Linear equations - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 31 exam-style questions on Linear 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.

31 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

Solve \(5(2x - 3.4) = 21\)

Model answer

\(2x - 3.4 = 21 \div 5 = 4.2\)
\(2x = 7.6\)
\(x = 3.8\)

Explanation

You can either divide both sides by 5 first, or expand the bracket to get \(10x - 17 = 21\). Both lead to \(x = 3.8\). When expanding, remember to multiply both terms in the bracket by 5.

Mark scheme

1 mark
M1 for \(10x - 17 = 21\) or \(2x - 3.4 = 4.2\)
oe
1 mark
A1 for \(x = 3.8\)
cao; correct answer with no working scores full marks

Question 2

3 marks

Solve \(4(3x - 2) = 5(x + 6.2)\)
Give your answer correct to 3 significant figures.

Model answer

\(12x - 8 = 5x + 31\)
\(12x - 5x = 31 + 8\)
\(7x = 39\)
\(x = \frac{39}{7} = 5.5714...\)
\(x = 5.57\) (3 s.f.)

Explanation

Expand both brackets, then collect the x terms on one side and the numbers on the other. \(5 \times 6.2 = 31\), so the numbers stay simple until the final division. The exact answer is \(\frac{39}{7}\); the question asks for 3 significant figures.

Mark scheme

1 mark
M1 for expanding both brackets correctly: \(12x - 8 = 5x + 31\)
oe; correct answer with no working scores full marks
1 mark
M1 for collecting terms correctly, e.g. \(7x = 39\)
ft their expansion if only one error
1 mark
A1 for 5.57
awrt 5.57 (5.5714...); accept \(\frac{39}{7}\)

Question 3

3 marks

Priya thinks of a number.
She multiplies her number by 3.5 and then subtracts 12.
She gets the same answer when she adds 9 to her number and then multiplies by 1.5

Work out Priya's number.

Model answer

Let the number be \(n\).
\(3.5n - 12 = 1.5(n + 9)\)
\(3.5n - 12 = 1.5n + 13.5\)
\(2n = 25.5\)
\(n = 12.75\)

Explanation

Turn each set of instructions into an expression and set them equal. Take care with the order: "adds 9 and then multiplies by 1.5" is \(1.5(n + 9)\), not \(1.5n + 9\). Check: both give 32.625.

Mark scheme

1 mark
P1 for one correct expression, e.g. \(3.5n - 12\) or \(1.5(n + 9)\)
oe; correct answer with no working scores full marks
1 mark
P1 for forming the equation \(3.5n - 12 = 1.5(n + 9)\)
oe
1 mark
A1 for 12.75
cao

Question 4

5 marks

Two mobile phone plans charge as follows.

Plan A: £12 each month plus 8p for each minute of calls
Plan B: £25 each month plus 3.5p for each minute of calls

Work out the least number of whole minutes of calls in a month for which Plan B is cheaper than Plan A.
You must show your working.

Model answer

Let \(m\) be the number of minutes.
Plan A: \(12 + 0.08m\); Plan B: \(25 + 0.035m\)
\(12 + 0.08m = 25 + 0.035m\)
\(0.045m = 13\)
\(m = 288.88...\)
The plans cost the same at 288.88... minutes, and Plan B is cheaper for more minutes than this.
Check: 288 minutes: A = £35.04, B = £35.08 (A cheaper); 289 minutes: A = £35.12, B = £35.115 (B cheaper)
Least number of whole minutes = 289

Explanation

Write each cost in pounds (8p = £0.08, 3.5p = £0.035) and set them equal to find where the costs are the same. Plan B has the higher fixed charge but the lower rate per minute, so it becomes cheaper above 288.88... minutes. The answer must be a whole number of minutes, so round up to 289, not to the nearest.

Mark scheme

1 mark
P1 for an expression for the cost of Plan A, e.g. \(12 + 0.08m\) (pounds) or \(1200 + 8m\) (pence)
oe
1 mark
P1 for an expression for the cost of Plan B, e.g. \(25 + 0.035m\) or \(2500 + 3.5m\)
oe; units must be consistent
1 mark
P1 for forming an equation (or inequality), e.g. \(12 + 0.08m = 25 + 0.035m\)
oe
1 mark
P1 for \(m = 288.8...\)
oe, e.g. \(0.045m = 13\) leading to 288.8...
1 mark
A1 for 289
cao; 288 or 288.9 scores P1 P1 P1 P1 A0

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