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Solving simple simultaneous equations - Exam Questions.docx

Built from the lesson script on 30 September 2026.

EDEXCEL GCSE MATHS · FOUNDATION & HIGHER

Exam Practice: Solving Simple Simultaneous Equations

Solving simple simultaneous equations · Equations and inequalities · Lesson 5 of 7 · 18 marks · 30 minutes

Name

Date

 

Instructions

• Answer all the questions.

• Write your answers in the spaces provided.

• The marks for each question are shown in brackets - use this as a guide to how much to write.

• The answers are on separate pages at the back. Attempt every question before you look at them.

• Answer all questions. Show your working.

Question 1 NON-CALCULATOR (3 marks)

Solve the simultaneous equations x + y = 7 and x − y = 1.

(Total for Question 1 = 3 marks)

Question 2 NON-CALCULATOR (3 marks)

Solve the simultaneous equations 2x + y = 11 and x + y = 7.

(Total for Question 2 = 3 marks)

Question 3 NON-CALCULATOR (3 marks)

Solve the simultaneous equations y = 2x − 1 and x + y = 8.

(Total for Question 3 = 3 marks)

Question 4 NON-CALCULATOR (4 marks)

2 adult tickets and 3 child tickets cost £27 in total. 1 adult ticket and 2 child tickets cost £15 in total. Work out the cost of an adult ticket and the cost of a child ticket.

(Total for Question 4 = 4 marks)

Question 5 NON-CALCULATOR (2 marks)

The graph shows the lines y = 2x − 3 and x + y = 6. Use the graph to solve the simultaneous equations y = 2x − 3 and x + y = 6.

(Total for Question 5 = 2 marks)

Question 6 NON-CALCULATOR (3 marks)

Solve the simultaneous equations 3x + 2y = 16 and x − y = 2.

(Total for Question 6 = 3 marks)

TOTAL FOR PAPER = 18 MARKS

 

Answers and mark scheme

Check your answer only once you have written one.

Question 1 (3 marks)

Adding gives 2x = 8, so x = 4. Then y = 3.

• Adding or a correct elimination M1

• x = 4 A1

• y = 3 A1

Question 2 (3 marks)

Subtracting gives x = 4. Then y = 7 − 4 = 3.

• Subtracting the equations M1

• x = 4 A1

• y = 3 A1

Question 3 (3 marks)

x + 2x − 1 = 8, so 3x = 9, x = 3 and y = 5.

• Substituting M1

• x = 3 A1

• y = 5 A1

Question 4 (4 marks)

2a + 3c = 27 and a + 2c = 15. a = 15 − 2c, so 30 − 4c + 3c = 27, c = 3, a = 9. Adult £9, child £3.

• Forming both equations M1

• A correct elimination or substitution M1

• One value correct A1

• Both values with units A1

Question 5 (2 marks)

The lines cross at (3, 3), so x = 3 and y = 3.

• Reading the crossing point M1

• x = 3 and y = 3 A1

Question 6 (3 marks)

x = y + 2, so 3(y + 2) + 2y = 16, 5y = 10, y = 2 and x = 4.

• Substituting or multiplying to match M1

• y = 2 A1

• x = 4 A1