Changing the subject of a formula · Expressions, formulae & change of subject

The subject of a formula is the letter on its own on one side. In $v = u + at$ , the subject is $v$ . Changing the subject means rearranging so a different letter stands alone. WAEC Paper 2 asks for it almost every year, usually followed by “hence find its value when…”.

The idea: undo, in reverse order

Think about what happens to the letter you want, starting from the letter itself. In $v = u + at$ , start with $t$ : it is multiplied by $a$ , then $u$ is added. To get $t$ back, undo those steps in reverse order: take away $u$ first, then divide by $a$ . And like an equation, whatever you do to one side, you do to the other.

Build it, then undo itOpposite operations, last one first

Undo it, in reverse orderMake t the subject

v = u + at

t
× a
+ u

v = u + at

Start from t and follow the chain: that's the order the operations were done. To get t back, undo them last first, like taking off your shoes before your socks.

Key idea

Follow what happens to the new subject, in order.
Undo those operations last one first, doing each to both sides.

Undo this	with this
$+$	$-$
$\times$	$\div$
square	square root
square root	square
power $\frac23$	power $\frac32$

Roots and powers

If the letter is inside a square root, get the root on its own first, then square both sides. Square the whole side, not just part of it: $\left(\frac{k}{m}\right)^2 = \frac{k^2}{m^2}$ .

Worked example · WAEC 2022

WAEC 2022 · Paper 1 · Q18

Make $t$ the subject of $k = m\sqrt{\dfrac{t - p}{r}}$ .

Get the root alone
$t$ is inside a square root, which is multiplied by $m$ .
Think first. The root is multiplied by $m$ . What do you do to both sides?
Divide both sides by m
$\frac{k}{m} = \sqrt{\frac{t - p}{r}}$
Square both sides
$\frac{k^2}{m^2} = \frac{t - p}{r}$
Think first. Now $t - p$ is divided by $r$ . What comes next?
Multiply by r, then add p
$\begin{aligned} \frac{k^2 r}{m^2} &= t - p \\ t &= \frac{k^2 r}{m^2} + p = \frac{k^2 r + p m^2}{m^2} \end{aligned}$
The answer is B.

More: formulas with roots and powers

When the letter appears twice

If the new subject appears in two places, you can’t undo straight away. Gather it up:

Clear any fractions (multiply both sides by the denominator).
Expand brackets.
Move every term with the new subject to one side, everything else to the other.
Take the new subject out as a common factor.
Divide by the bracket that’s left.

For $y = \dfrac{3w + 1}{2w - 5}$ , making $w$ the subject:

\begin{aligned} y(2w - 5) &= 3w + 1 \\ 2yw - 5y &= 3w + 1 \\ 2yw - 3w &= 5y + 1 \\ w(2y - 3) &= 5y + 1 \\ w &= \frac{5y + 1}{2y - 3} \end{aligned}

More: when the letter appears twice

Your turn

WAEC 2021 · Paper 2 · Q2

Expressions, formulae & change of subject

Given that $S = \sqrt{\dfrac{5d(L - d)}{7}}$ ,

(a)
make $L$ the subject of the relation;
(b)
find, correct to two decimal places, the value of $L$ when $d = 27$ and $S = 0.7$ .

Worked solution (try it first)

(a)

$L$ is inside a square root, so square both sides: $S^2 = \frac{5d(L - d)}{7}$ .
Multiply both sides by 7: $7S^2 = 5d(L - d) = 5dL - 5d^2$ .
Add $5d^2$ to both sides: $7S^2 + 5d^2 = 5dL$ .
Divide by $5d$ : $L = \frac{7S^2 + 5d^2}{5d}$ .

(b)

Substitute $d = 27$ and $S = 0.7$ : $L = \frac{7 \times 0.49 + 5 \times 729}{5 \times 27}$
$= \frac{3.43 + 3645}{135}$
$= \frac{3648.43}{135}$
$\approx 27.025$ .
Correct to two decimal places: $L = 27.03$ .

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