Practise 36 exam-style questions on Simultaneous equations for Edexcel GCSE Maths (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
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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
The grid shows the circle \(x^2 + y^2 = 16\) and the line \(y = x - 1\)
Use the graphs to find estimates for the solutions of the simultaneous equations \(x^2 + y^2 = 16\) \(y = x - 1\)
Model answer
The solutions are the points where the line crosses the circle. They cross at about (3.3, 2.3) and (−2.3, −3.3). \(x \approx 3.3, y \approx 2.3\) or \(x \approx -2.3, y \approx -3.3\)
Explanation
Each point where the line crosses the circle satisfies both equations. Read both coordinates of each crossing point, using the small squares (0.2 each). Every point on the line has \(y = x - 1\), which is a useful check on your readings.
Mark scheme
1 mark
B1 for one solution: \(x\) in the range 3.1 to 3.5 with \(y\) in the range 2.1 to 2.5
exact values \(\frac{1 + \sqrt{31}}{2} = 3.28...\), y = 2.28..; correct answer with no working scores full marks
1 mark
B1 for the other solution: \(x\) in the range −2.5 to −2.1 with \(y\) in the range −3.5 to −3.1
exact values x = −2.28..., y = −3.28...
Question 2
3 marks
Two numbers have a sum of 23.4 and a difference of 6.8
By forming and solving simultaneous equations, find the two numbers.
Model answer
\(x + y = 23.4\) \(x - y = 6.8\) Add: \(2x = 30.2\), so \(x = 15.1\) \(y = 23.4 - 15.1 = 8.3\) The numbers are 15.1 and 8.3
Explanation
Write one equation for the sum and one for the difference. Adding them eliminates \(y\). Check: \(15.1 + 8.3 = 23.4\) and \(15.1 - 8.3 = 6.8\).
Mark scheme
1 mark
P1 for forming both equations, e.g. \(x + y = 23.4\) and \(x - y = 6.8\)
oe; correct answer with no working scores full marks
1 mark
P1 for a correct method to eliminate one variable, e.g. \(2x = 30.2\)
oe
1 mark
A1 for 15.1 and 8.3
cao
Question 3
4 marks
At a theatre, 4 adult tickets and 3 child tickets cost a total of £71.10 2 adult tickets and 5 child tickets cost a total of £66
Work out the cost of one adult ticket and the cost of one child ticket.
Use a letter for each unknown price and write an equation for each sentence. Doubling the second equation makes the adult terms equal, so subtracting eliminates \(a\). Always check: \(4 \times 11.25 + 3 \times 8.70 = 71.10\).
Mark scheme
1 mark
P1 for forming both equations, e.g. \(4a + 3c = 71.1\) and \(2a + 5c = 66\)
oe; may use pence; correct answer with no working scores full marks
1 mark
P1 for a correct method to eliminate one variable, e.g. \(4a + 10c = 132\) and subtracting to get \(7c = 60.9\)
Replace \(y\) in the linear equation with \(3x^2 - 2\), then rearrange to a quadratic equal to 0. Find \(y\) from \(y = 6 - 2x\). Exact fractions are better than rounded decimals here; check \(3 \times \left(\frac{4}{3}\right)^2 - 2 = \frac{16}{3} - 2 = \frac{10}{3}\).
Mark scheme
1 mark
M1 for substituting, e.g. \(2x + 3x^2 - 2 = 6\)
oe, e.g. \(3x^2 - 2 = 6 - 2x\); correct answer with no working scores full marks
1 mark
M1 for \(3x^2 + 2x - 8 = 0\)
oe
1 mark
M1 for a correct method to solve, e.g. \((3x - 4)(x + 2) = 0\)
or a correct substitution into the formula
1 mark
A1 for \(x = \frac{4}{3}\) and \(x = -2\)
oe, accept 1.33 or better
1 mark
A1 for \(y = \frac{10}{3}\) and \(y = 10\), correctly paired
oe, accept 3.33 or better
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