Rearranging Formulae (Changing the Subject) — AQA GCSE Maths Revision Notes

What you'll learn

This topic covers rewriting a formula so that a different letter stands alone. By the end of this guide you should be able to change the subject of a simple formula, one containing brackets or fractions, and one where the new subject appears under a power or a root.

You should also be able to handle formulae where the new subject appears more than once, which is the hardest case on the specification, and know when rearranging is worth doing rather than simply substituting.

The organising idea is that rearranging a formula is solving an equation whose answer happens to be a letter. Every technique is one you already use: do the same to both sides, and undo the operations in the reverse of the order they were applied. Nothing new is required. What makes it feel harder is that the answer cannot be simplified into a single number, so you have to be willing to stop with an expression — and that the letter you want may be buried deeper than the letter currently in front. Recognising it as the same skill removes most of the difficulty.

Key terms and definitions

Formula — a rule connecting quantities, written with an equals sign.

Subject — the letter standing alone, usually on the left.

Change the subject — rewrite the formula so a different letter stands alone.

Inverse operation — the operation undoing another: + and −, × and ÷, powers and roots.

Factorise — take a common factor outside a bracket, which is the key step when the subject appears twice.

Term — a part of an expression separated by a plus or minus sign.

Core concepts

The method

Decide which letter you want as the subject, then peel away everything attached to it, undoing operations in the reverse of BIDMAS order — additions and subtractions first, then multiplications and divisions, then powers and roots last.

Whatever you do must be done to both sides, exactly as when solving an equation.

To make x the subject of y = 3x + 2: subtract 2 from both sides to get y − 2 = 3x, then divide both sides by 3 to get x = (y − 2)/3.

Note that the answer is an expression rather than a number, and that is the finished answer. Students sometimes keep going, looking for something tidier, and end up spoiling a correct result.

Which operation to undo first

Look at what is happening to the letter you want, from the outside in.

In y = 3x + 2, the x has been multiplied by 3 and then 2 has been added. The addition is the outermost operation, so it comes off first.

In y = 3(x + 2), the bracket means the addition happened first and the multiplication second, so the multiplication comes off first — divide both sides by 3 to get y/3 = x + 2, then subtract 2.

Reading the order of operations correctly is what decides the sequence of steps, and it is why two formulae that look similar are unpicked differently.

Formulae with fractions

Multiply both sides by the denominator to clear the fraction, remembering that the bar acts as a bracket around everything above it.

For y = (x + 4)/5: multiplying both sides by 5 gives 5y = x + 4, then subtracting 4 gives x = 5y − 4.

Where the letter you want is in the denominator, multiply by that denominator to bring it up to the top first. For y = 12/x, multiplying both sides by x gives xy = 12, then dividing by y gives x = 12/y.

That first move — multiplying by the letter you want — feels backwards but is exactly right: a letter cannot be isolated while it is stuck underneath.

Powers and roots

A power is undone by a root, and a root by a power.

For A = πr², divide both sides by π to get A/π = r², then square-root both sides to get r = √(A/π).

For y = √(x + 1), square both sides to get y² = x + 1, then subtract 1 to get x = y² − 1.

Two cautions. Square-rooting strictly gives a plus-or-minus answer, though in practical formulae — a radius, a length — only the positive root makes sense and is the expected answer. And squaring both sides applies to the whole of each side: for y = √x + 3, the 3 is outside the root, so subtract it before squaring.

When the new subject appears twice

This is the hardest case, and it always follows the same three steps.

Collect every term containing the wanted letter on one side, and everything else on the other. Factorise to take that letter outside a bracket. Divide by the bracket.

For y = ax + bx, both terms on the right contain x, so factorise to get y = x(a + b), then divide to get x = y/(a + b).

For a harder case, make x the subject of y = (x + 1)/(x − 2). Multiply both sides by (x − 2) to get y(x − 2) = x + 1, expand to get xy − 2y = x + 1, collect the x terms on one side to get xy − x = 2y + 1, factorise to get x(y − 1) = 2y + 1, and divide to get x = (2y + 1)/(y − 1).

Factorising is the step people fail to see. If the wanted letter appears in two terms and you are stuck, factorising is almost certainly the move.

Negative coefficients

If the letter ends up with a negative coefficient, either divide by that negative number or rearrange to avoid it.

For y = 5 − 2x: adding 2x to both sides gives y + 2x = 5, then subtracting y gives 2x = 5 − y, so x = (5 − y)/2.

Collecting the x terms on the side where they are positive avoids a sign error, and it is worth the extra half-line.

Checking a rearrangement

Substituting numbers checks the whole thing in seconds.

Take the original formula, pick easy values, and work out the other letter. Then put those numbers into your rearranged version and confirm it returns the value you started from.

For x = (y − 2)/3 from y = 3x + 2: if x = 4 then y = 14, and (14 − 2)/3 = 4. ✓

This catches almost every sign and order error, and it is far quicker than re-reading the algebra.

Rearrange or substitute?

Rearranging is worth doing when you need the same letter repeatedly, or when the question asks for it explicitly.

For a one-off value, substituting everything known and solving the resulting equation is usually easier, because you are then working with numbers rather than letters.

For A = ½bh with A = 36 and b = 9, substituting gives 36 = 4.5h and h = 8 — quicker than rearranging to h = 2A/b and then substituting.

Read the question: "make h the subject" wants the algebra, while "find h" usually does not.

Worked examples

Example 1: Brackets and the order of undoing

Make x the subject of y = 4(x − 3).

The bracket means 3 was subtracted first and the multiplication by 4 came second, so the multiplication is undone first.

Divide both sides by 4: y/4 = x − 3.

Add 3 to both sides: x = y/4 + 3.

Expanding first is also valid — y = 4x − 12, then y + 12 = 4x, giving x = (y + 12)/4 — and the two answers agree, since y/4 + 3 and (y + 12)/4 are the same expression.

Example 2: A root

Make h the subject of the formula for the period of a pendulum, T = 2π√(h/g).

Divide both sides by 2π: T/(2π) = √(h/g).

Square both sides — the whole of each side: T²/(4π²) = h/g.

Multiply both sides by g: h = gT²/(4π²).

Squaring only part of the left-hand side is the error to watch for; the bracket must take the whole expression.

Example 3: The subject appearing twice

Make t the subject of v = (3t + 5)/(t − 2).

Multiply both sides by (t − 2): v(t − 2) = 3t + 5.

Expand: vt − 2v = 3t + 5.

Collect the t terms on one side and everything else on the other: vt − 3t = 2v + 5.

Factorise out t: t(v − 3) = 2v + 5.

Divide by the bracket: t = (2v + 5)/(v − 3).

Every one of those five steps is routine; the only insight required is seeing that factorising is what frees the t.

Common mistakes and how to avoid them

Undoing operations in the wrong order. Work from the outside in, reversing BIDMAS.

Doing something to only one side. Write the operation beside the line and apply it to both.

Losing a sign when moving a term. A term changes sign as it crosses the equals sign, or simply add or subtract from both sides and let the sign look after itself.

Squaring or rooting only part of a side. The operation applies to the whole side.

Failing to factorise when the subject appears twice. Collect, factorise, divide.

Treating a fraction bar as covering one term. It acts as a bracket around everything above it.

Not stopping when finished. An expression is a valid answer; it will not reduce to a number.

Exam technique for "Rearranging Formulae"

Write each step on its own line with the operation noted at the side. Method marks are awarded for the sequence, and a visible sequence earns credit even when one step slips.

Identify the order the operations were applied in before starting, since that determines the order you undo them.

When the wanted letter appears twice, write "collect, factorise, divide" at the side and work through those three stages explicitly.

Check the finished rearrangement by substituting easy numbers into both versions.

Leave the answer as an expression, and do not attempt to simplify it into something it cannot become.

Read whether the question wants the rearrangement or just a value — for a single value, substituting first is usually faster.

Quick revision summary

Rearranging is solving an equation whose answer is a letter — same rules, same order, and the answer is an expression rather than a number.

Undo the operations in the reverse of BIDMAS order, doing the same to both sides each time.

Read the order the operations were applied in: y = 3x + 2 needs the addition off first, but y = 3(x + 2) needs the multiplication off first.

Clear a fraction by multiplying by the denominator, treating the bar as a bracket. If the wanted letter is in the denominator, multiply by it to bring it up first.

Undo a power with a root and a root with a power, applying the operation to the whole of each side.

When the wanted letter appears twice: collect its terms on one side, factorise it out, then divide by the bracket.

Check by substituting numbers into both the original and the rearranged version.