Rearranging Formulae
The Skill
Rearranging a formula means making a different variable the subject. The subject is the variable on its own on one side of the formula.
In $A = lw$, the subject is $A$. To make $l$ the subject: $l = \frac{A}{w}$
The Golden Rule
Whatever you do to one side, you must do to the other.
Rearranging with Addition/Subtraction
$$v = u + at$$
To make $u$ the subject, subtract $at$ from both sides: $$u = v - at$$
Rearranging with Multiplication/Division
$$A = lw$$
To make $l$ the subject, divide both sides by $w$: $$l = \frac{A}{w}$$
Two-Step Rearranging
$$C = 2\pi r$$
To make $r$ the subject:
- Divide both sides by $2\pi$:
$$r = \frac{C}{2\pi}$$
Rearranging with Powers (Higher Tier)
$$A = \pi r^2$$
To make $r$ the subject:
- Divide by $\pi$: $\frac{A}{\pi} = r^2$
- Square root: $r = \sqrt{\frac{A}{\pi}}$
The Traps
Common misconceptions and how to avoid them.
Dividing the wrong way round "The Fraction Flip"
The Mistake in Action
Make $v$ the subject of $t = \frac{d}{v}$
Wrong: $tv = d$ $v = td$
Why It Happens
Students correctly multiply both sides by $v$ but then don't complete the rearrangement properly. They multiply again instead of dividing.
The Fix
$$t = \frac{d}{v}$$
Step 1: Multiply both sides by $v$: $$tv = d$$
Step 2: Divide both sides by $t$: $$v = \frac{d}{t}$$
Think of it like this: If $t = \frac{d}{v}$, then $v = \frac{d}{t}$ (the non-subject swaps position).
Spot the Mistake
Make $v$ the subject of $t = \frac{d}{v}$
Multiply both sides by $v$: $tv = d$
$v = td$
Click on the line that contains the error.
Forgetting to square root or doing it incorrectly "The Root of the Problem"
The Mistake in Action
Make $r$ the subject of $A = \pi r^2$
Wrong: $\frac{A}{\pi} = r^2$ $r = \frac{A}{\pi} - 2$
Why It Happens
Students see $r^2$ and subtract 2 instead of taking the square root. They confuse the exponent with a term being added.
The Fix
The inverse of squaring is taking the square root, not subtracting 2.
$$A = \pi r^2$$
Divide by $\pi$: $$\frac{A}{\pi} = r^2$$
Take the square root: $$r = \sqrt{\frac{A}{\pi}}$$
Note: For GCSE, we typically only consider the positive root for measurements like radius.
Spot the Mistake
Make $r$ the subject of $A = \pi r^2$
$\frac{A}{\pi} = r^2$
$r = \frac{A}{\pi} - 2$
Click on the line that contains the error.
Not applying operations to entire expressions "The Partial Operation"
The Mistake in Action
Make $r$ the subject of $A = 2\pi r + 3$
Wrong:
$\frac{A}{2\pi} = r + 3$
$r = \frac{A}{2\pi} - 3$
Why It Happens
When dividing, students only divide the term with $r$, not the constant. They forget that division must apply to everything on that side.
The Fix
You must divide BOTH terms on the right side by $2\pi$:
Correct method: $$A = 2\pi r + 3$$
First subtract 3: $$A - 3 = 2\pi r$$
Then divide by $2\pi$: $$\frac{A - 3}{2\pi} = r$$
Key insight: Deal with addition/subtraction BEFORE multiplication/division (reverse of BIDMAS).
Spot the Mistake
Make $r$ the subject of $A = 2\pi r + 3$
$\frac{A}{2\pi} = r + 3$
$r = \frac{A}{2\pi} - 3$
Click on the line that contains the error.
The Deep Dive
Apply your knowledge with these exam-style problems.
Level 1: Fully Worked
Complete solutions with commentary on each step.
Question
Make $t$ the subject of $v = u + at$
Solution
$$v = u + at$$
Step 1: Subtract $u$ from both sides: $$v - u = at$$
Step 2: Divide both sides by $a$: $$\frac{v - u}{a} = t$$
Answer: $t = \frac{v - u}{a}$
Question
Make $h$ the subject of $V = \frac{1}{3}\pi r^2 h$
Solution
$$V = \frac{1}{3}\pi r^2 h$$
Step 1: Multiply both sides by 3: $$3V = \pi r^2 h$$
Step 2: Divide both sides by $\pi r^2$: $$\frac{3V}{\pi r^2} = h$$
Answer: $h = \frac{3V}{\pi r^2}$
Question
Make $u$ the subject of $v^2 = u^2 + 2as$
Solution
$$v^2 = u^2 + 2as$$
Step 1: Subtract $2as$ from both sides: $$v^2 - 2as = u^2$$
Step 2: Take the square root of both sides: $$\sqrt{v^2 - 2as} = u$$
Answer: $u = \sqrt{v^2 - 2as}$
(Note: In physical contexts, we take the positive root)
Level 2: Scaffolded
Fill in the key steps.
Question
Make $x$ the subject of $y = 3x - 7$
Level 3: Solo
Try it yourself!
Question
The formula for converting Celsius to Fahrenheit is $F = \frac{9C}{5} + 32$
Make $C$ the subject.
Show Solution
$$F = \frac{9C}{5} + 32$$
Subtract 32: $$F - 32 = \frac{9C}{5}$$
Multiply by 5: $$5(F - 32) = 9C$$
Divide by 9: $$C = \frac{5(F - 32)}{9}$$
Answer: $C = \frac{5(F - 32)}{9}$
Examiner's View
Mark allocation: Simple rearranging is 1-2 marks. Complex rearranging (squares, subject appears twice) is 3-4 marks.
Common errors examiners see:
- Using inverse operations incorrectly
- Forgetting to apply operations to the entire other side
- Errors with square roots
- Not fully isolating the subject
What gains marks:
- Clear step-by-step working
- Showing the inverse operation being applied
- Correct use of brackets when needed
AQA Notes
AQA often uses physics or geometry formulae. Be comfortable rearranging $v = u + at$, $A = \pi r^2$.