How to Solve Simultaneous Equations: A Step-by-Step GCSE Maths Guide

If you're studying for your GCSE Maths exam (AQA, Edexcel, or OCR), you'll definitely need to know how to solve simultaneous equations. They're a core algebra topic at KS4, often appearing in both the Foundation and Higher tier papers. This guide will walk you through exactly what they are and the two main methods for solving them, with clear examples you can follow.

What Are Simultaneous Equations?

Simultaneous equations are simply two (or more) equations that are true at the same time. They involve the same unknown variables, usually x and y. The "solution" to a pair of simultaneous equations is the values of x and y that satisfy both equations simultaneously.

Graphically, each equation represents a straight line. The solution is the coordinates of the point where the two lines cross.

flowchart TD
    A["Start: Pair of Simultaneous Equations"] --> B["Choose a Method"]
    B --> C["Elimination Method"]
    B --> D["Substitution Method"]
    C --> E["Make Coefficients of x or y Equal"]
    D --> F["Rearrange one equation, to make x or y the subject"]
    E --> G["Add or Subtract Equations, to Eliminate One Variable"]
    F --> H["Substitute Expression into, the Second Equation"]
    G --> I["Solve for Remaining Variable"]
    H --> I
    I --> J["Substitute Value Back, to Find Second Variable"]
    J --> K["Check Solution in, Both Original Equations"]
    K --> L["End: Solution Found"]

Method 1: Elimination

This method involves adding or subtracting the two equations to eliminate one of the variables. It's often the quickest method when both equations are in a similar form.

Step-by-Step Walkthrough

Let's solve:

( 3x + 2y = 8 )

( 5x - 2y = 16 )

Step 1: Check if coefficients match.
Look at the coefficients (the numbers in front) of x or y. Here, the y terms are +2y and -2y. They are already equal in size but opposite in sign. Perfect for elimination.

Step 2: Add the equations to eliminate y.
Because we have +2y and -2y, adding them will cancel y out.

(3x + 2y) + (5x - 2y) = 8 + 16
3x + 5x + 2y - 2y = 24
8x = 24

Step 3: Solve for x.
x = 24 ÷ 8
x = 3

Step 4: Substitute x = 3 into one of the original equations.
It's usually easier to pick the simpler equation. Let's use equation (1):
3(3) + 2y = 8
9 + 2y = 8
2y = 8 - 9
2y = -1
y = -0.5

Step 5: Check your solution.
Always check by substituting both values into the equation you didn't use in Step 4 (equation 2):
5(3) - 2(-0.5) = 15 - (-1) = 15 + 1 = 16 ✓ Correct.

Solution: ( x = 3, \; y = -0.5 ) or as a coordinate ( (3, -0.5) ).

What if the coefficients don't match?

You need to multiply one or both equations by a number to make the coefficients of one variable equal.

Example:

( 2x + 3y = 5 )

( 4x - y = 11 )

To eliminate x, multiply equation (1) by 2: ( 4x + 6y = 10 ).
Now subtract the new equation from equation (2):
(4x - y) - (4x + 6y) = 11 - 10
4x - y - 4x - 6y = 1
-7y = 1
y = -17
Then substitute back to find x.

Method 2: Substitution

This method is useful when one equation is easy to rearrange, or if one variable has a coefficient of 1. You substitute one equation into the other.

Step-by-Step Walkthrough

Let's solve:

( y = 2x - 1 )

( 3x + 4y = 17 )

Step 1: Rearrange one equation (if needed).
Equation (1) already has y as the subject: ( y = 2x - 1 ).

Step 2: Substitute this expression into the other equation.
Replace every y in equation (2) with ( (2x -
1) ):
3x + 4(2x -
1) = 17

Step 3: Solve for x.
3x + 8x - 4 = 17
11x - 4 = 17
11x = 21
x = 2111 or x = 11011

Step 4: Substitute this x value back into the rearranged equation.
y = 2 × 2111 - 1
y = 4211 - 1111
y = 3111 or y = 2911

Step 5: Check in equation (2).
3(2111) + 4(3111) = 6311 + 12411 = 18711 = 17 ✓ Correct.

Which Method Should I Use?

Use this simple flowchart to decide:

flowchart TD
    Start["Look at the two equations"] --> Decision1["Is one variable already, the subject e.g. y = ...?"]
    Decision1 -- Yes --> Sub["Use SUBSTITUTION"]
    Decision1 -- No --> Decision2["Are coefficients of x or y, identical or easy to match?"]
    Decision2 -- Yes --> Elim["Use ELIMINATION"]
    Decision2 -- No --> Sub2["Try SUBSTITUTION after, rearranging one equation"]

Practice Makes Perfect: Try These!

Question 1 (Elimination):
5x + y = 13
3x - y = 7

Question 2 (Substitution):
x = 4y + 3
2x - 3y = 32

Question 3 (Harder - you choose):
4x + 3y = 2.5
5x - 6y = 14

(Answers at the bottom of the post.)

Top Tips for GCSE Exam Success

Always label your equations (1) and (2) to keep your working clear.

Show every step of your working to secure method marks, even if your final answer is wrong.

Check your answer by substituting into both original equations.

Present your final answer clearly. Write "x = ..., y = ..." or as a coordinate pair.

Practice with worded problems. Exam questions often set simultaneous equations in a context, like "Two coffees and three teas cost £8.50...". Your first job is to write the equations: ( 2c + 3t = 8.5 ), etc.

Mastering these two methods will give you the confidence to tackle any simultaneous equation question that comes up in your GCSE Maths exam. Good luck with your revision!

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Answers:

( x = 2.5, \; y = 0.5 )

( x = 13, \; y = 2.5 )

( x = 2, \; y = -1.5 ) (Hint: Multiply the first equation by 2 to match the y coefficients.)