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hscscience Maths Std · Y11
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Module 1 · L4 of 13 ~45 min ⚡ +90 XP available

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.

Today's hook, The distance formula is $d = st$. If you know distance and time, how could you find speed? Rearranging lets you flip any formula to suit the question.
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Recall, your gut answer first

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Worksheets

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Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

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01
Recall, your gut answer first
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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?

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02
The rearranging principle you need to own
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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.

Make a different variable the subject.
Rearrange first, substitute second, it is almost always easier
Identify the subject first
The subject is the variable written by itself on one side. In $d = st$, the subject is $d$. In $s = \frac{d}{t}$, the subject is $s$.
Use inverse operations
To undo multiplication, divide. To undo division, multiply. Apply the same operation to both sides to maintain equality.
Check with numbers
Choose simple values and verify both forms give the same answer. If $s = 5$ and $t = 4$, then $d = 20$, and $s = 20/4 = 5$. It checks out.
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What you'll master
Know

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.
Understand

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.
Can do

Skills

  • Rearrange $d = st$ for $s$ and $t$.
  • Rearrange $C = 2\pi r$ for $r$.
  • Use a rearranged formula in context.
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Key terms
SubjectThe variable that stands alone on one side of a formula, e.g. $d$ in $d = st$.
RearrangeApply inverse operations to both sides of a formula to isolate a different variable as the new subject.
Inverse operationThe operation that undoes another: multiplication undoes division; addition undoes subtraction.
SubstituteReplace a variable with a known numerical value to calculate the remaining unknown.
FormulaAn equation that expresses a relationship between variables, e.g. $d = st$ or $C = 2\pi r$.
VerifyCheck that a rearranged formula is correct by substituting simple numbers and confirming both sides match.
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The subject is the variable by itself

In $d = st$, distance is the subject because $d$ is by itself.

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The subject is the variable by itself
core concept

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.

Rearrange d = s × t by covering the subject in the triangle.
Key idea: Rearranging is solving a formula for a variable rather than solving for a number.

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 $d$.

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Check a rearranged formula with numbers

We just saw that the subject of a formula is the variable alone on one side, and that rearranging uses the same inverse...

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Check a rearranged formula with numbers
core concept

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.

Communication habit: A rearranged formula should still give the same relationship when tested with the same values.

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?

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Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · REARRANGE d = st FOR s

Rearrange $d = st$ to make $s$ the subject.

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$d = st$
Write the original formula. Identify that $s$ is being multiplied by $t$. The inverse operation is to divide by $t$.
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PROBLEM 2 · REARRANGE A CIRCUMFERENCE FORMULA

The circumference of a circle is $C = 2\pi r$. Rearrange the formula to make $r$ the subject.

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$C = 2\pi r$
Write the original formula. The variable $r$ is being multiplied by $2\pi$. Divide both sides by $2\pi$.
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PROBLEM 3 · REARRANGE AND SUBSTITUTE

A rectangle has area $A = bh$. Its area is 84 cm² and its height is 7 cm. Find the base length.

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Make $b$ the subject: $b = \dfrac{A}{h}$
Divide both sides of $A = bh$ by $h$. Rearrange first before substituting.
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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 = $ .

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Common errors · the 3 traps that cost marks

Common errors · the 3 traps that cost marks

Common errors · the 3 traps that cost marks

Trap 01
Substituting before rearranging
If you substitute numbers into $C = 2\pi r$ before rearranging for $r$, you end up solving a numerical equation the hard way. Always make the required variable the subject first, then substitute.
Trap 02
Dividing by only part of the multiplier
In $C = 2\pi r$, you must divide by $2\pi$ (the whole multiplier), not just by 2. Leaving $\pi$ on the right gives $r = \frac{C}{2\pi}$ only if you divide the whole expression.
Trap 03
Thinking rearranging changes the relationship
Whether you write $d = st$, $s = \frac{d}{t}$, or $t = \frac{d}{s}$, all three express the same relationship. Rearranging chooses which variable to isolate; it does not change the formula's meaning.
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Revisit your thinking

Quick-fire practice · 4 rearrangements

Quick-fire practice · 4 rearrangements

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Rearrange $d = st$ to make $t$ the subject.

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Use your formula to find $t$ when $d = 240$ km and $s = 80$ km/h.

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Rearrange $A = bh$ to make $h$ the subject.

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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.

  • $s = \dfrac{d}{t}$
  • $t = \dfrac{d}{s}$
  • $r = \dfrac{C}{2\pi}$
  • $b = \dfrac{A}{h}$
  • Subject is $b$ (base)
  • Subject is $r$ (radius)
  • Subject is $t$ (time)
  • Subject is $s$ (speed)
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Revisit your thinking

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.

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Show what you have learned

Multiple choice, then short answer under exam conditions.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer
ApplyBand 43 marks

Q1. Rearrange $d = st$ to make $t$ the subject, then find $t$ when $d = 150$ km and $s = 50$ km/h. (3 marks)

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ApplyBand 44 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)

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AnalyseBand 52 marks

Q3. Explain why substituting before identifying the required subject can make a formula question harder. (2 marks)

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📖 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].