Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Algebra
Practise Changing the subject. 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.
Rearranging a formula so that a different letter is on its own, e.g. making r the subject of \(A = \pi r^2\). Simple rearrangements are on both tiers; Higher questions include harder powers and roots and the subject appearing twice.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
3
Rearrange a one-step formulaUse the inverse operation, e.g. \(P = x + 7\) gives \(x = P - 7\).
4
Rearrange a two-step formulaUndo the operations in reverse order, e.g. \(y = 3x - 5\) gives \(x = \frac{y + 5}{3}\).
5
Rearrange formulae with brackets or fractionsClear the fraction or bracket first, e.g. \(P = \frac{a + b}{2}\) gives \(a = 2P - b\).
5
Rearrange formulae with squares or square rootsUndo a square with a square root and a square root by squaring, e.g. \(A = \pi r^2\) gives \(r = \sqrt{\frac{A}{\pi}}\).
Notes
The method
The subject is the letter on its own on one side: in \(v = u + at\), v is the subject.
Rearranging is like solving an equation: do the same to both sides, using inverse operations.
Undo the operations in the reverse order to the order they were applied to the new subject.
Make x the subject of \(y = 4x - 9\): add 9 to get \(y + 9 = 4x\), then divide by 4 to get \(x = \frac{y + 9}{4}\).
Fractions, brackets, powers and roots
Clear fractions first by multiplying both sides by the denominator: \(T = \frac{k}{m}\) gives \(Tm = k\), so \(m = \frac{k}{T}\).
With a bracket, divide by the number outside or expand: \(P = 2(l + w)\) gives \(l = \frac{P}{2} - w\).
Undo a square with a square root: \(A = \pi r^2\) gives \(r^2 = \frac{A}{\pi}\), so \(r = \sqrt{\frac{A}{\pi}}\).
Undo a square root by squaring: \(t = \sqrt{\frac{h}{5}}\) gives \(t^2 = \frac{h}{5}\), so \(h = 5t^2\).
Cheatsheet
Do the same to both sides
Inverse pairs: + and −, × and ÷, squaring and square rooting
Clear fractions first: multiply both sides by the denominator
Undo operations in the reverse order
The final answer starts with the new subject on its own, e.g. \(r = \ldots\)
How to answer each type of question
Make a letter the subject (linear)
2 marks4
Move the terms without the new subject to the other side.
Divide by the number multiplying the new subject.
Example. Make t the subject of the formula \(w = 5t + 3k\)
Show the model answer
\(w - 3k = 5t\) M1 \(t = \frac{w - 3k}{5}\) A1
Rearrange a formula with a fraction
2 marks5
Multiply both sides by the denominator.
Divide by any letter multiplying the bracket.
Isolate the new subject.
Example. Make u the subject of \(s = \frac{(u + v)t}{2}\)
Check with numbers: choose a value for x, work out y from the original formula, then put y into your rearranged formula. You should get your x back.
Draw a function machine for the new subject, then run it backwards using inverse operations.
Higher: if the subject appears twice, say 'collect, factorise, divide'.
Where marks are lost
Only dividing one term: from \(y = 3x + 2\), writing \(\frac{y}{3} = x + 2\).
Square rooting term by term: \(\sqrt{a^2 + b^2}\) is not \(a + b\).
Leaving the subject on both sides, e.g. \(x = \frac{y - x}{2}\), which is not a finished rearrangement.
Dropping a minus sign when moving a term to the other side.
Exam technique
Write each step on a new line; each correct step can earn a method mark.
Your final answer must have the new subject on its own on one side and nowhere on the other side.
If a later part uses your rearranged formula, substitute into it carefully and keep brackets.
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 \(q = 7(p - 4r)\)[2]
Show the answer and mark scheme
Answer: \(p = \frac{q + 28r}{7}\)
M1 for \(\frac{q}{7} = p - 4r\) or \(q = 7p - 28r\)
A1 for \(p = \frac{q + 28r}{7}\) oe
Worked solution: \(\frac{q}{7} = p - 4r\), so \(p = \frac{q}{7} + 4r\) or \(p = \frac{q + 28r}{7}\)