Rearranging Formulas
Make a different variable the subject of a formula, then use the rearranged formula to solve practical problems. Rearranging is solving a formula for a variable rather than for a number — the same inverse-operation logic you already know.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
The distance formula is $d = st$. If you know distance and time, how could you find speed?
Without calculating — write your first rearrangement idea. What operation would you perform on both sides?
Rearranging a formula uses the same inverse-operation logic as solving an equation. The difference: you are isolating a variable symbol, not finding a number. The relationship between the variables stays the same.
In $d = st$, $d$ is the subject because it stands alone. To make $s$ the subject, divide both sides by $t$. Every step must keep both sides equal.
Key facts
- The subject of a formula is the variable written by itself.
- Rearranging uses inverse operations.
- A rearranged formula can be checked with a numerical example.
Concepts
- Changing the subject does not change the relationship.
- Rearrange before substituting when the required variable is not the subject.
- Each step must keep both sides equal.
Skills
- Rearrange $d = st$ for $s$ and $t$.
- Rearrange $C = 2\pi r$ for $r$.
- Use a rearranged formula in context.
In $d = st$, distance is the subject because $d$ is by itself. If a question asks for speed, it is often clearer to rearrange the formula to make $s$ the subject before substituting values.
What to write in your book
- Subject: the variable written alone on one side of the formula.
- Rearranging $d = st$ for $s$: divide both sides by $t$ to get $s = \frac{d}{t}$.
- Rearranging $d = st$ for $t$: divide both sides by $s$ to get $t = \frac{d}{s}$.
- Check: if $s = 5$ and $t = 4$, then $d = 20$. Rearranged: $s = 20/4 = 5$. Both sides agree.
Did you get this? True or false: in the formula $s = \frac{d}{t}$, the subject is $d$.
To check $s = \frac{d}{t}$ from $d = st$, choose simple values. If $s = 5$ and $t = 4$, then $d = 20$. The rearranged formula gives $s = \frac{20}{4} = 5$, so it matches.
What to write in your book
- Checking strategy: choose easy values (e.g. $s = 5$, $t = 4$, so $d = 20$), substitute into your rearranged formula, and confirm it gives back the original value.
- Rearranged forms: $d = st$ gives $s = \frac{d}{t}$ and $t = \frac{d}{s}$.
- $C = 2\pi r$ gives $r = \frac{C}{2\pi}$.
- $A = bh$ gives $b = \frac{A}{h}$ and $h = \frac{A}{b}$.
Quick check: Which is the correct rearrangement of $d = st$ to make $t$ the subject?
Worked examples · 3 in a row, reveal as you go
Rearrange $d = st$ to make $s$ the subject.
The circumference of a circle is $C = 2\pi r$. Rearrange the formula to make $r$ the subject.
A rectangle has area $A = bh$. Its area is 84 cm² and its height is 7 cm. Find the base length.
What to write in your book
- $d = st$ rearranged: $s = \frac{d}{t}$ and $t = \frac{d}{s}$.
- $C = 2\pi r$ rearranged: $r = \frac{C}{2\pi}$. Divide by the whole multiplier $2\pi$.
- $A = bh$ rearranged: $b = \frac{A}{h}$ and $h = \frac{A}{b}$.
- Strategy: rearrange first (get the required variable alone), then substitute numbers.
Fill the gap: To rearrange $C = 2\pi r$ for $r$, divide both sides by to get $r = \dfrac{C}{2\pi}$. If $C = 31.4$ and $\pi \approx 3.14$, then $r = $ .
Common errors · the 3 traps that cost marks
Quick-fire practice · 4 rearrangements
Rearrange $d = st$ to make $t$ the subject.
Use your formula to find $t$ when $d = 240$ km and $s = 80$ km/h.
Rearrange $A = bh$ to make $h$ the subject.
Use your formula to find $h$ when $A = 96$ m² and $b = 12$ m.
Odd one out: Three of these are correct statements about rearranging formulas. Which one is wrong?
Match it: Match each formula with the variable it makes the subject.
Earlier you thought about how to find speed from $d = st$. From $d = st$, speed is $s = \frac{d}{t}$ and time is $t = \frac{d}{s}$. The variable you need determines which version is most useful.
All three forms express the same relationship — dividing both sides by $t$ isolates $s$; dividing by $s$ isolates $t$. The choice depends on what the question asks for.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Rearrange $d = st$ to make $t$ the subject, then find $t$ when $d = 150$ km and $s = 50$ km/h. (3 marks)
Q2. Rearrange $C = 2\pi r$ to make $r$ the subject. Use it to find $r$ when $C = 31.4$ cm and $\pi \approx 3.14$. (4 marks)
Q3. Explain why substituting before identifying the required subject can make a formula question harder. (2 marks)
📖 Comprehensive answers (click to reveal)
Drill 1: $t = \frac{d}{s}$.
Drill 2: $t = \frac{240}{80} = 3$ h.
Drill 3: $h = \frac{A}{b}$.
Drill 4: $h = \frac{96}{12} = 8$ m.
Q1 (3 marks): Rearrange: $t = \frac{d}{s}$ [1]. Substitute: $t = \frac{150}{50}$ [1]. $t = 3$ h [1].
Q2 (4 marks): Rearrange: $r = \frac{C}{2\pi}$ [2]. Substitute: $r = \frac{31.4}{2 \times 3.14} = \frac{31.4}{6.28}$ [1]. $r = 5$ cm [1].
Q3 (2 marks): If numbers are substituted before rearranging, you are left solving a numerical equation (e.g. $31.4 = 2 \times 3.14 \times r$) rather than applying a clean formula directly [1]. Rearranging first isolates the required variable as an algebraic formula, making the substitution step straightforward [1].
Name the subject, choose the inverse operation, then check the rearranged formula with easy numbers. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering rearranging-formulas questions. Pool: lesson 4.
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