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Quadratic Simultaneous Equations (Cambridge (CIE) O Level Additional Maths): Revision Note

Exam code: 4037

Written by: Amber

Reviewed by: Dan Finlay

Updated on 2 July 2026

Quadratic simultaneous equations

What are quadratic simultaneous equations?

When there are two unknowns (say x and y) in a problem, we need two equations to be able to find them both: these are called simultaneous equations
If there is an x² or y² or xy in one of the equations then they are quadratic (or non-linear) simultaneous equations

How do I solve quadratic simultaneous equations?

Use the method of substitution
- Substitute the linear equation, y = ... (or x = ...), into the quadratic equation
  - Do not try to substitute the quadratic equation into the linear equation
Solve x² + y² = 25 and y - 2x = 5
- Rearrange the linear equation into y = 2x + 5
- Substitute this into the quadratic equation, replacing all y's with (2x + 5) in brackets
  - x² + (2x + 5)² = 25
- Expand and solve this quadratic equation (x = 0 and x = -4)
- Substitute each value of x into the linear equation, y = 2x + 5, to get their value of y
- Present your solutions in a way that makes it obvious which x belongs to which y
  - x = 0, y = 5 or x = -4, y = -3
Check your final solutions satisfy both equations

How do you use graphs to solve quadratic simultaneous equations?

Plot both equations on the same set of axes
- to do this, you can use a table of values (or, for straight lines, rearrange into y = mx + c if it helps)
Find where the lines intersect (cross over)
- The x and y solutions to the simultaneous equations are the x and y coordinates of the point of intersection
e.g. to solve y = x² + 3x + 1 and y = 2x + 1 simultaneously, first plot them both (see graph)
- find the points of intersection, (-1, -1) and (0, 1)
- the solutions are x = -1 and y = -1 or x = 0 and y = 1

Solution of quadratic simultaneous equations as the points of intersection of their graphs

Examiner Tips and Tricks

If the resulting quadratic has a repeated root then the line is a tangent to the curve (as there is only 1 solution)
If the resulting quadratic has no roots then the line does not intersect with the curve (as there are 0 solutions) – or you have made a mistake!
When giving your final answer, make sure you indicate which x and y values go together
- If you don’t make this clear you can lose marks for an otherwise correct answer

Worked Example

Solve the equations

x² + y² = 36
x = 2y + 6

Number the equations.

$\begin{array}{l} x^{2} + y^{2} = 36 (1) \\ x = 2 y + 6 (2) \end{array}$

There is one quadratic equation and one linear equation so this must be done by substitution.

Equation (2) is equal to $x$ so this can be eliminated by substituting it into the $x$ part for equation (1).
Substitute $x = 2 y + 6$ into equation (1).

${(2 y + 6)}^{2} + y^{2} = 36$

[1]

Expand the brackets, remember that a bracket squared should be treated the same as double brackets.

$\begin{array}{l} (2 y + 6) (2 y + 6) + y^{2} = 36 \\ 4 y^{2} + 6 (2 y) + 6 (2 y) + 6^{2} + y^{2} = 36 \end{array}$

[1]

Simplify.

$\begin{array}{rcl} 4 y^{2} + 12 y + 12 y + 36 + y^{2} & = & 36 \\ 5 y^{2} + 24 y + 36 & = & 36 \end{array}$

[1]

Rearrange to form a quadratic equation that is equal to zero.

$\begin{array}{rcl} 5 y^{2} + 24 y + 36 - 36 & = & 0 \\ 5 y^{2} + 24 y & = & 0 \end{array}$

The question does not give a specified degree of accuracy, so this can be factorised. Take out the common factor of $\begin{array}{rcl} y \end{array}$ .

$\begin{array}{rcl} y (5 y + 24) & = & 0 \end{array}$

Solve to find the values of $y$ .
Let each factor be equal to 0 and solve.

$\begin{array}{rcl} y_{1} & = & 0 \\ 5 y_{2} + 24 & = & 0 \Rightarrow y_{2} = - \frac{24}{5} = - 4.8 \end{array}$

[1]

Substitute the values of $y$ into one of the equations (the linear equation is easier) to find the values of $x$ .

$x_{1} = 2 (0) + 6 = 6 x_{2} = 2 (- \frac{24}{5}) + 6 = - 9.6 + 6$

$x_{1} = 6, y_{1} = 0$
$x_{2} = - 3.6, y_{2} = - 4.8$ [1]

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