GCSE Maths

Quadratic Simultaneous Equations

Edexcel
Practice Questions → Download PDF
Introduction Watch: Quadratic Simultaneous Equations Step by Step Solving Quadratic Simultaneous Equations Solved Examples

Introduction

  • Two or more equations that share variables and have set of values for variables that satisfy both the equations are called Simultaneous Equations.
  • If the maximum power raised to the variable in Simultaneous equations is one then these are called Linear Simultaneous Equations.
  • Similarly If the power raised to the variables is two they are called Quadratic Simultaneous Equations.

Watch: Quadratic Simultaneous Equations

https://youtu.be/BfYrxC2TAPc?si=642ODUCM1OL8KWs_

Step by Step Solving Quadratic Simultaneous Equations

Example:

$$ \begin{aligned} y &= 2x – 1 \\ y &= x^2 + 3x – 7 \end{aligned} $$
1
Substitute the value of one variable from linear equation(to eliminate one variable) –

Putting $y = 2x – 1$ in 2nd equation –

$$ \begin{aligned} 2x – 1 &= x^2 + 3x – 7 \\ x^2 + 3x – 7 – 2x + 1 &= 0 \\ x^2 + x – 6 &= 0 \\ x^2 + 3x – 2x – 6 &= 0 \\ x(x + 3) – 2(x + 3) &= 0 \\ (x + 3)(x – 2) &= 0 \\ x = -3 & \quad x = 2 \end{aligned} $$
2
Substitute the values obtained back into the original equation and find possible values of another variable-

For $x = -3$ from 1st Equation-

$$ \begin{aligned} y &= 2x – 1 \\ y &= 2(-3) – 1 \\ y &= -6 – 1 \\ y &= -7 \end{aligned} $$

For $x = 2$ from 1st Equation-

$$ \begin{aligned} y &= 2x – 1 \\ y &= 2(2) – 1 \\ y &= 4 – 1 \\ y &= 3 \end{aligned} $$
3
Final Answer: $$x = -3 \text{ and } y = -7 \quad \text{OR} \quad x = 2 \text{ and } y = 3$$

Need help with Quadratic Simultaneous Equations?

Our tutors explain it step by step, matched to your exam board.

Book a Consultation →

Solved Examples

Solved Example

Solve the following equation:

$$ \begin{aligned} 2x – y &= 7 \\ x^2 – 7x + y &= 7 \end{aligned} $$
SOLUTION
1
Substitute the value of one variable from linear equation(to eliminate one variable)-

From 1st Equation-

$$ y = 2x – 7 $$

Put in 2nd Equation-

$$ \begin{aligned} x^2 – 7x + (2x – 7) &= 7 \\ x^2 – 7x + 2x – 7 – 7 &= 0 \\ x^2 – 5x – 14 &= 0 \\ x^2 – 7x + 2x – 14 &= 0 \\ x(x – 7) + 2(x – 7) &= 0 \\ (x + 2)(x – 7) &= 0 \\ x = -2 & \quad x = 7 \end{aligned} $$
2
Substitute the values obtained back into the original equation and find possible values of another variable-

Put $x = -2$ in 1st Equation –

$$ \begin{aligned} 2(-2) – y &= 7 \\ -4 – y &= 7 \\ y &= -4 – 7 \\ y &= -11 \end{aligned} $$

Put $x = 7$ in 1st Equation –

$$ \begin{aligned} 2(7) – y &= 7 \\ 14 – y &= 7 \\ y &= 7 \end{aligned} $$

Final Answer: $$ x = -2 \text{ and } y = -11 \quad \text{OR} \quad x = 7 \text{ and } y = 7 $$

Solved Example

Solve the following equation:

$$ \begin{aligned} x + y &= 12 \\ x^2 + y^2 &= 80 \end{aligned} $$
SOLUTION
1
Substitute the value of one variable from linear equation(to eliminate one variable)-

From 1st Equation,

$$ x = 12 – y $$

Put in 2nd Equation,

$$ \begin{aligned} (y – 12)^2 + y^2 &= 80 \\ y^2 + 144 – 24y + y^2 &= 80 \\ 2y^2 – 24y + 144 – 80 &= 0 \\ 2y^2 – 24y + 64 &= 0 \end{aligned} $$

Divide by 2 on both sides and doing factorisation-

$$ \begin{aligned} y^2 – 12y + 32 &= 0 \\ y^2 – 8y – 4y + 32 &= 0 \\ y(y – 8) – 4(y – 8) &= 0 \\ (y – 4)(y – 8) &= 0 \\ y = 4 & \quad y = 8 \end{aligned} $$
2
Substitute the values obtained back into the original equation and find possible values of another variable-

Put $y = 4$ in 1st Equation –

$$ \begin{aligned} x + y &= 12 \\ x + 4 &= 12 \\ x &= 8 \end{aligned} $$

Put $y = 8$ in 1st Equation –

$$ \begin{aligned} x + y &= 12 \\ x + 8 &= 12 \\ x &= 4 \end{aligned} $$

Final Answer: $$ x = 8 \text{ and } y = 4 \quad \text{OR} \quad x = 4 \text{ and } y = 8 $$

Solved Example

Solve the following equation:

$$ \begin{aligned} 8x + y &= 42 \\ x^2 – y &= 6 \end{aligned} $$
SOLUTION
1
Substitute the value of one variable from linear equation(to eliminate one variable)-

From 1st Equation-

$$ y = 42 – 8x $$

Put in 2nd Equation-

$$ \begin{aligned} x^2 – (42 – 8x) &= 6 \\ x^2 – 42 + 8x – 6 &= 0 \\ x^2 + 8x – 48 &= 0 \\ x^2 + 12x – 4x – 48 &= 0 \\ x(x + 12) – 4(x + 12) &= 0 \\ (x + 12)(x – 4) &= 0 \\ x = -12 & \quad x = 4 \end{aligned} $$
2
Substitute the values obtained back into the original equation and find possible values of another variable-

Put $x = -12$ in 1st Equation –

$$ \begin{aligned} 8x + y &= 42 \\ 8(-12) + y &= 42 \\ -96 + y &= 42 \\ y &= 138 \end{aligned} $$

Put $x = 4$ in 1st Equation –

$$ \begin{aligned} 8x + y &= 42 \\ 8(4) + y &= 42 \\ 32 + y &= 42 \\ y &= 10 \end{aligned} $$

Final Answer: $$ x = -12 \text{ and } y = 138 \quad \text{OR} \quad x = 4 \text{ and } y = 10 $$

Solved Example

Solve the following equation:

$$ \begin{aligned} 2x^2 + y &= 5 \\ 2x + y &= 1 \end{aligned} $$
SOLUTION
1
Substitute the value of one variable from linear equation(to eliminate one variable)-

From 1st Equation,

$$ y = 1 – 2x $$

Put in 2nd Equation,

$$ \begin{aligned} 2x^2 + 1 – 2x &= 5 \\ 2x^2 – 2x – 4 &= 0 \end{aligned} $$

Divide by 2 on both sides-

$$ \begin{aligned} x^2 – x – 2 &= 0 \\ x^2 – 2x + x – 2 &= 0 \\ x(x – 2) + 1(x – 2) &= 0 \\ (x – 2)(x + 1) &= 0 \\ x = 2 & \quad x = -1 \end{aligned} $$
2
Substitute the values obtained back into the original equation and find possible values of another variable-

Put $x = 2$ in 2nd Equation –

$$ \begin{aligned} 2(2) + y &= 1 \\ y &= -3 \end{aligned} $$

Put $x = -1$ in 2nd Equation –

$$ \begin{aligned} 2(-1) + y &= 1 \\ y &= 3 \end{aligned} $$

Final Answer: $$ x = 2 \text{ and } y = -3 \quad \text{OR} \quad x = -1 \text{ and } y = 3 $$

Solved Example

Solve the following equation:

$$ \begin{aligned} y + x &= 8 \\ y &= x^2 + 2x + 4 \end{aligned} $$
SOLUTION
1
Substitute the value of one variable from linear equation(to eliminate one variable)-

From 1st Equation,

$$ y = 8 – x $$

Put in 2nd Equation-

$$ \begin{aligned} 8 – x &= x^2 + 2x + 4 \\ x^2 + 2x + 4 – 8 + x &= 0 \\ x^2 + 3x – 4 &= 0 \\ x^2 + 4x – x – 4 &= 0 \\ x(x + 4) – 1(x + 4) &= 0 \\ (x – 1)(x + 4) &= 0 \\ x = 1 & \quad x = -4 \end{aligned} $$
2
Substitute the values obtained back into the original equation and find possible values of another variable-

Put $x = 1$ in 1st Equation –

$$ \begin{aligned} y + x &= 8 \\ y + 1 &= 8 \\ y &= 8 – 1 \\ y &= 7 \end{aligned} $$

Put $x = -4$ in 1st Equation –

$$ \begin{aligned} y + x &= 8 \\ y + (-4) &= 8 \\ y &= 8 + 4 \\ y &= 12 \end{aligned} $$

Final Answer: $$ x = 1 \text{ and } y = 7 \quad \text{OR} \quad x = -4 \text{ and } y = 12 $$

Ready to test your knowledge?

You've reviewed the notes. Now try exam-style questions on Quadratic Simultaneous Equations.

Start Practice Questions →
Previous Topic ← Simultaneous Equations Next Topic Circle Theorems →

Want a Tutor for This Topic?

Sehaj
Sehaj Lead Maths Tutor Personalised lessons matched to your exam board.
Book a Consultation →
Or explore all tutoring options →

Ready to Get Started?

Book a consultation and let's build a plan for your child. £3.99 refundable deposit, deducted from your first lesson.

Book Your Consultation →