Rearranging Formulas

Changing the subject of a formula: two steps, fractions, squares and roots, and the subject appearing twice.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024 · 95 real past-paper questions below

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What Rearranging Means

“Make x the subject” means get x on its own on one side. The other side is the rest of the formula.

Example

You already do this with numbers. If $5 x + 2 = 17$ , you take 2 and divide by 5. Rearranging does the same moves with letters.

Rule

Undo what was done to x, last thing first. Whatever you do to one side, do to the other.

Key idea

The answer has letters in it, like $x = \frac{y - c}{5}$ . That is fine: it is still x on its own.

Key idea

A weather app that shows both °C and °F uses $F = 1.8 C + 32$ . To go back the other way it rearranges it, last thing first: $C = \frac{F - 32}{1.8}$ . So $98.6^{\circ} F$ is $\frac{66.6}{1.8} = 37^{\circ} C$ , normal body temperature.

Undo, Last Thing First

Most rearranging questions are two moves, worth 2 marks. Undo what was done to x, last thing first.

Worked example

1
$y = 5 x + c$ — Make x the subject. To build y, x was multiplied by 5, then c was added. Undo them in reverse order.
2
$y - c = 5 x$ — c was added last, so undo it first: take c from both sides.
3
$\frac{y - c}{5} = x$ — x was multiplied by 5, so divide both sides by 5. The whole of y − c goes on top.
4
$x = \frac{y - c}{5}$ — Swap the sides so x is written first. It means exactly the same.

Common mistake

Writing $x = y - \frac{c}{5}$ . The whole of $y - c$ is divided by 5, so it all goes on top of the fraction.

A Minus in Front of the Letter

Exams often put a minus in front of the letter you want, like $v = 8 - 3 t$ . Here is how to keep the signs right.

Worked example

1
$v = 8 - 3 t$ — Make t the subject. The t term is −3t, so it has a minus in front.
2
$v + 3 t = 8$ — Add 3t to both sides. Now the t term is positive, so there is no minus to lose.
3
$3 t = 8 - v$ — Take v from both sides, so only the t term is left on the left.
4
$t = \frac{8 - v}{3}$ — t was multiplied by 3, so divide both sides by 3. The whole of 8 − v goes on top.

Key idea

Another route also gets full marks: $v - 8 = - 3 t$ , so $t = \frac{v - 8}{- 3}$ . Dividing top and bottom by −1 gives the same $\frac{8 - v}{3}$ .

Common mistake

Writing $t = \frac{v - 8}{3}$ . Check with numbers: t = 2 gives v = 8 − 6 = 2, and $\frac{2 - 8}{3} = - 2$ , not 2. The sign was lost.

Brackets in the Formula

Some formulas have the letter inside a bracket. Treat the whole bracket as one block until it is on its own.

Worked example

1
$A = \frac{1}{2} (a + b) h$ — Make b the subject. This is the area of a trapezium. The bracket (a + b) was halved and multiplied by h.
2
$2 A = (a + b) h$ — Multiply both sides by 2. This clears the half.
3
$\frac{2 A}{h} = a + b$ — The whole bracket was multiplied by h, so divide both sides by h. The bracket is now on its own, so drop it.
4
$b = \frac{2 A}{h} - a$ — Take a from both sides, and write b first.

Key idea

You can also expand first: $2 A = ah + bh$ , so $bh = 2 A - ah$ and $b = \frac{2 A - ah}{h}$ . This is the same answer written another way.

Common mistake

Writing $b = \frac{2 A - a}{h}$ . The h multiplies the whole bracket, so divide by h first, then take a away.

Fractions and Squares

If the letter is inside a fraction, clear the fraction first. If it is squared, the square root comes last.

Worked example

1
$k = \frac{w - 3}{4}$ — Make w the subject. 3 was taken from w, then the result was divided by 4.
2
$4 k = w - 3$ — Dividing by 4 was done last, so undo it first: multiply both sides by 4. The fraction is gone.
3
$4 k + 3 = w$ — Add 3 to both sides to undo the −3.
4
$w = 4 k + 3$ — Swap the sides so w is written first.

Worked example

1
$y = 2 r^{2} - 5$ — Make r the subject. r was squared, then multiplied by 2, then 5 was taken away.
2
$y + 5 = 2 r^{2}$ — Undo the last move first: add 5 to both sides.
3
$\frac{y + 5}{2} = r^{2}$ — Divide both sides by 2, so r² is on its own.
4
$r = \pm \frac{y + 5}{2}$ — Squaring was the first thing done to r, so the square root comes last. Both + and − square to the same number, so write ±.

Key idea

A square root can be positive or negative. For lengths and times, the positive one is the answer.

Square Roots and Cubes (Extended/Higher)

If the letter is under a square root, square both sides first. That removes the root, then carry on as normal.

Worked example

1
$T = 2 d + e$ — Make d the subject. The root was the last thing done, so undo it first.
2
$T^{2} = 2 d + e$ — Square both sides. Squaring undoes a square root, so the root is gone.
3
$T^{2} - e = 2 d$ — Take e from both sides.
4
$d = \frac{T ^{2} - e}{2}$ — d was multiplied by 2, so divide both sides by 2. Write d first.

Cubed

V = 2 x^{3}

x = 3 \frac{V}{2}

Divide by 2, then cube root. No ± here: a cube root has only one answer.

Cube root

R = 3 p

p = R^{3}

Cube both sides. Cubing undoes a cube root.

Common mistake

Squaring too early. In $T = d + 3$ , first take 3: $T - 3 = d$ . Then square: $d = (T - 3)^{2}$ , which is not $T^{2} - 9$ .

Subject Appears Twice (Extended/Higher)

Harder formulas have x in two places, often in a fraction. These are 3–4 marks. Get every x term on one side, then take x out as a factor.

Worked example

1
$y = \frac{2 x + 3}{x - 1}$ — Make x the subject. x is in two places: on the top and on the bottom of the fraction.
2
$y (x - 1) = 2 x + 3$ — Multiply both sides by the bottom, (x − 1). This clears the fraction.
3
$x y - y = 2 x + 3$ — Expand the bracket: y × x = xy and y × (−1) = −y. Now the x terms are on both sides.
4
$x y - 2 x = y + 3$ — Take 2x from both sides and add y to both sides. Every x term is now on the left.
5
$x (y - 2) = y + 3$ — Both terms on the left contain x, so take x out as a factor. Now x appears only once.
6
$x = \frac{y + 3}{y - 2}$ — x is multiplied by (y − 2), so divide both sides by (y − 2).

Common mistake

Leaving x on both sides, like $x = \frac{y + 3 + 2 x}{y}$ . It is not finished until x appears only once.

Twice and Under a Root (Extended/Higher)

Top-grade questions add a square root or a square to the “appears twice” type. Remove the root first, then use the same collect-and-factorise method.

Worked example

1
$y = \frac{2 x + 1}{x - 3}$ — Make x the subject. x is in two places, and everything is under a square root.
2
$y^{2} = \frac{2 x + 1}{x - 3}$ — Square both sides. The root is gone and y becomes y².
3
$y^{2} (x - 3) = 2 x + 1$ — Multiply both sides by the bottom, (x − 3), to clear the fraction.
4
$x y^{2} - 3 y^{2} = 2 x + 1$ — Expand the bracket: y² × x = xy² and y² × (−3) = −3y².
5
$x y^{2} - 2 x = 3 y^{2} + 1$ — Take 2x from both sides and add 3y² to both sides, so the x terms are together on the left.
6
$x (y^{2} - 2) = 3 y^{2} + 1$ — Both terms on the left contain x, so take x out as a factor.
7
$x = \frac{3 y ^{2} + 1}{y ^{2} - 2}$ — Divide both sides by (y² − 2). x now appears only once.

Key idea

If the letter is squared in both places, like $p = \frac{3 k ^{2} + 1}{5 - k ^{2}}$ , treat $k^{2}$ as one letter. Make $k^{2}$ the subject this same way, then square root at the very end.

Which Method? Read the Question

Find the subject in the formula and see what surrounds it.

You see	You do
Added and multiplied	Undo the adding first, then the multiplying
A minus in front of it	Add that term to both sides so it is positive
Inside a bracket	Divide by what multiplies the bracket, or expand first
Inside a fraction	Multiply by the bottom first
Squared	Finish with a square root (±)
Under a root	Square both sides
Cubed or under a cube root	Cube root, or cube both sides
In two places	Collect the terms, then factorise

Key idea

Check by putting numbers in. If y = 5x + c with x = 3 and c = 2 gives y = 17, then (17 − 2) ÷ 5 should give 3 back.

Key idea

Show every move on its own line. Each correct move earns a mark, even if a slip spoils the final answer. An answer with no working scores nothing if it is wrong.

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