GCSEFurther MathsAQARevision Notes2. Algebra & FunctionsShapes of GraphsSolving Equations using Graphs

Solving Equations using Graphs (AQA GCSE Further Maths): Revision Note

Exam code: 8365

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on 11 August 2026

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Understand this topic with deeper explanations and understanding checks.

Solving equations using graphs

How do we use graphs to solve equations?

Solutions are always read off the x-axis
Solutions of f(x) = 0 are where the graph of y = f(x) crosses the x-axis
If asked to use the graph of y = f(x) to solve a different equation (the question will say something like “by drawing a suitable straight line”) then:
- Rearrange the equation to be solved into f(x) = mx + c and draw the line y = mx + c
- Solutions are the x-coordinates of where the line (y = mx + c) crosses the curve (y = f(x))
- E.g. if given the curve for y = x³ + 2x²+ 1 and asked to solve x³ + 2x² − x − 1 = 0, then;
  1. rearrange x³ + 2x² − x − 1 = 0 to x³ + 2x²+ 1 = x + 2
  2. draw the line y = x + 2 on the curve y = x³ + 2x²+ 1
  3. read the x-values of where the line and the curve cross (in this case there would be 3 solutions, approximately x = -2.2, x = -0.6 and x = 0.8);

Note that solutions may also be called roots

Examiner Tips and Tricks

If solving an equation, give the x values only as your final answer
If solving a pair of linear simultaneous equations give an x and a y value as your final answer
If solving a pair of simultaneous equations where one is linear and one is quadratic, give two pairs of x and y values as your final answer

Worked Example

The graph of $y = x^{3} + x^{2} - 3 x - 1$ is shown below.
Use the graph to estimate the solutions of the equation $x^{3} + x^{2} - 4 x = 0$ . Give your answers to 1 decimal place.

Cubic-Linear-Intersections-(before), IGCSE & GCSE Maths revision notes

We are given a different equation to the one plotted so we must rearrange it to $f (x) = m x + c$ (where $f (x)$ is the plotted graph)

$\begin{array}{rcl} x^{3} + x^{2} - 4 x & = & 0 \end{array}$

$\begin{array}{rcl} + x - 1 + x - 1 \end{array}$

$\begin{array}{rcl} x^{3} + x^{2} - 3 x - 1 & = & x - 1 \end{array}$

Now plot $y = x - 1$ on the graph- this is the solid red line on the graph below

Cubic-Linear-Intersections-(after), IGCSE & GCSE Maths revision notes

The solutions are the $x$ coordinates of where the curve and the straight line cross so

$x = - 2.6, x = 0, x = 1.6$

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