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 calculatingwrite 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.
The subject of a formula is the variable alone on one side. Rearrange by applying the same inverse operations used to solve equations. Rearranging to the required subject before substituting is usually cleaner than substituting and then solving.
Pause, copy the definition of "subject" (the variable alone on the left side), the rearranging method (apply inverse operations in reverse order, same as solving equations), and the tip to rearrange first when you need to find an input variable rather than an output into your book.
Did you get this? True or false: in the formula $s = \frac{d}{t}$, the subject is $d$.
We just saw that the subject of a formula is the variable alone on one side, and that rearranging uses the same inverse operations as solving equations. That raises a question: after rearranging a formula, how do you confirm the new form is correct before using it in a calculation? This card answers it → substitute simple numbers into both the original and rearranged formulas; if both give the same result, the rearrangement is correct.
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.
Check a rearranged formula: substitute simple values into the original formula to get a known output, then substitute the same values into the rearranged formula, if it gives the same result, the rearrangement is correct.
Pause, copy the number-substitution check: pick simple values, evaluate the original formula to get a known output, then substitute those same values into the rearranged formula, if both give the same result, the rearrangement is correct into your book.
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.
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 confidencethat 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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