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A1–A3Algebraic notation, substitution and identities

Edexcel GCSE Maths (1MA1), Higher tier · Algebra

Practise Algebraic notation, substitution and identities. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

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Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy2 marks
Show that \(4(2x - 3) - 2(x - 5) \equiv 2(3x - 1)\).[2]
Show the answer and mark scheme
Answer: \(8x - 12 - 2x + 10 = 6x - 2 = 2(3x - 1)\)
  • M1 for expanding both brackets: \(8x - 12 - 2x + 10\)
  • C1 for \(6x - 2 = 2(3x - 1)\) (or both sides shown equal to \(6x - 2\))

Worked solution: \(4(2x - 3) - 2(x - 5) = 8x - 12 - 2x + 10 = 6x - 2 = 2(3x - 1)\)

Question 2Medium1 mark
Simplify \(3p \times 3p^{2}\).[1]
Show the answer and mark scheme
Answer: \(9p^{3}\)
  • B1 for \(9p^{3}\)

Worked solution: In algebra, multiply the numbers and add the powers of the same letter: 3 × 3 = 9 and 1 + 2 = 3, so the answer is \(9p^{3}\).

Question 3Hard3 marks
Work out the value of \(20p^{2} - 24pq\) when \(p = \frac{1}{2}\) and \(q = \frac{1}{4}\).[3]
Show the answer and mark scheme
Answer: \(2\)
  • M1 for \(20 \times \left(\frac{1}{2}\right)^{2} = 5\)
  • M1 for \(24 \times \frac{1}{2} \times \frac{1}{4} = 3\)
  • A1 for 2 cao

Worked solution: \(20 \times \left(\frac{1}{2}\right)^{2} = 20 \times \frac{1}{4} = 5\) and \(24 \times \frac{1}{2} \times \frac{1}{4} = 3\).
\(5 - 3 = 2\).

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