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Changing the Subject

A formula is written to give one letter in terms of the rest. Rearranging it, so a different letter stands alone, lets you answer the question the formula was not originally built for.

MYP 4Standard MathsAlgebraCriteria A · C · D~10 min read

Written by Fayek Jawad Haider · MYP student

What "the subject" means

The subject of a formula is the letter on its own, usually to the left of the equals sign. In \(y = 3x + 4\) the subject is \(y\). Changing the subject to \(x\) means rearranging until it reads \(x = \ldots\).

Subject of a formula
The variable that stands alone on one side, expressed in terms of all the others.

The golden rule is balance: whatever you do to one side of the equals sign, you must do to the other. That keeps the two sides equal at every step.

Working with inverse operations

To free a letter, peel away everything attached to it using the inverse, the operation that undoes it. Addition is undone by subtraction, multiplication by division, and so on.

Worked example

Make \(x\) the subject of \(y = 3x + 4\).

1
Undo the \(+4\) by subtracting \(4\) from both sides: \(y - 4 = 3x\).
2
Undo the \(\times 3\) by dividing both sides by \(3\): \(\dfrac{y - 4}{3} = x\).
3
Write it with the subject on the left.
\(x = \dfrac{y - 4}{3}\)

Divide the whole side, not just one term

When you divide by \(3\), the entire left side becomes \(\dfrac{y - 4}{3}\). Writing \(y - \dfrac{4}{3}\) instead splits the side up and changes its value.

When the subject is multiplied

If the letter you want is multiplied by something, divide it out. If it is inside a product like \(2 \pi r\), divide by the whole coefficient in front of it.

Worked example

Make \(r\) the subject of \(C = 2 \pi r\).

1
The \(r\) is multiplied by \(2\pi\). Divide both sides by \(2\pi\).
2
This gives \(\dfrac{C}{2\pi} = r\).
\(r = \dfrac{C}{2\pi}\)

Order of undoing

Undo operations in the reverse order they were applied, the same way you take off a coat before a jumper. Deal with terms added or subtracted first, then the multiplying or dividing.

FormulaMake subjectResult
\(v = u + at\)\(t\)\(t = \dfrac{v - u}{a}\)
\(P = 2a + 2b\)\(a\)\(a = \dfrac{P - 2b}{2}\)
\(A = \tfrac{1}{2}bh\)\(h\)\(h = \dfrac{2A}{b}\)

Check yourself

1. Make \(x\) the subject of \(y = 5x - 2\). +

Add \(2\) to both sides: \(y + 2 = 5x\). Divide by \(5\). Answer: \(x = \dfrac{y + 2}{5}\).

2. Make \(h\) the subject of \(A = \tfrac{1}{2}bh\). +

Multiply both sides by \(2\): \(2A = bh\). Divide by \(b\). Answer: \(h = \dfrac{2A}{b}\).

3. Make \(t\) the subject of \(v = u + at\). +

Subtract \(u\): \(v - u = at\). Divide by \(a\). Answer: \(t = \dfrac{v - u}{a}\).

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