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Edexcel GCSE Maths (Non-calculator) Smart practice A mixed round aimed at your weak spots
Unit 1
Number
Unit 2
Algebra
Unit 3
Ratio, proportion and rates of change
Unit 4
Geometry and measures
Unit 5
Probability
Unit 6
Statistics
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Expressions, equations, formulae and identities

Practise 19 exam-style questions on Expressions, equations, formulae and identities 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.

  • 19 questions
  • 1–5 marks each
  • Instant marking
  • No sign-up needed
19 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

Here is an expression.
\[6x^2 + 9x\]

Which of the following are factors of \(6x^2 + 9x\)?
\(3\) \(6\) \(x\) \(3x\) \(9x\) \(2x + 3\)

Model answer

\(6x^2 + 9x = 3x(2x + 3)\)
So the factors in the list are 3, \(x\), \(3x\) and \(2x + 3\).
6 is not a factor because \(9x\) is not a multiple of 6; \(9x\) is not a factor because \(6x^2\) is not a multiple of \(9x\).

Explanation

Factorise fully first: the highest common factor is \(3x\), giving \(3x(2x + 3)\). Any factor of \(3x\), and the bracket itself, divides the expression exactly. A factor must divide every term, so 6 and \(9x\) do not work.

Mark scheme
  1. 2 marks

    B2 for 3, \(x\), \(3x\) and \(2x + 3\) only

    B1 for two or three of these with no incorrect answers, or all four with one incorrect answer (6 or 9x)

Question 2

2 marks

Sophie is asked to solve
\[7x + 3 - 2x = 5x + 3\]
She cannot find a value for \(x\).

Explain why.

Model answer

\(7x + 3 - 2x = 5x + 3\)
So the statement is \(5x + 3 = 5x + 3\).
Both sides are identical, so it is true for every value of \(x\). It is an identity, not an equation, so there is no single value to find.

Explanation

Simplify both sides before trying to solve. If the two sides become exactly the same, the \(x\) terms cancel and you are left with something always true, like \(3 = 3\). That means every value of \(x\) works, so the statement is an identity.

Mark scheme
  1. 1 mark

    M1 for simplifying the left-hand side to \(5x + 3\)

  2. 1 mark

    C1 for explaining that both sides are the same, so the statement is true for every value of \(x\) (it is an identity)

    oe, e.g. "it simplifies to 3 = 3, so any value of x works"

Question 3

3 marks

\[(x + a)(x + 5) \equiv x^2 + bx + 15\]
where \(a\) and \(b\) are constants.

Work out the value of \(a\) and the value of \(b\).

Model answer

\((x + a)(x + 5) = x^2 + 5x + ax + 5a = x^2 + (a + 5)x + 5a\)
Constant terms: \(5a = 15\), so \(a = 3\)
Coefficients of \(x\): \(b = a + 5 = 8\)

Explanation

Because it is an identity, the two sides must match term by term. Compare the constant terms first to find \(a\), then compare the coefficients of \(x\) to find \(b\). Check: \((x + 3)(x + 5) = x^2 + 8x + 15\).

Mark scheme
  1. 1 mark

    M1 for expanding to \(x^2 + 5x + ax + 5a\)

    oe, e.g. \(x^2 + (a + 5)x + 5a\); correct answer with no working scores full marks

  2. 1 mark

    A1 for \(a = 3\)

    from 5a = 15

  3. 1 mark

    A1 for \(b = 8\)

    ft \(b = a + 5\) for their a

Question 4

4 marks

\[2(3x + a) - b(x - 2) \equiv x + 14\]
where \(a\) and \(b\) are constants.

Work out the value of \(a\) and the value of \(b\).

Model answer

\(2(3x + a) - b(x - 2) = 6x + 2a - bx + 2b = (6 - b)x + (2a + 2b)\)
Coefficients of \(x\): \(6 - b = 1\), so \(b = 5\)
Constants: \(2a + 2b = 14\), so \(2a + 10 = 14\) and \(a = 2\)

Explanation

Expand, then group the terms in \(x\) and the constant terms. For an identity, the coefficients of \(x\) must be equal on both sides, and so must the constants. Note \(-b \times -2 = +2b\). Check: \(2(3x + 2) - 5(x - 2) = 6x + 4 - 5x + 10 = x + 14\).

Mark scheme
  1. 1 mark

    M1 for expanding both brackets correctly, \(6x + 2a - bx + 2b\)

    correct answer with no working scores full marks

  2. 1 mark

    M1 for equating coefficients, e.g. \(6 - b = 1\) or \(2a + 2b = 14\)

  3. 1 mark

    A1 for \(b = 5\)

  4. 1 mark

    A1 for \(a = 2\)

    ft their b in \(2a + 2b = 14\)

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