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A5Changing the subject

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

Practise Changing the subject. 2 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
Zara makes \(x\) the subject of \(y = \frac{3x + 2}{5}\). Here is her working.
Line 1: \(5y = 3x + 2\)
Line 2: \(5y + 2 = 3x\)
Line 3: \(x = \frac{5y + 2}{3}\)
(a) Write down the line in which Zara made a mistake, and explain the mistake.[1]
(b) Write down the correct answer.[1]
Show the answer and mark scheme
(a) Answer: Line 2: she should subtract 2 from both sides, not add 2.
  • C1 for Line 2 with a correct explanation (2 should be subtracted from both sides)

Worked solution: To remove + 2 from the right-hand side, subtract 2 from both sides: \(5y - 2 = 3x\). Zara added 2.

(b) Answer: \(x = \frac{5y - 2}{3}\)
  • B1 for \(x = \frac{5y - 2}{3}\) oe

Worked solution: \(5y - 2 = 3x\), so \(x = \frac{5y - 2}{3}\).

Question 2Medium2 marks
Make \(p\) the subject of the formula \(q = 2(p - 2r)\)[2]
Show the answer and mark scheme
Answer: \(p = \frac{q + 4r}{2}\)
  • M1 for \(\frac{q}{2} = p - 2r\) or \(q = 2p - 4r\)
  • A1 for \(p = \frac{q + 4r}{2}\) oe

Worked solution: \(\frac{q}{2} = p - 2r\), so \(p = \frac{q}{2} + 2r\) or \(p = \frac{q + 4r}{2}\)

Question 3Hard5 marks
\(v = \frac{2t + 1}{t - 3}\)
(a) Make \(t\) the subject of the formula.[4]
(b) Explain why \(v\) can never be equal to 2.[1]
Show the answer and mark scheme
(a) Answer: \(t = \frac{3v + 1}{v - 2}\)
  • M1 for \(v(t - 3) = 2t + 1\)
  • M1 for collecting the \(t\) terms on one side: \(vt - 2t = 3v + 1\)
  • M1 for \(t(v - 2) = 3v + 1\)
  • A1 for \(t = \frac{3v + 1}{v - 2}\) oe

Worked solution: \(v(t - 3) = 2t + 1\)
\(vt - 3v = 2t + 1\)
\(vt - 2t = 3v + 1\)
\(t(v - 2) = 3v + 1\)
\(t = \frac{3v + 1}{v - 2}\)

(b) Answer: If \(v = 2\) then \(2(t - 3) = 2t + 1\), so \(-6 = 1\), which is impossible (equivalently, the denominator \(v - 2\) would be 0).
  • C1 for a correct reason, e.g. the denominator \(v - 2\) would be zero (you cannot divide by 0), or substituting \(v = 2\) gives \(-6 = 1\)

Worked solution: With \(v = 2\): \(2t - 6 = 2t + 1\) gives \(-6 = 1\), which is false for every \(t\). So \(v\) can never be 2.

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