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GCSE simultaneous equations practice and worked examples

Find values that make two linear equations true at the same time. Work from matching coefficients to elimination and a context problem.

Foundation and Higher · Useful near the top of Foundation and for Higher revision. Later questions include more demanding arithmetic.

Core pathway. Core GCSE methods are useful across Foundation and Higher. Start with the core examples and questions.

Higher extension. More demanding selections are marked Higher extension: practice questions 6, 7. Use these after the core work, following your teacher’s guidance.

2 worked examples and 8 original practice questions · Allow 25–40 minutes · Read online or print. Higher extensions are labelled. No account or email needed.

Learn the methodTry the questions

Printing includes the method, examples and space for working, followed by a separate answer section.

A useful approach

  1. Choose a variable to eliminate; multiply a whole equation if needed to make matching coefficients.
  2. Add or subtract the equations to remove that variable, then solve the remaining equation.
  3. Substitute into an original equation to find the other variable. Check both equations.

A mistake to watch for

Multiply every term, including the constant. When subtracting equations, subtract negative terms carefully.

Learn each method, then practise it

Read each line of working and explain why it follows from the previous line.

Eliminate matching coefficients

Add or subtract the two equations to remove one unknown. Solve for the other and substitute back.

Worked example 1

Simultaneous Equations · Core GCSE skill · Calculator allowed · 3 marks

Solve the simultaneous equations

x + y = 11

x - y = 3

  1. Add the two equations: 2x = 14, so x = 7
  2. Substitute into x + y = 11: 7 + y = 11, so y = 4
  3. Check in the second equation: 7 - 4 = 3 ✓
Answer: x = 7, y = 4
Report a problem with this question

Now practise this method: Question 1 · Question 2 · Question 3

Make coefficients match first

Multiply every term of an equation by the same number before eliminating a variable. Check both original equations.

Worked example 2

Simultaneous Equations · Core GCSE skill · No calculator · 3 marks

Solve the simultaneous equations

3x + 2y = 19

x + 4y = 13

  1. Multiply the first equation by 2: 6x + 4y = 38
  2. Subtract the second equation: 5x = 25, so x = 5
  3. Substitute into x + 4y = 13: 5 + 4y = 13, so y = 2
  4. Check in the first equation: 15 + 4 = 19 ✓
Answer: x = 5, y = 2
Report a problem with this question

Now practise this method: Question 4 · Question 5 · Question 6 · Question 7 · Question 8

Your practice questions

Write your working on paper. Marks indicate how much working to show; these questions are self-marked and do not change saved practice results. Use squared paper for drawing questions.

Worked answers: Simultaneous equations

Compare the reasoning as well as the final answer. Another correct method is valid. If a step is unclear, revisit an example before trying a similar question.

Answer 1

Show answer and working for question 1
  1. Subtract the second equation from the first: 2x = 10, so x = 5
  2. Substitute into x + y = 7: 5 + y = 7, so y = 2
  3. Check in the first equation: 15 + 2 = 17 ✓
Answer: x = 5, y = 2
Back to question 1

Answer 2

Show answer and working for question 2
  1. Subtract the second equation from the first: 2y = 4, so y = 2
  2. Substitute into 2x + y = 8: 2x + 2 = 8, so 2x = 6 and x = 3
  3. Check in the first equation: 6 + 6 = 12 ✓
Answer: x = 3, y = 2
Back to question 2

Answer 3

Show answer and working for question 3
  1. Add the two equations to eliminate y: 6x = 24, so x = 4
  2. Substitute into 2x + y = 10: 8 + y = 10, so y = 2
  3. Check in the first equation: 16 - 2 = 14 ✓
Answer: x = 4, y = 2
Back to question 3

Answer 4

Show answer and working for question 4
  1. Multiply the first equation by 3: 3x + 6y = 12
  2. Subtract the second equation: y = 3
  3. Substitute into x + 2y = 4: x + 6 = 4, so x = -2
  4. Check in the second equation: -6 + 15 = 9 ✓
Answer: x = -2, y = 3
Back to question 4

Answer 5

Show answer and working for question 5
  1. Multiply the first equation by 5 and the second by 2: 15x + 10y = 120 and 4x + 10y = 54
  2. Subtract: 11x = 66, so x = 6
  3. Substitute into 3x + 2y = 24: 18 + 2y = 24, so y = 3
  4. Check in the second equation: 12 + 15 = 27 ✓
Answer: x = 6, y = 3
Back to question 5

Answer 6

Show answer and working for question 6
  1. Multiply the first equation by 2 and the second by 3: 8x + 6y = 10 and 18x - 6y = 42
  2. Add: 26x = 52, so x = 2
  3. Substitute into 4x + 3y = 5: 8 + 3y = 5, so 3y = -3 and y = -1
  4. Check in the second equation: 12 + 2 = 14 ✓
Answer: x = 2, y = -1
Back to question 6

Answer 7

Show answer and working for question 7
  1. Multiply the first equation by 3 and the second by 5: 18x + 15y = 30 and 20x - 15y = 65
  2. Add: 38x = 95, so x = = 2.5
  3. Substitute into 6x + 5y = 10: 15 + 5y = 10, so 5y = -5 and y = -1
  4. Check in the second equation: 10 + 3 = 13 ✓
Answer: x = 2.5, y = -1
Back to question 7

Answer 8

Show answer and working for question 8
  1. Let t be the cost of a tea and c the cost of a coffee, in pounds: 3t + 2c = 7.80 and 2t + 3c = 8.20
  2. Multiply the first equation by 3 and the second by 2: 9t + 6c = 23.40 and 4t + 6c = 16.40
  3. Subtract: 5t = 7.00, so t = 1.40
  4. Substitute into 3t + 2c = 7.80: 4.20 + 2c = 7.80, so 2c = 3.60 and c = 1.80
  5. Check in the second equation: 2 × 1.40 + 3 × 1.80 = 2.80 + 5.40 = £8.20 ✓
Answer: (tea) £1.40   (coffee) £1.80
Back to question 8

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