𝑥 Algebra — Collect & Simplify

Algebra is the language of engineering. Every formula — from Ohm’s law to beam stress — is algebra. Master it here and every other topic in engineering maths becomes much more straightforward.

T Level Core 4.1Expressions & equations Indices lawsExpanding & factorising Rearranging formulae

What is algebra?Using letters to represent unknown values

In arithmetic we work with known numbers: 3 + 5 = 8. In algebra we use letters (variables) to represent unknown or changing values.

V = I × R (Ohm’s law)

F = m × a (Newton’s second law)

E = m × c² (Einstein’s energy equation)

💡 Every engineering formula is algebra. Once you can rearrange V = IR, you can find any one of V, I or R when you know the other two.

What you need to master

① Expressions: simplify, collect like terms

② Indices: the 5 laws

③ Expand and simplify brackets

④ Factorise: common factor and quadratics

⑤ Solve linear and quadratic equations

⑥ Rearrange formulae (change the subject)

✅ All of these use the same golden rule: whatever you do to one side, do the same to the other!

📝Expressionsterms & collecting

🔧Indices5 laws

🔵Expandingbrackets

✂Factorisingreverse expand

⚖Equationssolve for x

📊Formulaerearranging

⚙Engineeringreal uses

📝

Expressions & Terms

An expression is a collection of terms combined with + or −. Simplifying means collecting like terms together.

Like TermsOnly collect terms with exactly the same variable AND power

✅ Like terms: 3x and 7x (both x¹)
✅ Like terms: 4x² and −x² (both x²)
✅ Like terms: 5 and −3 (both constants)

❌ NOT like terms: 3x and 3x² (different powers)
❌ NOT like terms: 4x and 4y (different variables)

Simplify: 5x + 3y − 2x + 7y − 1

x terms: 5x − 2x = 3x

y terms: 3y + 7y = 10y

Answer: 3x + 10y − 1 ✅

SubstitutionReplacing variables with numbers to evaluate an expression

Engineers do this every time they use a formula — substitute known values in, get the answer out.

If x=3, y=2: find 4x² − 3y + 1

Evaluate 4x² − 3y + 1 where x=3, y=2

x²: 3² = 9

4x²: 4 × 9 = 36

3y: 3 × 2 = 6

Answer: 36 − 6 + 1 = 31 ✅

📝 Try it — Expression Evaluatorax²+bx+c at any x

ℹ️ How to use: Type numbers into the boxes for a, b, c and x. The result updates instantly as you type.

a (x²)

b (x)

c (const)

x value

👆Try: a=4,b=−3,c=1,x=3 · a=1,b=0,c=0,x=5 (just x²) · a=0,b=2,c=7,x=4 (linear)

🔧

Indices Laws

The five laws that govern how powers (indices/exponents) work. These appear in every engineering formula involving areas, volumes, and electrical/mechanical quantities.

ℹ️ How to use: Type a base number and two powers (m and n). All five index laws calculate automatically.

Fractional IndicesRoots written as powers

x¹ᐟ² = √x (square root)

x¹ᐟ³ = ∛x (cube root)

x²ᐟ³ = (∛x)² (cube root then square)

💡 xᵐᐟⁿ means “take the nth root of x, then raise to the power m”

Evaluate 27²ᐟ³

Cube root: ∛27 = 3

Square: 3² = 9 ✅

Worked ExamplesIndices laws in action

Simplify: x³ × x⁵ ÷ x´

Multiply: x³ × x⁵ = x⁸ (add powers: 3+5)

Divide: x⁸ ÷ x´ = x´ (subtract: 8−4)

Answer: x´ ✅

Simplify: (2x²)³

Coefficient: 2³ = 8

Variable: (x²)³ = x⁶ (multiply: 2×3)

Answer: 8x⁶ ✅

⚠️ Most common mistake: Using Law 1 when multiplying DIFFERENT bases. x³ × y² ≠ (xy)⁵ — the laws only work when the base is the SAME!

🔧 Try it — Indices LawsAll 5 laws at once

Base x

Power m

Power n

👆x=2: m=3,n=2 (Law 1: 2³×2²=32) · see Laws 2–5 auto-calculate · Try x=3, m=4, n=2

🔵

Expanding Brackets

Expanding means multiplying out the brackets. Every term inside the bracket is multiplied by what is outside (or by every term in a second bracket).

Singlea(b + c) = ab + ac

Single bracketMultiply each term inside by the term outside

Multiply the outside term by the first term inside
Multiply the outside term by the second term inside
Write the results with the correct sign

Expand: 3(2x + 5)

3 × 2x: = 6x

3 × 5: = 15

Answer: 6x + 15 ✅

Expand: −4(3x − 2)

−4 × 3x: = −12x

−4 × −2: = +8 (neg × neg = pos)

Answer: −12x + 8 ✅

Double bracketsFOIL method: First, Outer, Inner, Last

First: a×c Outer: a×d Inner: b×c Last: b×d

(a+b)(c+d) = ac + ad + bc + bd

Expand: (x + 3)(x + 4)

First: x × x = x²

Outer: x × 4 = 4x

Inner: 3 × x = 3x

Last: 3 × 4 = 12

Collect: x² + 7x + 12 ✅

Grid MethodExpand (x + 3)(x + 4) visually — every combination multiplied out

⚠️ Difference of Two Squares: (x+a)(x−a) = x² − a² e.g. (x+5)(x−5) = x² − 25. The middle terms cancel!

🔵 Try it — Bracket ExpanderSingle or double brackets

ℹ️ How to use: Choose Single or Double from the dropdown, then type your coefficients and constants.

Single: a(bx+c) Double: (ax+b)(cx+d)

a (outside)

b (x coefficient)

c (const)

👆Single: 3(2x−1) · −2(x+5) · Double: (x+3)(x+4) · (2x−1)(x+3)

✂

Factorising

Factorising is the reverse of expanding — you write an expression as a product of its factors. Think of it as “putting brackets back in”.

Common FactorFind the HCF of all terms, take it outside

Find the HCF of all the coefficients
Find the highest power of each variable common to ALL terms
Place the HCF outside the bracket
Divide each term by the HCF to find what goes inside

Factorise: 6x² + 9x

HCF: HCF(6,9)=3, common variable: x

Outside: 3x

Inside: 6x²÷3x=2x 9x÷3x=3

Answer: 3x(2x + 3) ✅

QuadraticsFactorise x² + bx + c — find two numbers

For x² + bx + c, find two numbers that multiply to give c and add to give b.

x² + bx + c = (x + p)(x + q) where p×q=c and p+q=b

Factorise: x² + 7x + 12

Need: two numbers × to 12, + to 7

Pairs ×12: 1×12, 2×6, 3×4

3+4=7: ✅ found them!

Answer: (x + 3)(x + 4) ✅

Factorise: x² − 5x − 14

Need: two numbers × to −14, + to −5

Try: −7 and +2 → (−7)×(+2)=−14 ✅ (−7)+(+2)=−5 ✅

Answer: (x − 7)(x + 2) ✅

📌 Difference of two squares: a² − b² = (a+b)(a−b) e.g. x² − 25 = (x+5)(x−5)

✂ Try it — HCF FactoriserCommon factor method

ℹ️ How to use: Type the coefficient and power for each term. The tool finds the highest common factor and writes out the factorised form.

Enter coefficients of ax² + bx to find the common factor.

a (x² coefficient)

b (x coefficient)

👆Try: 6x²+9x · 4x²−8x · 15x²+10x — HCF and factored form shown step by step.

⚖

Solving Equations

Finding the value of the unknown. Always use the balance method — whatever you do to one side, do exactly the same to the other.

Linear EquationsOne unknown, power of 1

Expand any brackets first
Collect all x terms on one side
Collect all number terms on the other side
Divide both sides to find x
Always check your answer by substituting back in

Solve: 5x − 3 = 2x + 9

−2x both: 3x − 3 = 9

+3 both: 3x = 12

÷3 both: x = 4 ✅

Quadratic Equationsax² + bx + c = 0 — use the formula

x = (−b ± √(b² − 4ac)) / 2a

Solve: x² − 5x + 6 = 0

Factorise: (x−2)(x−3) = 0

Solutions: x−2=0 or x−3=0

Answer: x=2 or x=3 ✅

Solve: 2x² + x − 3 = 0 (using formula)

a=2,b=1,c=−3

Disc: 1²−4(2)(−3) = 1+24 = 25

x= (−1 ± 5) / 4

Answer: x=1 or x=−1.5 ✅

⚖ Try it — Linear Equation Solverax + b = cx + d

ℹ️ How to use: Enter values for a, b, c and d in ax + b = cx + d. The step-by-step solution appears below.

a (left x)

b (left c)

=

c (right x)

d (right c)

👆Try: 5x−3=2x+9 · 3x+7=22 (c=0,d=22) · 2x=8 (b=0,c=0,d=8)

📊

Rearranging Formulae

Changing the subject of a formula — the most important engineering algebra skill. Used every time you need to find a different variable from a known formula.

Golden RuleDo the same operation to BOTH sides. Work backwards through BODMAS.

Step-by-StepRearrange to make the target variable the subject

Identify what you want to make the subject
Move everything else to the other side (reverse BODMAS order)
If the variable appears twice, factorise it out first
Check by substituting numbers into both versions

📌 Reverse BODMAS: undo operations in reverse order — undo + and − first, then × and ÷, then powers and roots.

Ohm’s LawV = IR — rearrange for I and R

V = I × R

Make I the subject

Start: V = I × R

÷R both: V/R = I

Answer: I = V/R ✅

Make R the subject

Start: V = I × R

÷I both: R = V/I ✅

Make u the subject of: v = u + at

Start: v = u + at

−at both: v − at = u

Answer: u = v − at ✅

⚙️ Make r the subject of: A = πr²

Start: A = πr²

÷π both: A/π = r²

√ both: r = √(A/π) ✅

⚙️ 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 ✅

Make v the subject of KE = ½mv²

Start: KE = ½mv²

×2 both: 2KE = mv²

÷m both: 2KE/m = v²

√ both: v = √(2KE/m) ✅

📊 Try it — Engineering Formula CalculatorRearrange & solve

ℹ️ How to use: Use the dropdown to pick a formula, then type the known values. The tool rearranges and solves for the missing quantity.

Formula

Find

👆Try: V=IR, find I V=12, R=4 · σ=F/A, find A F=6600, σ=120 · KE, find v KE=100, m=2

⚙️

Applications

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

⚡Electrical: Simplify: R=R₁+R₁+R₂ = 2R₁+R₂.
🔧Mechanical: Net force: F_net=F₁+F₂−F₃. Collect terms.
🏗️Construction: Total bags: 3x+2x+x = 6x. Simplify before ordering.
🌡️Daily Life: Total cost at checkout: 3a+2b+a = 4a+2b.

⚙

Algebra in Engineering

Every engineering formula is algebra in action. Here’s where each skill appears in real T Level engineering contexts.

⚡ Electrical — Ohm’s Law & Power

Rearrange V=IR and P=IV to find any variable. This is the most fundamental engineering algebra skill.

V = IR → I = V/R → R = V/I P = IV → I = P/V → V = P/I P = I²R → I = √(P/R)
Example: V=12V, R=4Ω I = 12/4 = 3A P = 3 × 12 = 36W

🔨 Mechanics — SUVAT Equations

The 5 equations of motion all need rearranging. Algebra lets you find whichever variable you need.

v = u + at → a = (v−u)/t s = ut + ½at² → t needs quadratic formula v² = u² + 2as → a = (v²−u²)/2s
Example: u=0, a=9.81, t=3s v = 0 + 9.81×3 = 29.43 m/s

🏗 Structural — Stress & Strain

σ = F/A and E = σ/ε both need rearranging to find different quantities from measured values.

σ = F/A → F = σA → A = F/σ E = σ/ε → ε = σ/E → σ = Eε
Example: F=4500N, A=50mm² σ = 4500/50 = 90 N/mm² = 90 MPa

⚙ Gears — Velocity Ratio & MA

Gear ratios, mechanical advantage and torque equations all use algebraic manipulation of simple formulae.

VR = N₂/N₁ → N₂ = VR × N₁ T = F × r → F = T/r P = T × ω → T = P/ω
ω = 2πn/60 (n in rpm)

🔧 T Level Exam — Rearrange and solve using stress formula

Problem: A bolt must withstand a tensile stress no greater than 120 MPa. Find the minimum cross-sectional area if the load is 6.6 kN.

Formula: σ = F/A → A = F/σ

Convert: F = 6.6 kN = 6600 N σ = 120 N/mm²

Calculate: A = 6600 / 120

Answer: A = 55 mm² ✅

⚙️ Try it — Quadratic Formulaax²+bx+c=0

ℹ️ How to use: Type values for a, b and c in ax² + bx + c = 0. The roots are calculated step by step.

x = (−b ± √(b²−4ac)) / 2a

a

b

c

👆Try: a=1,b=−5,c=6 (neat roots 2&3) · a=1,b=2,c=1 (repeated) · a=1,b=0,c=−9

🎮

Quick Fire Quiz

Test your knowledge — 10 questions, instant feedback.

⚡ Algebra 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 in 3 tiers — T Level / Level 3 Engineering style with full working shown.

🟢 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.

Q1

Simplify: 7x + 3y − 4x + 2y

[2]

x terms: 7x − 4x = 3x
y terms: 3y + 2y = 5y
Answer: 3x + 5y

Q2

Evaluate 3x² − 2x + 5 when x = 4

[2]

3(4²) − 2(4) + 5 = 3(16) − 8 + 5 = 48 − 8 + 5 = 45

Q3

Simplify: x´ × x³

[1]

Add powers: 4+3 = x⁷

Q4

Expand: 4(3x − 5)

[2]

4×3x = 12x 4×(−5) = −20
= 12x − 20

Q5

Solve: 4x + 3 = 19

[2]

4x = 19 − 3 = 16
x = 16 ÷ 4 = x = 4

🟡 Tier 2 — Core Skills

Q6

Simplify: (3x³y²)²

[2]

3²=9 (x³)²=x⁶ (y²)²=y⁴
= 9x⁶y⁴

Q7

Expand and simplify: (x + 5)(x − 3)

[3]

F: x² O: −3x I: 5x L: −15
Collect: x² + 2x − 15
= x² + 2x − 15

Q8

Factorise fully: 12x²y − 8xy²

[2]

HCF = 4xy
12x²y ÷ 4xy = 3x 8xy² ÷ 4xy = 2y
= 4xy(3x − 2y)

Q9

Solve: 3x − 7 = x + 9

[3]

3x − x = 9 + 7
2x = 16
x = 8 Check: 3(8)−7=17 = 8+9=17 ✔

Q10

Make t the subject of: v = u + at

[2]

v − u = at
t = (v − u) / a

🔴 Tier 3 — T Level Challenge

Q11

Factorise: x² − 3x − 18

[2]

Need: × to −18, + to −3 → −6 and +3
(x − 6)(x + 3)

Q12

Solve: 2x² − 7x + 3 = 0. Give exact answers.

[4]

Factorise: (2x−1)(x−3)=0
2x−1=0 → x=1/2 x−3=0 → x=3
x = 0.5 or x = 3

Q13

Make r the subject of: V = (4/3)πr³

[3]

×3/4π both: 3V/4π = r³
Cube root: r = ∛(3V/4π)

Q14

A resistor network has R_total = R₁R₂/(R₁+R₂). Make R₂ the subject.

[4]

Hint: multiply both sides by (R₁+R₂) first

R_T(R₁+R₂) = R₁R₂
R_T⋅R₁ + R_T⋅R₂ = R₁R₂
R_T⋅R₁ = R₁R₂ − R_T⋅R₂
R_T⋅R₁ = R₂(R₁ − R_T)
R₂ = R_T⋅R₁ / (R₁ − R_T)

Q15

A beam deflects by δ = FL³/(3EI). A steel beam (E=200 GPa, I=4.5×10⁶ mm⁴) is 2 m long and must not deflect more than 5 mm. Find the maximum load F in kN.

[5]

Rearrange: F = 3δEI/L³
δ=5mm, E=200,000 N/mm², I=4.5×10⁶mm⁴, L=2000mm
L³ = 8×10⁹ mm³
F = (3×5×200000×4.5×10⁶) / (8×10⁹)
= (1.35×10ʳ) / (8×10⁹)
= 1.6875×10³ N
= 1.69 kN ✅

🎯 Score guide: Q1–5 = foundations (9 marks) · Q6–10 = core skills (12 marks) · Q11–15 = T Level challenge (18 marks)