Practise 36 exam-style questions on Simultaneous equations for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
36 questions
1–5 marks each
Instant marking
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20 questions, marked as you go
Sample questions with model answers
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Question 1
2 marks
Leah is solving the simultaneous equations \(3x + 2y = 16\) \(x + 2y = 8\) Here is her working. Subtract: \(2x = 24\), so \(x = 12\) \(12 + 2y = 8\), so \(y = -2\)
Leah's answer is wrong. Explain her mistake and find the correct solution.
Model answer
When she subtracted, Leah worked out \(16 + 8 = 24\) on the right instead of \(16 - 8 = 8\). \(2x = 8\), so \(x = 4\) \(4 + 2y = 8\), so \(y = 2\)
Explanation
When you subtract one equation from another, you must subtract every term, including the numbers on the right-hand side. Always check a solution in both original equations: \(3 \times 4 + 2 \times 2 = 16\).
Mark scheme
1 mark
C1 for identifying the mistake, e.g. she added the right-hand sides instead of subtracting: \(16 - 8 = 8\), not 24
From the second equation, \(2y = 4x\), so \(y = 2x\) \(6x + 4(2x) = 7\) \(14x = 7\) \(x = \dfrac{1}{2}\) \(y = 2 \times \dfrac{1}{2} = 1\)
Explanation
The second equation rearranges easily to \(y = 2x\), so substitution is quick here (elimination works too). Solutions do not have to be whole numbers. Check: \(6 \times 0.5 + 4 \times 1 = 7\).
Mark scheme
1 mark
M1 for a correct method to eliminate one variable, e.g. doubling the second equation and adding to get \(14x = 7\)
or substituting \(y = 2x\) into the first equation; correct answer with no working scores full marks
1 mark
M1 for substituting their value to find the other variable
1 mark
A1 for \(x = \dfrac{1}{2},\ y = 1\)
oe, e.g. \(x = 0.5\)
Question 3
3 marks
In a café, 2 coffees and 3 muffins cost £10.30 3 coffees and 2 muffins cost £10.70
Maria has £20. Does she have enough money to buy 4 coffees and 4 muffins? You must show your working.
Model answer
\(2c + 3m = 10.30\) \(3c + 2m = 10.70\) Add: \(5c + 5m = 21.00\) So \(c + m = 4.20\) 4 coffees and 4 muffins cost \(4 \times 4.20 = 16.80\), which is £16.80. Yes, £20 is enough (she would have £3.20 left).
Explanation
You do not always need to find each value: adding the equations gives the cost of 5 coffees and 5 muffins straight away. Solving fully (a coffee costs £2.30 and a muffin £1.90) also works. Finish with a clear yes or no and the total cost.
Mark scheme
1 mark
P1 for forming two equations, e.g. \(2c + 3m = 10.30\) and \(3c + 2m = 10.70\)
oe, in pounds or pence
1 mark
P1 for a correct method, e.g. adding to get \(5c + 5m = 21\), or solving to get \(c = 2.30\) and \(m = 1.90\)
1 mark
C1 for "Yes" with £16.80 as the cost of 4 coffees and 4 muffins
Question 4
4 marks
A rectangle has a perimeter of 22 cm and an area of 30 cm².
Use algebra to find the length and the width of the rectangle.
Model answer
Let the length be \(l\) cm and the width \(w\) cm. \(2l + 2w = 22\), so \(w = 11 - l\) \(lw = 30\) \(l(11 - l) = 30\) \(11l - l^2 = 30\) \(l^2 - 11l + 30 = 0\) \((l - 5)(l - 6) = 0\) \(l = 5\) or \(l = 6\) The rectangle is 6 cm long and 5 cm wide.
Explanation
The perimeter gives a linear equation and the area gives a non-linear one, so this is a linear/quadratic pair. Substituting from the linear equation gives a quadratic; its two solutions are the length and the width. Check: \(2(6 + 5) = 22\) and \(6 \times 5 = 30\).
Mark scheme
1 mark
P1 for forming two equations, e.g. \(2l + 2w = 22\) and \(lw = 30\)
oe, e.g. \(l + w = 11\)
1 mark
P1 for substituting to get a quadratic, e.g. \(l(11 - l) = 30\) leading to \(l^2 - 11l + 30 = 0\)
1 mark
P1 for solving their quadratic, e.g. \((l - 5)(l - 6) = 0\)
1 mark
A1 for 6 cm and 5 cm
answer only with no algebra scores 0
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