Rearranging formulas and repeated subjects
Isolate a chosen variable while respecting whole groups, repeated occurrences and roots.
Core: the subject occurs once without its powers or roots. Extended: the subject can occur twice, in a denominator, or under a power or root.
Before you begin
- Solve a linear equation and factorise a common term.
- Interpret a square root and both solutions of a square equation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Make x the subject of y = 2x + 5.
What you will learn
- Change the subject of a formula with inverse operations.
- Factor out a subject that appears twice in Extended work.
- Handle powers, roots and denominator restrictions in Extended formulas.
Isolate the complete group containing the subject
To make x the subject of y = 3x − 7, add seven to both sides and divide by three: x = (y + 7)/3. The fraction bar must cover the whole sum. Rearranging a formula gives a relationship for many possible inputs; it is not the same as substituting numbers to calculate one value.
For P = 2l + 2w, making w the subject gives P − 2l = 2w and w = (P − 2l)/2. You can also simplify this as P/2 − l. Check the rearrangement by substituting a known rectangle into both forms, with compatible units throughout.
Extended: collect repeated occurrences
For y = ax + bx, the subject x appears in both terms. Factorise to y = x(a + b), then divide to get x = y/(a + b), provided a + b ≠ 0. Dividing each term separately before collecting would leave x on both sides and obscure the common structure.
For y = (x + a)/(x + b), first require x ≠ −b. Multiply by x + b: yx + yb = x + a. Collect x terms: x(y − 1) = a − yb, giving x = (a − yb)/(y − 1) when y ≠ 1. The exceptional value y = 1 must be checked in the original formula; it may be impossible or may permit many x values depending on a and b.
Extended: reverse powers carefully
For A = πr², a physical radius is r = √(A/π), with A ≥ 0 and r ≥ 0. If r were an unrestricted real variable in a pure equation, both square-root signs could be relevant. State the context that selects the positive value rather than silently discarding a sign.
For v = √(u² + 2as), square the whole equality to get v² = u² + 2as. Hence s = (v² − u²)/(2a) for a ≠ 0. The original principal square root requires v ≥ 0; squaring alone could otherwise allow a negative v that does not satisfy the original expression. Recheck domains when the formula involves roots or division.
Pause and explain
Extended: from p = qx + rx, what should you do before dividing?
Worked example
Extended: make x the subject of y = (3x − 2)/(x + 4).
Show the worked solution
- The original denominator requires x ≠ −4. Multiply to get yx + 4y = 3x − 2.
- Collect: x(y − 3) = −2 − 4y.
- Divide to obtain x = (−2 − 4y)/(y − 3), for y ≠ 3. The original formula cannot give y = 3.
Answer x = (−2 − 4y)/(y − 3), y ≠ 3; original x ≠ −4
From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Make h the subject of A = bh/2, for b ≠ 0.
Give me a hint
Undo the division by two before dividing by b.
Compare my reasoning
- Multiply both sides by two: 2A = bh.
- Divide both sides by b.
- h = 2A/b; substituting b = 4 and h = 3 gives A = 6 and recovers h = 3.
h = 2A/b
Look for these in your work
- Applied inverses to complete groups.
- Kept the non-zero divisor condition.
Extended: make x the subject of p = 5x − qx.
Give me a hint
The two x terms have a shared factor.
Compare my reasoning
- Factorise the right side as x(5 − q).
- Divide by 5 − q when q ≠ 5.
- x = p/(5 − q). If q = 5, the original equation requires p = 0 and does not determine a unique x.
x = p/(5 − q), q ≠ 5
Look for these in your work
- Collected the repeated subject by factorisation.
- Identified the exceptional coefficient.
Extended: a cylindrical tank has V = πr²h. Find r in terms of V and h, then calculate it for V = 72π m³ and h = 8 m.
Give me a hint
A physical radius is non-negative; isolate its square first.
Compare my reasoning
- For h > 0, divide to get r² = V/(πh).
- Take the non-negative root: r = √[V/(πh)].
- Here r = √(72/8) = 3 m; π × 3² × 8 = 72π checks the volume.
r = √[V/(πh)]; 3 m
Look for these in your work
- Selected the root justified by a physical radius.
- Checked dimensions and substituted volume.
Common mistakes
- Dividing only one term of a sum.
- Leaving repeated subject terms unfactorised.
- Ignoring values that make a new divisor zero.
What can you explain now?
Make t the subject of v = u + at, for a ≠ 0.
Compare with the explanation
t = (v − u)/a
Subtract u from both sides, then divide the complete difference by a.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Rearrange a formula with its subject once, then one with its subject twice. Explain any restrictions before testing with numbers.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs contains 21; Coordinate geometry contains 10 teaching lessons and a mixed checkpoint. Nine earlier overviews support selected topics across the syllabus. The other six syllabus areas, full cumulative assessment and human teacher review are not yet complete. Written lesson practice is self-checked, not automatically graded. The checkpoint samples skills and does not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources