Simultaneous Equations — Elimination Method

Simultaneous equations involve two equations with two unknowns, and you need to find values that satisfy both equations at the same time. The elimination method works by adding or subtracting the equations to remove one variable, leaving a single equation in one unknown that you can solve directly. This topic appears on both Foundation and Higher tiers across AQA, Edexcel, and OCR, and it is one of the most commonly examined algebra skills after basic equation solving. On this page you will learn when and how to use elimination, see fully worked examples for both straightforward and trickier cases, and practise with exam-style questions.

What Are Simultaneous Equations?

Simultaneous equations are a pair (or set) of equations that share the same unknowns. The solution is the pair of values that makes both equations true at the same time.

For example: 2x + y = 7 and 3x − y = 8.

Graphically, solving simultaneous equations means finding the point where two straight lines intersect.

To eliminate a variable, make the coefficients of that variable the same in both equations, then add or subtract.

If the matching coefficients have the same sign, subtract the equations. If they have opposite signs, add them.

A helpful mnemonic: Same Signs Subtract, Different signs Do add (SSSD).

Step-by-Step Method

Label the equations ① and ② for easy reference.
Choose which variable to eliminate. Pick the one whose coefficients are easiest to match.
Multiply one or both equations so that the chosen variable has the same coefficient in both.
Add or subtract the equations to eliminate that variable.
Solve the resulting single equation.
Substitute the value you found back into one of the original equations to find the other variable.
Check by substituting both values into the other original equation.

When coefficients already match

Sometimes the coefficients already match (e.g., both equations have 2y). In this case, skip straight to step 4.

When you need to multiply both equations

If the coefficients do not share a simple multiple, multiply each equation by a different number to create matching coefficients. For instance, if one equation has 3x and the other has 5x, multiply the first by 5 and the second by 3 to get 15x in both.

Worked Example 1 — Foundation Level

Question: Solve simultaneously: 3x + 2y = 16 and x + 2y = 10.

Working:

Label: ① 3x + 2y = 16 and ② x + 2y = 10.

Step 1: The y coefficients are already the same (both 2y), with the same sign.

Step 2: Subtract ② from ①.

(3x + 2y) − (x + 2y) = 16 − 10

2x = 6

x = 3

Step 3: Substitute x = 3 into ②.

3 + 2y = 10

2y = 7

y = 3.5

Check using ①: 3(3) + 2(3.5) = 9 + 7 = 16 ✓

Answer: x = 3, y = 3.5

Worked Example 2 — Higher Level

Question: Solve simultaneously: 4x + 3y = 23 and 5x − 2y = 10.

Working:

Label: ① 4x + 3y = 23 and ② 5x − 2y = 10.

Step 1: Eliminate y. Multiply ① by 2 and ② by 3 to make the y coefficients equal (both 6y).

①×2: 8x + 6y = 46 ②×3: 15x − 6y = 30

Step 2: The y terms have opposite signs (+6y and −6y), so add the equations.

8x + 15x + 6y − 6y = 46 + 30

23x = 76

x = 76/23

Hmm — let us re-examine. Actually:

(8x + 6y) + (15x − 6y) = 46 + 30

23x = 76

x = 76 ÷ 23. This does not simplify neatly. Let us re-check the original question: 4(76/23) + 3y = 23 gives a fractional y. This is fine — not all simultaneous equations have whole-number answers, especially on Higher papers.

x = 76/23 ≈ 3.30 (to 2 d.p.)

Substitute into ②: 5(76/23) − 2y = 10 → 380/23 − 2y = 10 → 2y = 380/23 − 230/23 = 150/23 → y = 75/23 ≈ 3.26.

Let us instead use a cleaner example. Restarting with: 3x + 2y = 19 and 5x − 2y = 13.

Label: ① 3x + 2y = 19 and ② 5x − 2y = 13.

Step 1: The y coefficients are already matched: +2y and −2y (opposite signs), so add.

(3x + 2y) + (5x − 2y) = 19 + 13

8x = 32

x = 4

Step 2: Substitute into ①: 3(4) + 2y = 19 → 12 + 2y = 19 → 2y = 7 → y = 3.5.

Check using ②: 5(4) − 2(3.5) = 20 − 7 = 13 ✓

Answer: x = 4, y = 3.5

Higher extension — requiring multiplication of both equations:

Solve: 2x + 3y = 12 ① and 5x + 4y = 23 ②.

Eliminate x: multiply ① by 5 and ② by 2.

①×5: 10x + 15y = 60 ②×2: 10x + 8y = 46

Subtract: 7y = 14 → y = 2.

Substitute y = 2 into ①: 2x + 6 = 12 → 2x = 6 → x = 3.

Check using ②: 5(3) + 4(2) = 15 + 8 = 23 ✓

Answer: x = 3, y = 2

Common Mistakes

Subtracting when you should add (or vice versa). Use the SSSD rule: same signs subtract, different signs add. Write the signs explicitly before deciding.
Forgetting to multiply every term. When you multiply an equation by 3, every term — including the constant on the right — must be multiplied.
Substituting back into the wrong equation. It does not matter which you use, but always check using the other equation to verify your answer.
Arithmetic errors in the addition/subtraction step. Write the equations directly above each other and line up the terms. This makes addition and subtraction much clearer.
Giving only one value. You must state both x and y. A single value does not fully solve the system.

Exam Tips

Label your equations ① and ② and refer to them by number throughout. Examiners appreciate structured working.
Show the multiplication step clearly. Write "①×2:" before the new equation. This earns explicit method marks.
Always check in the equation you did not use for substitution. If both values satisfy both equations, you can be confident in your answer.
If the answer is a fraction or decimal, that is fine. Not every simultaneous equation has neat whole-number answers, especially on Higher papers.

Practice Questions

Q1 (Foundation): Solve: 2x + y = 11 and 2x − y = 5.

Answer: Add the equations: 4x = 16, x = 4. Substitute: y = 3. Solution: x = 4, y = 3.

Q2 (Foundation/Higher): Solve: 3x + 4y = 26 and x + 4y = 14.

Answer: Subtract: 2x = 12, x = 6. Substitute: 6 + 4y = 14, y = 2. Solution: x = 6, y = 2.

Q3 (Higher): Solve: 4x + 3y = 29 and 6x + y = 27.

Answer: Multiply ② by 3: 18x + 3y = 81. Subtract ①: 14x = 52, x = 26/7. However, for neater working multiply ② by 3 and subtract from it: 18x + 3y − 4x − 3y = 81 − 29, 14x = 52, x = 52/14 = 26/7. Then y = 27 − 6(26/7) = 27 − 156/7 = 33/7. Solution: x = 26/7, y = 33/7.

Ready to sharpen your simultaneous equations skills? Start revising with GCSEMathsAI — our AI tutor generates unlimited practice pairs and shows you every step of the elimination process.

For more on key algebra formulas, see GCSE Maths Formulas You Must Know.

Summary

Simultaneous equations have two unknowns and require two equations to solve.
The elimination method removes one variable by adding or subtracting the equations.
Match the coefficients by multiplying one or both equations as needed.
Same signs: subtract. Different signs: add.
Substitute the found value back into either original equation to find the second unknown.
Always check your solution in the equation you did not use for substitution.
This method complements the substitution method, and you should be comfortable with both for the exam.

Simultaneous Equations: Elimination –