Simultaneous equations are a pair of equations in two unknowns, usually x and y, that must both be true at the same time. Free notes and 11 questions answered.
Updated 21 Mar 2026 · Revise Right Team
Solving simultaneous equations algebraically is part of Solving equations and inequalities in GCSE AQA Higher Maths.
4 min
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GCSE AQA Higher
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What this lesson covers
Simultaneous equations are a pair of equations in two unknowns, usually x and y, that must both be true at the same time. Free notes and 11 questions answered.
Solve two simultaneous linear equations in two variables using the elimination method
Solve two simultaneous linear equations in two variables using the substitution method
Interpret the solution of a pair of simultaneous equations as the coordinates of the intersection point of two straight lines
Apply simultaneous equation methods to exam-style problems, showing full algebraic working
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Use this lesson to stay inside Solving equations and inequalities while connecting the detail back to the wider Algebra topic.
Simultaneous equations: A pair of equations in two variables that must both be satisfied at the same time. The solution is the unique pair of values that makes both equations true simultaneously.
Elimination method: An algebraic technique for solving simultaneous equations by adding or subtracting multiples of the equations to remove one variable, leaving a single equation in one unknown.
Substitution method: An algebraic technique for solving simultaneous equations by rearranging one equation to express one variable in terms of the other, then substituting that expression into the second equation.
Solution of simultaneous equations: The pair of values $(x, y)$ that satisfies both equations. Geometrically, this is the point where the two straight lines intersect on a coordinate grid.
Revision notes
Read the core explanations from Solving simultaneous equations algebraically before testing yourself from memory.
1. Why Two Equations at Once?
Imagine you know that two numbers add to 10, and their difference is 4. Either fact alone gives you infinitely many possibilities. Put both facts together and there is exactly one answer. That is the power of simultaneous equations: two constraints, one unique solution.
In GCSE Maths, you will meet pairs of linear equations in two unknowns — typically $x$ and $y$. Your job is to find the single pair of values that satisfies both equations at the same time. There are two reliable algebraic methods: elimination and substitution. This lesson covers both.
The graph makes the idea concrete: the solution $(3, 2)$ sits on both lines. Algebraic methods let you find that point precisely, without drawing anything — which is exactly what the exam expects.
2. The Elimination Method
Elimination works by combining the two equations so that one variable cancels out. Once you are left with a single equation in one unknown, you can solve it directly, then substitute back to find the other variable.
The key decision is whether to add or subtract. If the coefficients of one variable are equal in magnitude and the same sign, subtract. If they are equal in magnitude but opposite in sign, add. If neither coefficient matches, multiply one or both equations first to create a match.
Worked example: Elimination — Equal Coefficients (No Scaling Needed)
Solve the simultaneous equations: $3x + 2y = 16$ and $x + 2y = 8$
Equation (1): $3x + 2y = 16$
Equation (2): $x + 2y = 8$ — Labelling keeps your working organised and earns method marks.
$(3x + 2y) - (x + 2y) = 16 - 8$
$2x = 8$ — The $2y$ terms are identical, so subtracting removes $y$ entirely. [M1]
$x = \frac{8}{2} = 4$ — Divide both sides by 2. [A1]
$4 + 2y = 8$
$2y = 4$
$y = 2$ — Use the simpler equation to find $y$. [A1]
$3(4) + 2(2) = 12 + 4 = 16$ ✓ — Substituting back into the equation you did NOT use to find $y$ confirms the answer is correct.
Answer: $x = 4$, $y = 2$
That example was clean because the $y$ coefficients already matched. Most exam questions require you to scale one or both equations first — here is how that works.
Worked example: Elimination — Scaling One Equation First
Solve the simultaneous equations: $5x + 2y = 19$ and $3x - y = 8$
Equation (1): $5x + 2y = 19$
Equation (2): $3x - y = 8$ — Identify which variable to eliminate. The $y$ coefficients are 2 and $-1$. Multiply (2) by 2 to match.
$2 \times (3x - y) = 2 \times 8$
$6x - 2y = 16$ — call this equation (3) — Every term must be multiplied by 2. [M1]
$(5x + 2y) + (6x - 2y) = 19 + 16$
$11x = 35$ — $+2y$ and $-2y$ cancel. We add because the signs are opposite. [M1]
$x = \frac{35}{11}$ — Leave as a fraction — do not round mid-calculation. [A1]
$3 \times \frac{35}{11} - y = 8$
$\frac{105}{11} - y = 8$
$y = \frac{105}{11} - 8 = \frac{105}{11} - \frac{88}{11} = \frac{17}{11}$ — Convert 8 to elevenths before subtracting. [A1]
Substitution takes a different route: rearrange one equation to write one variable in terms of the other, then plug that expression into the second equation. You end up with one equation in one unknown — which you can solve directly.
Substitution is particularly efficient when one equation already has a variable with a coefficient of 1 (or $-1$), because the rearrangement is straightforward and avoids fractions. Look for that before you start.
Worked example: Substitution — Standard Exam Example
Solve the simultaneous equations: $y = 2x - 1$ and $3x + 4y = 18$
Equation (1): $y = 2x - 1$
Equation (2): $3x + 4y = 18$ — Equation (1) already expresses $y$ in terms of $x$. Substitute directly into (2). [M1]
$3x + 4(2x - 1) = 18$ — Replace every $y$ in equation (2) with the expression $(2x - 1)$. Use brackets to avoid sign errors.
Both elimination and substitution reach the same answer — the method you choose does not affect the marks awarded, as long as your working is clear and complete. Choose whichever feels more natural for the structure of the equations in front of you.
4. Choosing Your Method
There is no single rule that always picks the faster method, but the table below gives you a reliable decision framework for the types of equations you will see at GCSE.
Equation structure
Suggested method
Reason
One equation already has $y =...$ or $x =...$
Substitution
No rearrangement needed; substitute directly.
One variable has coefficient 1 in either equation
Substitution
Rearranging gives a whole-number expression, avoiding fractions.
Coefficients of one variable are equal (or easily made equal)
Elimination
One multiply and one add/subtract removes a variable cleanly.
Both equations in $ax + by = c$ form with no obvious coefficient of 1
Elimination
Scale both equations to match one variable's coefficient.
5. Deliberate Practice
Now it is your turn. Work through each pair of equations using the method you judge most appropriate. Show all steps — method marks are available even if your final answer contains an arithmetic slip.
6. Exam Technique and Common Mistakes
Simultaneous equation questions typically carry 3–4 marks. The mark scheme awards marks for the correct method (even with an arithmetic error), for finding one correct value, and for finding both correct values. Showing clear, labelled working protects your method marks even when you slip on arithmetic.
The most reliable habit is to always verify your solution by substituting both values back into the equation you did not use to find the second variable. If the two sides do not balance, you know to recheck before moving on — not after the exam.
Simultaneous equations also appear in worded problems. The skill there is setting up the equations correctly from the context — once you have two equations in two unknowns, the algebraic method is identical to everything covered above.
Worked example: Exam-Style Worded Problem
Two friends, Aisha and Ben, buy snacks at a café. Aisha buys 3 sandwiches and 2 drinks for £11.50. Ben buys 1 sandwich and 4 drinks for £9.50. Find the cost of one sandwich and one drink. [4 marks]
Let $s$ = cost of one sandwich (£) and $d$ = cost of one drink (£).
Equation (1): $3s + 2d = 11.50$
Equation (2): $s + 4d = 9.50$ — Defining variables clearly earns the first method mark. [M1]
$3 \times (s + 4d) = 3 \times 9.50$
$3s + 12d = 28.50$ — call this equation (3) — Multiply every term, including the right-hand side. [M1]
$3(2.70) + 2(1.70) = 8.10 + 3.40 = 11.50$ ✓ — Both values satisfy equation (1). State the answer in context.
Answer: One sandwich costs £2.70 and one drink costs £1.70.
Every simultaneous equation problem at GCSE — whether presented as bare algebra or as a worded context — reduces to the same core skill: form two equations, apply elimination or substitution systematically, and verify. Master the method on straightforward pairs first, and the worded versions become a translation exercise rather than a new challenge.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-21.
Simultaneous equations are a pair of equations in two unknowns, usually x and y, that must both be true at the same time. Two constraints give one unique solution, which is also the intersection point of the two straight lines.
How do you solve simultaneous equations word problems?
Define variables, form two equations, then solve. If 3 sandwiches and 2 drinks cost £11.50 and 1 sandwich and 4 drinks cost £9.50, then 3s + 2d = 11.50 and s + 4d = 9.50, giving one sandwich at £2.70 and one drink at £1.70.
Do you add or subtract simultaneous equations?
In elimination, subtract the equations when the matching coefficients have the same sign, and add them when they have opposite signs. If no coefficients match, multiply one or both equations first, for example multiplying 3x - y = 8 by 2 to match 5x + 2y = 19.
Should you use elimination or substitution?
Use substitution when one equation already has y = or x = form, or a variable has a coefficient of 1. Use elimination when the coefficients of one variable are equal or easily made equal; both methods reach the same answer.
What do you do when the coefficients do not match in simultaneous equations?
When no coefficients match in simultaneous equations, multiply one or both equations so that one variable has equal coefficients. For 5x + 2y = 19 and 3x - y = 8, multiplying the second by 2 gives 6x - 2y = 16, and adding removes y.
How do you check the answer to simultaneous equations?
The answer to simultaneous equations is checked by substituting both values into the equation not used to find the second variable. For x = 4 and y = 2, 3(4) + 2(2) = 12 + 4 = 16 confirms the solution of 3x + 2y = 16.
Can simultaneous equations have fraction answers?
Simultaneous equations can have fraction answers, which should be left as fractions rather than rounded mid-calculation. For example, 5x + 2y = 19 and 3x - y = 8 give x = 35/11 and y = 17/11.
How do you substitute when y = 2x - 1?
When one equation is y = 2x - 1, replace every y in the other equation with (2x - 1) in brackets. For 3x + 4(2x - 1) = 18, this gives 11x - 4 = 18, so x = 2 and then y = 2(2) - 1 = 3.
How many marks are simultaneous equations questions worth?
Simultaneous equations questions typically carry 3–4 marks, given for the correct method, for one correct value and for both correct values. Clear, labelled working keeps the method marks even if an arithmetic slip happens.
Why should you label simultaneous equations?
Labelling simultaneous equations as (1) and (2) keeps the working organised and earns method marks. It makes clear which equation is being multiplied, added or subtracted at each step.
How do you solve 2x + y = 7 and x - 3y = 1?
2x + y = 7 and x - 3y = 1 are solved by rearranging the second to x = 1 + 3y and substituting: 2(1 + 3y) + y = 7 gives 7y = 5, so y = 5/7. Then x = 1 + 15/7 = 22/7.