3The solutions are , and 3, 1.x y x y

K S Sandhu

Substitute for x from (2) into (1)

Factorise and solve for y

Substitute for x in (2)

Solve the simultaneous equations y = x2 (1) x + y = 6 (2)

y = (6 – y)2

y = 36 – 12y + y2

(y – 4)(y - 9) = 0

y2 – 13y + 36 = 0

When 4, 4 6 2y x x When 9, 9 6 3y x x

The solutions are 2, 4 and 3, 9.x y x y

y = 4 or y = 9