Year 10 Maths · Unit 2 · Lesson 10

Formulas and Rearrangement

Learn to change the subject of any formula using inverse operations, then substitute values with confidence.

40 min 8 cards 130 XP
A = pi r^2 r = sqrt(A/pi) Change the subject A = 50.27, r = 4 r = 4, A = 50.27
Think First
warm-up

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.

Record your answer in your workbook.
1
The Big Idea
+5 XP to read

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.

F = ma m = F/a m = F/a
same rules as equations
2
Lesson Objectives
what you will learn
  • 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
3
Key Vocabulary
terms to know
Subject of a formulaThe variable that stands alone on one side of the equals sign, e.g. r in r = C/(2pi).
RearrangementThe process of rewriting a formula so a different variable becomes the subject.
Inverse operationThe operation that undoes another: addition/subtraction, multiplication/division, powers/roots.
FormulaA rule written using symbols that shows the relationship between quantities, e.g. A = pi r^2.
SubstitutionReplacing variables with known values to calculate a result.
Linear modelA relationship that changes at a constant rate, often written as y = mx + c.
4
Spot the Trap
heads-up

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: $r = \\sqrt{\\frac{A}{\\pi}}$.

5
Changing the Subject
+5 XP

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. $x = \\frac{y-5}{3}$.

y = 3x + 5 y - 5 = 3x x = (y-5)/3
undo operations in reverse
Write the operation chain
For y = 3x + 5: x --x3-> 3x --+5-> y. Reverse it: -5, then /3.
Treat it like an equation
Every rule for solving linear equations applies to rearranging formulas.
Check by substituting back
Plug x = (y-5)/3 back into y = 3x + 5 to verify it works.
6
Reverse BIDMAS for Formulas
+5 XP

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.

P = 2a + 2b P - 2a = 2b b = (P-2a)/2
inverse order
One operation per line
Each line of working should perform exactly one inverse operation.
Watch grouped terms
In P = 2(a + b), the 2 multiplies the entire bracket, not just a.
Factorise when helpful
If the same variable appears multiple times, factorise it out first.
7
Powers and Roots
+5 XP

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.

A = pi r^2 A/pi = r^2 r = sqrt(A/pi)
isolate before rooting
Divide first, root second
Never take a square root while other operations still apply to the squared term.
Two solutions
r^2 = k gives r = sqrt(k) and r = -sqrt(k). In geometry, usually take the positive root.
Cube roots too
For r^3, isolate r^3 first, then take the cube root. Cube roots have only one real solution.
8
Substituting After Rearranging
+5 XP

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.

A = lw w = A/l w = 48/8 = 6
rearrange, then substitute
Rearrange first
Get the formula into the shape you need before plugging in numbers.
Watch your units
Make sure all values use consistent units before substituting.
Check with the original
Plug your answer back into the original formula to verify.

Watch Me Solve It · 3 examples

Watch Me Solve It · Linear formula
+15 XP per step
Q1
PROBLEM
Make $r$ the subject of $C = 2\\pi r$.
  1. 1
    Identify 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.
  2. 2
    Divide both sides by 2pi
    $\\frac{C}{2\\pi} = \\frac{2\\pi r}{2\\pi}$
    The 2pi on the right cancels, leaving r alone.
  3. 3
    Write the final formula and check
    $r = \\frac{C}{2\\pi}$
    Check: if C = 2pi(5) = 10pi, then r = 10pi/(2pi) = 5. Correct.
Answer$r = \\frac{C}{2\\pi}$
Watch Me Solve It · Two-step formula
+15 XP per step
Q2
PROBLEM
Make $h$ the subject of $V = \\frac{1}{3}Bh$.
  1. 1
    Write the operation chain
    $h \\xrightarrow{\\times B} Bh \\xrightarrow{\\div 3} V$
    To isolate h, reverse the chain: undo /3 first, then undo xB.
  2. 2
    Multiply both sides by 3
    $3V = 3 \\times \\frac{1}{3}Bh = Bh$
    The 3 and 1/3 cancel on the right, leaving Bh.
  3. 3
    Divide both sides by B
    $\\frac{3V}{B} = \\frac{Bh}{B}$
    Undo multiplication by B with division by B.
  4. 4
    Simplify 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.
Answer$h = \\frac{3V}{B}$
Watch Me Solve It · Power formula
+15 XP per step
Q3
PROBLEM
Make $r$ the subject of $A = \\pi r^2$.
  1. 1
    Identify 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.
  2. 2
    Divide both sides by pi
    $\\frac{A}{\\pi} = \\frac{\\pi r^2}{\\pi} = r^2$
    Isolating the squared term before taking the root is critical.
  3. 3
    Take the square root of both sides
    $\\sqrt{\\frac{A}{\\pi}} = \\sqrt{r^2} = r$
    In geometric contexts, we take the positive root only.
  4. 4
    Write 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.
Answer$r = \\sqrt{\\frac{A}{\\pi}}$

Brain Trainer · 4 problems

D
Brain Trainer · Formula Rearrangement
4 problems

Four formulas to rearrange. Show each step clearly, then reveal the answer to check your working.

  1. 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. 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. 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. 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: $2(a + \frac{P}{2} - a) = 2(\frac{P}{2}) = P$.
Complete in your workbook.

Quick Check · 5 questions

1
Make $x$ the subject of $y = 3x + 5$.
+10 XP
2
Make $r$ the subject of $A = \\pi r^2$.
+10 XP
3
If $v = u + at$, then $a =$
+10 XP
4
Make $b$ the subject of $P = 2a + 2b$.
+10 XP
5
The formula $F = \\frac{9C}{5} + 32$ converts Celsius to Fahrenheit. Make $C$ the subject.
+10 XP

Show Your Working · 3 questions

Show Your Working
9 marks total
Apply Medium 3 MARKS

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 $\\pi = \\frac{22}{7}$.

Answer in your workbook.
Apply Medium 3 MARKS

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.

Answer in your workbook.
Evaluate Medium 3 MARKS

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)

Answer in your workbook.
Comprehensive Answers

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].

Stretch Challenge · +25 XP, +10 coins

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: $t = \frac{-u + \sqrt{u^2 + 2as}}{a}$.

R
Quick Review

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

Practise changing the subject step by step with instant feedback. Try linear, fractional and power formulas.

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