🔄 Rearranging Formulae — Algebra 5

Rearranging (or transposing) a formula means changing which variable is the subject — isolating the variable you want to find. Engineers do this every single day: Ohm’s Law V=IR can find V, I or R depending on what you rearrange it to.

Change the subject Inverse operations Reverse BODMAS Squares & roots V=IR, F=ma, s=ut+½at²

What does “subject” mean?

The subject of a formula is the variable on its own on one side of the equals sign. In V = IR, V is the subject — it’s expressed in terms of I and R.

V = IR → V is the subject

I = V/R → I is the subject

R = V/I → R is the subject

💡 Rearranging = making a different variable the subject. The formula’s meaning never changes — only which variable you’re expressing everything else in terms of.

Why engineers rearrange formulae

Find different unknowns — F=ma can find force, mass or acceleration depending on what you rearrange to
Design work — “what radius gives me area = 50 cm²?” requires rearranging A=πr² to find r
Checking units — rearranging reveals the units of a quantity
Exam questions — T Level papers give you a formula and ask you to find a specific variable

📌 In T Level / Level 3 Engineering, the most-tested rearrangements involve: V=IR, F=ma, P=IV, s=ut+½at², E=½mv², A=πr², and stress/strain formulae.

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The Golden Rule & Reverse BODMAS

Every rearrangement follows one unbreakable rule, and one strategic guide. Master these two ideas and you can rearrange any formula ever written.

The Golden RuleWhatever you do to one side, you MUST do to the other

Golden Rule: Whatever operation you apply to one side of the equation, you must apply exactly the same operation to the other side. This keeps the equation balanced and true.

Reverse BODMASThe order to undo operations — work from bottom of BODMAS upwards

💡 In 3x + 7 = 22, the 7 was added last (outermost), so undo it first (−7). The 3 was multiplied (inner), so undo it second (÷3). Always work from the outside inward!

⚖ Try it — Balance CheckerVerify both sides stay equal

ℹ️ How to use: Type values for a, b and c in the equation ax + b = c. Both sides update together so you can see how operations keep the balance.

Start with ax + b = c. Apply an operation — see both sides update simultaneously.

c (right-hand side)

👆Try 3x+7=22 · 5x−3=12 (b=−3) · 2x+0=14 — watch the step-by-step balance operations appear.

1️⃣

Simple One & Two-Step Rearrangements

Start with the basics — formulae where you only need one or two inverse operations to isolate the subject. The step-diagram shows every move clearly.

VisualMake I the subject of V = IR — one operation needed

One-step examples

Make m the subject of: F = ma

Start: F = m × a

÷ a both: F/a = m

Answer: m = F/a ✅

Make C the subject of: V = IR + C

Start: V = IR + C

− IR both: V − IR = C

Answer: C = V − IR ✅

Two-step examples

Make x the subject of: y = 3x + 5

Start: y = 3x + 5

− 5 both: y − 5 = 3x

÷ 3 both: (y − 5)/3 = x

Answer: x = (y − 5)/3 ✅

Make u the subject of: v = u + at

Start: v = u + at

− at both: v − at = u

Answer: u = v − at ✅

Three-step exampleMake a the subject of: s = ut + ½at²

Final answer: a = 2(s − ut) / t²
Check: substitute a back into s=ut+½at² to verify.

1️⃣ Try it — One & Two-Step RearrangingChoose a formula & find any variable

ℹ️ How to use: Use the dropdown to choose a formula, then enter the known values. The tool shows each rearranging step in order.

Formula

Find

👆Try: V=IR, find I V=12, R=4 · F=ma, find a F=1500, m=60 · v=u+at, find t v=30, u=10, a=4

Formulae with Powers & Roots

When the subject is squared or cubed, you need to take a root to undo it. When it’s inside a root, you square (or cube) to undo that. Inverse pairs: square ↔ square root, cube ↔ cube root.

Inverse operations for powers

x² → undo with √x (square root)

x³ → undo with ³√x (cube root)

√x → undo with x² (square both sides)

³√x → undo with x³ (cube both sides)

💡 Always perform the root AFTER you have isolated the power term on its own. Don’t rush — square root before dividing is the most common ordering error.

Worked Examples

Make v the subject of: KE = ½mv²

Start: KE = ½mv²

×2 both: 2KE = mv²

÷m both: 2KE/m = v²

√ both: v = √(2KE/m) ✅

Make L the subject of: T = 2π√(L/g)

Start: T = 2π√(L/g)

÷2π both: T/2π = √(L/g)

Square both: (T/2π)² = L/g

×g both: L = g(T/2π)² ✅

² Try it — Powers & Roots RearrangingSquares, cubes & roots

ℹ️ How to use: Select a formula from the dropdown and enter your values. The tool applies square roots or powers and shows the working.

Formula

Find

👆Try: A=πr², find r A=78.5 · KE=½mv², find v KE=100, m=2 · E=mc², find c E=9×10¹⁶, m=1

Formulae with Fractions

When the subject is part of a fraction, multiply by the denominator first to clear the fraction. This is the key move that unlocks everything else.

Key move: Multiply both sides by the denominator to clear fractions immediately.

Subject in the numerator

Make F the subject of: P = F/A

Start: P = F/A

×A both: PA = F

Answer: F = PA ✅

Make u the subject of: v² = u² + 2as

Start: v² = u² + 2as

−2as: v² − 2as = u²

√ both: u = √(v² − 2as) ✅

Subject in the denominatorTrickiest type — two moves needed

Make R the subject of: I = V/R

Start: I = V/R

×R both: IR = V (clears the R from denominator)

÷I both: R = V/I

Answer: R = V/I ✅

Make t the subject of: v = s/t

Start: v = s/t

×t both: vt = s

÷v both: t = s/v ✅

Parallel resistors — hardest fraction typeMake R⊂2; the subject of: 1/R = 1/R⊂1; + 1/R⊂2;

Make R⊂2; the subject of: 1/R⊂T; = 1/R⊂1; + 1/R⊂2;

Start: 1/R⊂T; = 1/R⊂1; + 1/R⊂2;

−1/R⊂1;: 1/R⊂2; = 1/R⊂T; − 1/R⊂1;

Common denom: 1/R⊂2; = (R⊂1; − R⊂T;) / (R⊂T;·R⊂1;)

Flip both: R⊂2; = R⊂T;·R⊂1; / (R⊂1; − R⊂T;) ✅

💡 When you have 1/x = something, flip both sides: x = 1/something.

÷ Try it — Fractions RearrangingNumerator & denominator subjects

ℹ️ How to use: Pick a formula from the dropdown, enter the known quantities, and see how cross-multiplying removes the fraction step by step.

Solve P = F/A style formulae (anything of the form y = a/b).

Formula

Find

👆Try: σ=F/A, find A F=9000, σ=150 · I=V/R, find R V=24, I=0.3 · v=s/t, find t s=120, v=30

🔀

When the Subject Appears Twice

The trickiest type — the variable you want appears in two different terms. You must collect those terms together, then factorise to isolate it.

Strategy: Move all terms containing the subject to one side → factorise out the subject → divide by the bracket.

Method

Move all terms containing the subject to one side of the equation
Move all other terms to the other side
Factorise the side with the subject — take the subject out as a common factor
Divide both sides by the bracket that remains

💡 Think: “collect, factorise, divide”. Those three words describe the whole method!

Worked: make x the subject of ax + b = cx + d

Make x subject: ax + b = cx + d

Collect x: ax − cx = d − b

Factorise: x(a − c) = d − b

÷(a−c): x = (d − b)/(a − c) ✅

⚙ Make R⊂1; subject of: R⊂T; = R⊂1;·R⊂2; / (R⊂1; + R⊂2;)

Start: R⊂T;(R⊂1; + R⊂2;) = R⊂1;·R⊂2; (multiply both × denominator)

Expand: R⊂T;·R⊂1; + R⊂T;·R⊂2; = R⊂1;·R⊂2;

Collect R⊂1;: R⊂T;·R⊂1; − R⊂1;·R⊂2; = −R⊂T;·R⊂2;

Factorise: R⊂1;(R⊂T; − R⊂2;) = −R⊂T;·R⊂2;

Answer: R⊂1; = −R⊂T;·R⊂2; / (R⊂T; − R⊂2;) = R⊂T;·R⊂2; / (R⊂2; − R⊂T;) ✅

Make v subject: P = mv − mu

Start: P = mv − mu (v only appears once here)

+mu both: P + mu = mv

÷m both: v = (P + mu)/m ✅

⚠️ Most common error: Forgetting that you can’t just “move” a variable from one term without dividing. ax + bx = x(a+b) requires factorising — you can’t write ax = bx = x!

🔀 Try it — Subject Appears TwiceCollect, factorise, divide

ℹ️ How to use: Choose a formula where the subject appears on both sides. Enter the values and watch how factorising collects the subject before dividing.

Solve ax + b = cx + d for x — where x appears on both sides.

a (left x)

b (left const)

c (right x)

d (right const)

👆Try: 5x+3=2x+18 · 4x−1=x+8 · 3x+5=3x+5 (identity — infinite solutions!)

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Applications

Where this topic is used in engineering, manufacturing, maintenance and daily life.

⚡Electrical: V=IR → R=V/I. P=IV → I=P/V.
🌡️Thermal: Q=mcΔT → ΔT=Q/(mc). Find temperature rise.
🔧Mechanical: σ=F/A → A=F/σ. Minimum cross-section for given stress.
🚗Kinematics: v²=u²+2as → s=(v²−u²)/(2a). Stopping distance.

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Engineering Formulae — Rearranging in Context

Every T Level / Level 3 Engineering formula can be rearranged. Here are the six most-tested ones, each shown rearranged for every variable.

⚡ Ohm’s Law — V = IR

V = IR → given I and R, find V

I = V/R → given V and R, find I

R = V/I → given V and I, find R

Find R when V=24V, I=0.3A

Rearrange: R = V/I

Substitute: R = 24 / 0.3

Answer: R = 80 Ω ✅

🔨 Newton’s Second Law — F = ma

F = ma → given m and a, find F

m = F/a → given F and a, find m

a = F/m → given F and m, find a

Find a when F=1500N, m=60kg

Rearrange: a = F/m

Substitute: a = 1500/60

Answer: a = 25 m/s² ✅

⚡ Electrical Power — P = IV

P = IV → find P

I = P/V → find I

V = P/I → find V

💡 Combined with V=IR: P = I²R and P = V²/R — make R or I the subject of these too!

I = √(P/R) ← from P=I²R

V = √(PR) ← from P=V²/R

🚗 SUVAT — v = u + at

v = u + at → find v

u = v − at → find u

a = (v−u)/t → find a

t = (v−u)/a → find t

Find t when v=30, u=10, a=4

Rearrange: t = (v−u)/a

Substitute: t = (30−10)/4 = 20/4

Answer: t = 5 s ✅

🏗 Stress — σ = F/A

σ = F/A → find stress

F = σA → find force

A = F/σ → find area

Find A when F=9000N, σ=150 N/mm²

Rearrange: A = F/σ

Substitute: A = 9000/150

Answer: A = 60 mm² ✅

⚙ Rotational Power — P = Tω

P = Tω → find power (W)

T = P/ω → find torque (N·m)

ω = P/T → find angular velocity (rad/s)

ω = 2πn/60 ← convert rpm to rad/s

Find T when P=2200W, n=1450rpm

ω: 2π×1450/60 = 151.8 rad/s

T = P/ω: 2200/151.8

Answer: T = 14.5 N·m ✅

⚙️ Try it — Engineering Formula SolverAll six formulae, any variable

Formula

Find

👆Try: V=IR, find R V=24, I=0.3 · σ=F/A, find A F=9000, σ=150 · P=Tω, find T P=2200, ω=151.8

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Quick Fire Quiz

Test your knowledge — 10 questions, instant feedback.

⚡ Rearranging Blitz

ℹ️ How to play: A question appears on screen. Click the answer you think is correct from the four options. Your score and streak update after each question. Click Next to move on.

Score: 0 Streak: 0 🔥 Q 1/10

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Practice Questions

15 questions across 3 tiers. Show every step — in T Level exams, working carries marks even if the final answer is wrong.

🟢 Tier 1 — Foundations

ℹ️ How to use: Work through each question and write down your answer. When you are ready, click Show Answer to reveal the full worked solution and mark scheme.

Make a the subject of: F = ma

[1]

F = ma
Divide both sides by m:
a = F/m

Make R the subject of: V = IR

[1]

V = IR
Divide both sides by I:
R = V/I

Make u the subject of: v = u + at

[2]

v = u + at
Subtract at from both sides:
u = v − at

Make F the subject of: P = F/A

[2]

P = F/A
Multiply both sides by A:
F = PA

Make x the subject of: y = 4x − 3

[2]

y = 4x − 3
+3 both sides: y + 3 = 4x
÷4 both sides: x = (y + 3)/4

🟡 Tier 2 — Core Skills

Make r the subject of: A = πr²

[3]

Hint: divide by π first, then square root

A = πr²
÷π: A/π = r²
√ both sides: r = √(A/π)

Make v the subject of: KE = ½mv²

[3]

KE = ½mv²
×2: 2KE = mv²
÷m: 2KE/m = v²
√: v = √(2KE/m)

Make a the subject of: s = ut + ½at²

[3]

s = ut + ½at²
−ut: s − ut = ½at²
×2: 2(s−ut) = at²
÷t²: a = 2(s−ut)/t²

Make t the subject of: s = vt − ½at²

[3]

Hint: this is a quadratic in t — rearrange to at² − 2vt + 2s = 0, then use the quadratic formula

s = vt − ½at²
Rearrange: ½at² − vt + s = 0
Multiply ×2: at² − 2vt + 2s = 0
Quadratic formula (A=a, B=−2v, C=2s):
t = (2v ± √(4v²−8as)) / 2a = (v ± √(v²−2as)) / a

Q10

Make L the subject of: T = 2π√(L/g)

[4]

T = 2π√(L/g)
÷2π: T/2π = √(L/g)
Square: (T/2π)² = L/g
×g: L = g(T/2π)² = gT²/(4π²)

🔴 Tier 3 — T Level Challenge

Q11

Make I the subject of: P = I²R

[3]

P = I²R
÷R: P/R = I²
√: I = √(P/R)

Q12

Make R the subject of: P = V²/R

[3]

P = V²/R
×R both sides: PR = V²
÷P both sides: R = V²/P

Q13

Make R⊂1; the subject of: 1/R⊂T; = 1/R⊂1; + 1/R⊂2;

[4]

1/R⊂1; = 1/R⊂T; − 1/R⊂2;
= (R⊂2; − R⊂T;)/(R⊂T;·R⊂2;)
Flip both sides:
R⊂1; = R⊂T;·R⊂2; / (R⊂2; − R⊂T;)

Q14

Make x the subject of: 3x + 2y = 5x − 4 (subject appears twice)

[3]

Collect x terms: 3x − 5x = −4 − 2y
−2x = −4 − 2y
Divide by −2:
x = (4 + 2y)/2 = 2 + y

Q15

A shaft delivers power P = 4.5 kW at 960 rpm. Using P = Tω and ω = 2πn/60, find the torque T. Then rearrange to find n if T remains the same but P increases to 6 kW.

[5]

ω = 2π×960/60 = 100.5 rad/s
T = P/ω = 4500/100.5 = 44.8 N·m

Now P=6000W, T=44.8 N·m:
ω = P/T = 6000/44.8 = 133.9 rad/s
n = 60ω/(2π) = 60×133.9/(2π)
n = 1279 rpm ✅

🎯 Score guide: Q1–5 = foundations (8 marks) · Q6–10 = core skills (16 marks) · Q11–15 = T Level challenge (18 marks)

👉 Next topic: Solving Equations — using rearranging plus factorising to find exact values.