Algebra

Changing the subject of a formula: the algebra skill O-Level students underestimate

By Mr Gan · O-Level E Maths · Updated June 2026 · 7 min read

To change the subject of a formula, isolate the target variable on one side of the equation using inverse operations in reverse order: undo addition/subtraction first, then multiplication/division, then powers/roots last. If the target variable appears more than once, group all its terms on one side, factorise it out, then divide. This single method handles every O-Level "make x the subject" question.

Why this topic quietly costs marks

Changing the subject of a formula doesn't look hard in isolation — most students who are comfortable solving linear equations can rearrange y = mx + c to find x without thinking twice. The problem is that O-Level questions disguise it inside mensuration, kinematics, and finance formulas, and the target variable is often buried inside a square root, a squared term, or a fraction. Students who only practised the simple version freeze when the formula looks unfamiliar.

It also shows up as a hidden requirement inside larger questions — for example, a question asks for the radius of a sphere given its volume, which secretly requires rearranging V = (4/3)πr³ before you can substitute. Missing this step means the whole question collapses, even if the student understood the geometry perfectly.

The subject of a formula — what it actually means

In a formula like V = πr²h, V is the subject. The subject is a variable that appears alone, on one side, and nowhere else in the equation. "Changing the subject" means rearranging the formula so a different variable — say, h — takes that position instead, giving h = V / (πr²).

The order of operations rule — and why it reverses

When you evaluate an expression, you follow BODMAS: brackets, powers, then multiplication/division, then addition/subtraction. When you're rearranging to isolate a variable, you peel away operations in the opposite order — undo addition and subtraction first, because they were applied last in BODMAS, and they're now the outermost layer wrapped around your target variable.

Step order when isolating

1. Undo + / −
2. Undo × / ÷
3. Undo powers / roots
4. Undo brackets (expand or factor)

Inverse operation pairs

+ ↔ −
× ↔ ÷
squared ↔ square root
cubed ↔ cube root

Watch the whole method in about a minute.

Worked example 1 — basic linear formula

Make x the subject of y = 3x − 7.

Solution

x is wrapped in two layers: it's multiplied by 3, then 7 is subtracted. Undo the subtraction first — add 7 to both sides:
y + 7 = 3x

Undo the multiplication — divide both sides by 3:
(y + 7) / 3 = x

Write x on the left, as is conventional:
x = (y + 7) / 3

Worked example 2 — target variable inside a square root

The period of a pendulum is given by T = 2π√(L/g). Make L the subject.

Solution

Isolate the square root first — divide both sides by 2π:
T / (2π) = √(L/g)

Undo the square root by squaring both sides:
(T / 2π)² = L/g

Undo the division by g — multiply both sides by g:
L = g(T / 2π)²

This can also be written as L = gT² / 4π² by expanding the square — either form is accepted.

Common mistake

Students often square only the T, not the entire fraction T/(2π). You must square everything that sits on that side of the equation: (T/2π)² = T²/4π², not T²/2π. Use brackets to keep the whole expression together until you're ready to expand.

Worked example 3 — target variable appears twice

Make x the subject of ax + 3 = bx − 5.

Solution

x appears on both sides. Collect all x-terms on one side and all constants on the other:
ax − bx = −5 − 3

Simplify the right side:
ax − bx = −8

Factorise x out from the left side — this is the step students most often forget:
x(a − b) = −8

Divide both sides by (a − b):
x = −8 / (a − b)

Recognise this pattern instantly: any time the target variable appears more than once in the formula, the method is always collect → factorise → divide. There is no shortcut around factorising — attempting to divide before grouping the terms will not isolate the variable correctly.

Worked example 4 — target variable in the denominator

Make r the subject of A = πr² + 2πrh, where the formula represents total surface area of a cylinder including one circular base. (Simplified version: make r the subject of 1/u + 1/v = 1/f for the variable u — a common O-Level lens formula style question.)

Solution — for 1/u + 1/v = 1/f, make u the subject

Isolate the term containing u:
1/u = 1/f − 1/v

Combine the right side into a single fraction using a common denominator (fv):
1/u = (v − f) / fv

Both sides are now reciprocals of each other — flip both sides upside down (this is the fastest way to isolate u when it's in a denominator):
u = fv / (v − f)

Common mistake

Students often try to "cross multiply" immediately without combining the right-hand side into one fraction first. Combine into a single fraction before flipping — trying to manipulate two separate fractions at once leads to errors almost every time.

A 4-question checklist before you start

Does the target variable appear once or more than once? If more than once, you will need to factorise — expect this step.

Is the target variable inside a root or a power? Plan to undo that last, after isolating it.

Is the target variable in a denominator? You'll likely need to cross-multiply or combine fractions before isolating it.

After rearranging, substitute your answer back into the original formula with simple numbers to check it works.

Frequently asked questions

What's the difference between "evaluate" and "make x the subject"?

Evaluating means substituting numbers into a formula to get a numerical answer. Making x the subject means rearranging the formula algebraically so x stands alone — no numbers are substituted, the answer is still in terms of other letters.

Do I always divide by the coefficient last?

Not always — it depends on what's wrapped around the variable. The rule is to undo operations in reverse BODMAS order: addition/subtraction first, then multiplication/division, then powers/roots. Work outward to inward, based on what's closest to your target variable.

What if the target variable is negative after rearranging, like −x = 5?

Multiply (or divide) both sides by −1 to flip the sign: x = −5. This is a final cleanup step and is required for full marks — leaving the answer as −x = 5 is considered incomplete.

Is there a difference between O-Level Sec 3 and Sec 4 versions of this topic?

The method is identical throughout. What changes is the complexity of the formulas — Sec 3 typically uses linear formulas, including the ones behind direct and inverse proportion, while Sec 4 introduces formulas with the variable inside roots, squares, or appearing twice, often disguised inside mensuration or science-context questions.

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