Simultaneous equations can also be solved when one of the equations is in a quadratic form. In this instance:

• Only one of the variables will be a squared term: replace the other one;

• Substitute into the quadratic equation;

• Solve the quadratic equation, using factorisation, completing the square, or the quadratic formula;

• Solve the second variable by substitution into the linear equation;

Remember there will probably be two answers (although there may be one (twice) or none. For each answer, find the corresponding unknown of the other variable.

## Example 1

Solve y=2x+10 and y=-x^2-4x+5

Replace the y in the quadratic with 2x + 10 from the linear equation and rearrange to have zero on one side:

 Replace y  2x +10 = -x^2 -4x +5 Add x^2 both sides x^2 +2x +10 =  -4x +5 Add 4x both sides x^2 +6x +10 =   +5 Subtract 5 both sides x^2 +6x +5 = 0  

Factorise the quadratic: (x+1)(x+5)=0

Solutions for x are -1 and -5, as that is what makes each bracket equal to zero

using the linear equation:

When x=-1 then y = 2(-1)+10=8

When x=-5 then y = 2(-5)+ 10 = 0

Solutions are x=-1, y=8 and x=-5, y=0

Check:

8 = -(-1)2 - 4(-1) + 5 ✔

0 = -(-5)2 - 4(-5) + 5 ✔

Answer: x=-1, y=8 and x=-5, y=0

## Example 2

Solve x-2y=4 and 2y=x^2-8x+4.

Rearrange x-2y=4 to make 2y the subject:

2y=x-4

 Using the quadratic: 2y = x^2 -8x +4 Replace 2y: x -4 = x^2 -8x +4 Subtract x from both sides:  -4 = x^2 -9x +4 Add 4 to both sides:  0 = x^2 -9x +8

Factorise x^2-9x+8 to (x-1)(x-8)

Gives solutions of x=1 and x=8 as this makes each bracket zero

Substitute into the linear equations to get y for each value

when x=1, then (1)-2y=4 and y=-frac(3)(2)

when x=8, then (8)-2y=4 and y=-2

Solutions are: x=1, y=-frac(3)(2)

and: x=8, y=2

Check:

2(-frac(3)(2))=(1)^2-8(1)+4

2(2)=(8)^2-8(8)+4

Answer: x=1, y=-1frac(1)(2) and x=8, y=2