Before we begin: The formula for the area of a rectangle is $A = l \times w$. If you know the area and the width, how would you find the length? Write down your method.
Changing the subject of a formula is identical to solving an equation. The only difference: instead of finding a numerical answer, you express one variable in terms of the others. The same rules apply: whatever you do to one side, do to the other, and undo operations in reverse BIDMAS order.
$F = ma$ becomes $m = F/a$. Same steps, different goal.
- Change the subject of a linear formula using inverse operations
- Rearrange formulas containing brackets and powers
- Take square roots correctly after isolating the squared term
- Substitute values into a rearranged formula
- Recognise and avoid common rearrangement errors
Wrong: Rearranging $C = 2\pi r$ to $r = \frac{C}{2}$ and forgetting the $\pi$.
Right: Divide by both $2$ and $\pi$: $r = \frac{C}{2\pi}$. The entire product $2\pi$ is multiplied by $r$.
Wrong: Taking the square root before isolating the squared term. In $A = \pi r^2$, writing $r = \sqrt{A} - \pi$.
Right: First divide by $\pi$, then take the square root: $r = \sqrt{\frac{A}{\pi}}$.
To change the subject of a formula, treat it exactly like solving an equation. Identify the operations applied to the desired variable, then undo them in reverse BIDMAS order. The goal is to get that variable alone on one side.
$y = 3x + 5$: subtract 5, then divide by 3. $x = \frac{y-5}{3}$.
When a variable has multiple operations applied to it, undo them in the reverse order of BIDMAS. Undo addition and subtraction first, then multiplication and division, then powers and roots. This keeps every step valid.
$P = 2a + 2b$: subtract $2a$, then divide by 2.
When the subject is squared (or cubed), you must isolate the squared term before taking the square root. Taking the root too early is one of the most common errors in formula rearrangement. And remember: a square root has two solutions, positive and negative.
$A = \pi r^2$: divide by $\pi$ first, then square root.
Once you have rearranged a formula, you can substitute known values to calculate the unknown. Rearrange first, substitute second. This is more reliable than trying to substitute into the original formula and then solving.
$A = lw$ → $w = A/l$ → $w = 48/8 = 6$ cm.
Watch Me Solve It · 3 examples
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1Identify the operation on r$C = 2\pi r$ means $r$ is multiplied by $2\pi$.To undo multiplication by 2pi, divide both sides by 2pi.
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2Divide both sides by 2pi$\frac{C}{2\pi} = \frac{2\pi r}{2\pi}$The 2pi on the right cancels, leaving r alone.
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3Write the final formula and check$r = \frac{C}{2\pi}$Check: if C = 2pi(5) = 10pi, then r = 10pi/(2pi) = 5. Correct.
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1Write the operation chain$h \xrightarrow{\times B} Bh \xrightarrow{\div 3} V$To isolate h, reverse the chain: undo /3 first, then undo xB.
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2Multiply both sides by 3$3V = 3 \times \frac{1}{3}Bh = Bh$The 3 and 1/3 cancel on the right, leaving Bh.
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3Divide both sides by B$\frac{3V}{B} = \frac{Bh}{B}$Undo multiplication by B with division by B.
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4Simplify and check$h = \frac{3V}{B}$Check: if h = 6, B = 5, then V = (1/3)(5)(6) = 10. And 3(10)/5 = 6. Correct.
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1Identify the operations on r$A = \pi r^2$ means r is squared, then multiplied by $\pi$.Reverse BIDMAS: undo xpi first, then undo the square.
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2Divide both sides by pi$\frac{A}{\pi} = \frac{\pi r^2}{\pi} = r^2$Isolating the squared term before taking the root is critical.
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3Take the square root of both sides$\sqrt{\frac{A}{\pi}} = \sqrt{r^2} = r$In geometric contexts, we take the positive root only.
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4Write the final formula and check$r = \sqrt{\frac{A}{\pi}}$Check: if r = 4, A = pi(16) = 16pi. Then sqrt(16pi/pi) = sqrt(16) = 4. Correct.
Brain Trainer · 4 problems
Four formulas to rearrange. Show each step clearly, then reveal the answer to check your working.
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1 Make $x$ the subject of $y = 2x + 4$.
Subtract 4: $y - 4 = 2x$. Divide by 2: $x = \frac{y-4}{2}$. Check: $2(\frac{y-4}{2}) + 4 = y - 4 + 4 = y$. -
2 Make $a$ the subject of $v = u + at$.
Subtract $u$: $v - u = at$. Divide by $t$: $a = \frac{v-u}{t}$. Check: $u + (\frac{v-u}{t})t = u + v - u = v$. -
3 Make $h$ the subject of $V = \pi r^2 h$.
Divide by $\pi r^2$: $h = \frac{V}{\pi r^2}$. Check: $\pi r^2(\frac{V}{\pi r^2}) = V$. -
4 Make $b$ the subject of $P = 2(a + b)$.
Divide by 2: $\frac{P}{2} = a + b$. Subtract $a$: $b = \frac{P}{2} - a$. Check: $2(a + \frac{P}{2} - a) = 2(\frac{P}{2}) = P$.
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. Make $h$ the subject of $V = \pi r^2 h$ (volume of a cylinder). Then find $h$ when $V = 616$ cm³ and $r = 7$ cm. Use $\pi = \frac{22}{7}$.
Q7. Make $m$ the subject of $E = mc^2$. A nuclear reaction releases $E = 1.8 \times 10^{14}$ joules of energy. If $c = 3 \times 10^8$ m/s, calculate the mass $m$ that was converted to energy.
Q8. A car rental company uses the formula $C = 50 + 0.25d$ where $C$ is the total cost in dollars and $d$ is the distance driven in kilometres.
(a) Make $d$ the subject of the formula. (1 mark)
(b) A customer was charged $185. How many kilometres did they drive? (1 mark)
(c) Explain why this formula is a linear model and identify what the gradient and y-intercept represent in this context. (1 mark)
Quick Check
1. B Subtract 5: $y - 5 = 3x$. Divide by 3: $x = \frac{y-5}{3}$.
2. C Divide by $\pi$: $\frac{A}{\pi} = r^2$. Square root: $r = \sqrt{\frac{A}{\pi}}$.
3. B Subtract $u$: $v - u = at$. Divide by $t$: $a = \frac{v-u}{t}$.
4. C Subtract $2a$: $P - 2a = 2b$. Divide by 2: $b = \frac{P-2a}{2}$.
5. B Subtract 32: $F - 32 = \frac{9C}{5}$. Multiply by 5: $5(F-32) = 9C$. Divide by 9: $C = \frac{5(F-32)}{9}$.
Show Your Working Model Answers
Q6 (3 marks): Divide by $\pi r^2$: $h = \frac{V}{\pi r^2}$ [1]. Substitute: $h = \frac{616}{\frac{22}{7} \times 7^2} = \frac{616}{22 \times 7} = \frac{616}{154} = 4$ cm [2].
Q7 (3 marks): $m = \frac{E}{c^2}$ [1]. $m = \frac{1.8 \times 10^{14}}{(3 \times 10^8)^2} = \frac{1.8 \times 10^{14}}{9 \times 10^{16}} = 0.2 \times 10^{-2} = 2 \times 10^{-3}$ kg [2].
Q8 (3 marks): (a) $C = 50 + 0.25d$ → $C - 50 = 0.25d$ → $d = \frac{C-50}{0.25} = 4(C-50)$ [1]. (b) $d = 4(185-50) = 4 \times 135 = 540$ km [1]. (c) This is a linear model because $C$ changes at a constant rate with respect to $d$. The gradient (0.25) represents the cost per kilometre (25 cents per km). The y-intercept (50) represents the fixed base cost regardless of distance driven [1].
The Kinematic Challenge
Make $t$ the subject of $s = ut + \frac{1}{2}at^2$. This is a quadratic in $t$. Rearrange to standard form $at^2 + bt + c = 0$ and use the quadratic formula to solve for $t$. Show every step.
Reveal solution
Rearrange: $s = ut + \frac{1}{2}at^2$ → $\frac{1}{2}at^2 + ut - s = 0$.
Multiply by 2: $at^2 + 2ut - 2s = 0$.
Use quadratic formula with $A = a$, $B = 2u$, $C = -2s$:
$t = \frac{-2u \pm \sqrt{(2u)^2 - 4(a)(-2s)}}{2a} = \frac{-2u \pm \sqrt{4u^2 + 8as}}{2a}$
$t = \frac{-2u \pm 2\sqrt{u^2 + 2as}}{2a} = \frac{-u \pm \sqrt{u^2 + 2as}}{a}$
Since time must be positive, we take the positive root: $t = \frac{-u + \sqrt{u^2 + 2as}}{a}$.
Same as Equations
Every rule for solving applies to rearranging
Reverse BIDMAS
Undo operations in the opposite order
Isolate Before Rooting
Divide first, then take square roots
Rearrange Then Substitute
Get the formula right before plugging in numbers
One Step Per Line
Clear working earns method marks
Always Check
Substitute back into the original formula
Interactive: Formula Rearranger
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