Edexcel · GCSE Maths · 1MA1 · Foundation and Higher
M9 · Formulae and rearranging
Revision notes, worked examples and methods for formulae and rearranging.
Notes and quizzes ready · 50 questions · Video coming soon.
Foundation shows shared notes and questions. Higher includes the shared content and labelled Higher extensions. Changing tier starts the notes and a new quiz set.
Revise the key ideas
Using and rearranging formulae
The subject of a formula is the variable on its own, usually on the left. In A = lw, A is the subject. Changing the subject rearranges the relation without changing its meaning.
Treat both sides equally and undo operations in reverse order. From y = 3x + 5, subtract 5 from both sides, then divide by 3: x = y − 53.
Worked example: From v = u + at, subtract u and divide by t: a = v − ut, where t ≠ 0. Check the result by multiplying both sides by t and adding u.
If the subject is inside a bracket, undo any outside operation first. For P = 2(l + w), l = P2 − w.
If A = πr2, divide by π, then take the positive root for a radius: r = √(Aπ). Geometrical lengths cannot be negative.
Modelling and examples
Worked example: A taxi charges £4 plus £1.80 per kilometre. C = 4 + 1.8d. If C = £22, then d = (22 − 4) ÷ 1.8 = 10 km.
Use consistent units in formulae. In distance = speed × time, a speed in km/h requires time in hours if distance is to be in km.
Write a formula by identifying fixed and changing quantities. For n identical tickets costing £p each with one £q booking fee, total cost is T = np + q.
Substitute into the original formula to check a rearrangement. If y = 3x + 5 with x = 4 gives y = 17, the rearranged formula should recover x = 4 from y = 17.
Higher — the subject appearing more than once
When the required variable appears in several terms, collect its terms on one side and factorise it out. Moving only one occurrence does not finish the rearrangement.
Worked example: y = ax + bx gives y = x(a + b), so x = ya + b, provided a + b ≠ 0.
Worked example: y = 3x + 2x − 1. Multiply by x − 1: yx − y = 3x + 2. Collect x terms: x(y − 3) = y + 2, so x = y + 2y − 3, where y ≠ 3 and the original x ≠ 1.
With squared variables, algebraic solutions may need ±: y = x2 + 4 gives x = ±√(y − 4), for y ≥ 4. Context may select a positive solution.
State restrictions introduced by division or roots. Dividing by a quantity that could be zero can lose valid cases or create undefined expressions.
Test yourself
50 questions · Sets of 10 from the selected tier. For fractions, use / when typing; for powers, use superscripts or ^. Follow each question's answer format. These quick checks support revision; practise full written solutions and proofs too.
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