Changing the Subject of a Formula
Changing the subject rearranges a formula so that a different pronumeral stands alone, using the same inverse steps as solving an equation. This NSW Year 9 Advanced maths subtopic includes powers, roots and subjects that appear twice (NESA outcome MA5‑EQU‑P‑02).
You will rearrange formulas such as \(v = u + at\), \(C = \frac{5}{9}(F - 32)\) and \(A = \pi r^2\), and handle roots, as in \(T = 2\pi\sqrt{\frac{L}{g}}\). Harder questions gather the subject from two terms, as in \(b = \frac{a + 1}{a - 1}\), use a rearranged formula for a tin's height or a car's speed, and correct a wrong rearrangement of \(v^2 = u^2 + 2as\). It is part of Year 9 Advanced Equations.
Theory
To change the subject of a formula, rearrange it so one chosen letter is on its own on one side. Use inverse operations in reverse order, doing the same to both sides, just as when solving an equation. This Year 9 Advanced maths page covers fractions and brackets, squares and roots, formulas where the subject appears twice, reciprocal formulas, and using the new formula.
The subject of a formula is the letter on its own on one side. In \(A=\pi r^2\), \(A\) is the subject.
To change the subject, treat the formula like an equation to solve for the new letter. Every step is done to both sides.
Use inverse operations: \(+\) and \(-\) undo each other, \(\times\) and \(\div\) undo each other, and squaring and square-rooting undo each other (as do cubing and cube-rooting).
Undo the operations on the new subject in reverse order: the last thing done to it is the first thing undone.
If the new subject appears in two places, collect those terms on one side and factorise it out. Then divide by the bracket.
| Operation on the subject | Undo it by |
|---|---|
| \(+a\) | \(-a\) |
| \(\times a\) | \(\div a\) |
| divided by \(a\) | \(\times a\) |
| squared | \(\pm\sqrt{\ }\) (or \(+\sqrt{\ }\) if it must be positive) |
| square root | square both sides |
| cubed | \(\sqrt[3]{\ }\) |
Subject in a reciprocal: combine the other side into one fraction, then flip both sides.
How to change the subject
- Clear fractions by multiplying both sides by the denominator, and square both sides to remove a square root containing the subject.
- Expand brackets if the subject is inside one and appears elsewhere too; otherwise divide by the bracket's multiplier.
- Isolate the subject's term by moving every other term to the other side.
- Divide by the coefficient (or factorised bracket), then take a root if the subject is squared or cubed.
- Check by substituting simple numbers into the old and new formulas.
Using and checking a rearranged formula
- Substitute into the rearranged formula once, rather than rearranging with numbers each time (Examples 2 and 4).
- Square every factor: \((3k)^2 = 9k^2\), not \(3k^2\).
- Units: convert where needed after substituting, for example m/s to km/h by multiplying by \(3.6\).
- Spotting an error: test a rearrangement with numbers. If the values do not satisfy the original formula, find the faulty step. A square root cannot be taken term by term: \(\sqrt{a^2-b^2}\ne a-b\).
- Restrictions: a division by a bracket needs that bracket not to be zero; state conditions such as \(y\ne 2\) when asked.
Undo the operations on \(r\) in reverse order: \(\div 3\), then \(\times\pi h\), then squaring.
| \(3V\) | \(=\) | \(\pi r^2h\) |
| \(\dfrac{3V}{\pi h}\) | \(=\) | \(r^2\) |
| \(r\) | \(=\) | \(\boldsymbol{\sqrt{\dfrac{3V}{\pi h}}}\) |
Only the positive root is needed, because a radius is positive.
(i) Make \(h\) the subject.
(ii) With \(g=9.8\), find the height that gives a speed of \(14\) m/s.
(i) To undo a square root, square both sides.
| \(v^2\) | \(=\) | \(2gh\) |
| \(h\) | \(=\) | \(\boldsymbol{\dfrac{v^2}{2g}}\) |
(ii) Substitute into the new formula.
| \(h\) | \(=\) | \(\dfrac{14^2}{2\times 9.8} = \dfrac{196}{19.6}\) |
| \(\) | \(=\) | \(\mathbf{10}\text{ m}\) |
Check: \(\sqrt{2\times 9.8\times 10}=\sqrt{196}=14\) ✓
Clear the fraction, expand, collect every \(x\)-term on one side, then factorise out \(x\).
| \(y(x+4)\) | \(=\) | \(2x-3\) |
| \(xy+4y\) | \(=\) | \(2x-3\) |
| \(xy-2x\) | \(=\) | \(-3-4y\) |
| \(x(y-2)\) | \(=\) | \(-(3+4y)\) |
| \(x\) | \(=\) | \(\dfrac{-(3+4y)}{y-2} = \boldsymbol{\dfrac{3+4y}{2-y}}\) |
Check with \(x=1\): \(y=\dfrac{-1}{5}=-0.2\), and \(\dfrac{3+4(-0.2)}{2-(-0.2)}\) \(=\dfrac{2.2}{2.2}\) \(=1\) ✓. Multiplying top and bottom by \(-1\) tidies the signs.
(i) Make \(r\) the subject.
(ii) \(\$5000\) grows to \(\$5600\) in \(3\) years. Find the interest rate.
(i) Divide by \(P\) to remove the bracket's multiplier, then undo the \(+1\) and the \(\times n\).
| \(\dfrac{A}{P}\) | \(=\) | \(1+rn\) |
| \(\dfrac{A}{P}-1\) | \(=\) | \(rn\) |
| \(r\) | \(=\) | \(\dfrac{A-P}{Pn}\) |
\(\dfrac{A}{P}-1=\dfrac{A-P}{P}\), so dividing by \(n\) gives \(\boldsymbol{r=\dfrac{A-P}{Pn}}\).
(ii)
| \(r\) | \(=\) | \(\dfrac{5600-5000}{5000\times 3} = \dfrac{600}{15\,000}\) |
| \(\) | \(=\) | \(0.04 = \mathbf{4\%}\text{ a year}\) |
Common pitfalls
Frequently asked questions
How do you change the subject of a formula?
Treat it like an equation and solve for the new letter. Undo the operations on that letter in reverse order, doing the same thing to both sides each time, until it is on its own.
What do you do when the subject appears twice?
Clear any fractions and expand brackets. Move every term containing the subject to one side and everything else to the other, factorise the subject out, then divide by the bracket.
How do you make a squared letter the subject?
Get the squared term on its own, then take the square root of both sides. Write \(\pm\), unless the letter stands for something that must be positive, such as a length.
How do you remove a square root when changing the subject?
Get the square root on its own on one side, then square both sides. Square every factor on the other side too.
How can I check my rearranged formula?
Choose simple values, work out the original subject with the old formula, then put the values into your new formula. You should get back the value you started with.