What this page covers
Equations, inequalities and formulae are three of the most consistently tested areas in GCSE Maths. Together they appear on almost every paper at both Foundation and Higher tier. This page covers:
- Linear equations — solving equations with one unknown
- Quadratic equations — solving by factorising, completing the square and the formula
- Inequalities — solving and representing linear and quadratic inequalities
- Rearranging formulae — changing the subject of a formula
Each section includes worked examples followed by graded practice questions. Full answers are at the bottom of the page.
Linear equations
A linear equation has one unknown (usually x) with no powers higher than 1. The goal is to isolate the unknown on one side by performing the same operation to both sides.
Key rule: whatever you do to one side, you must do to the other.
Worked Example 1 — One-step equation
Solve: 4x − 3 = 17
1
Add 3 to both sides: 4x = 20
2
Divide both sides by 4: x = 5
Answerx = 5
Worked Example 2 — Unknown on both sides
Solve: 5x + 2 = 2x + 14
1
Subtract 2x from both sides: 3x + 2 = 14
2
Subtract 2 from both sides: 3x = 12
3
Divide both sides by 3: x = 4
Answerx = 4
Worked Example 3 — Equation with brackets
Solve: 3(2x − 1) = 2(x + 7)
1
Expand both sides: 6x − 3 = 2x + 14
2
Subtract 2x from both sides: 4x − 3 = 14
3
Add 3 to both sides: 4x = 17
Answerx = 4.25
Quadratic equations
A quadratic equation has the form ax² + bx + c = 0. At GCSE you need three methods: factorising, completing the square, and the quadratic formula.
Worked Example 4 — Solving by factorising
Solve: x² + 5x + 6 = 0
1
Find two numbers that multiply to 6 and add to 5: +2 and +3
2
Factorise: (x + 2)(x + 3) = 0
3
Set each bracket to zero: x = −2 or x = −3
Answerx = −2 or x = −3
Worked Example 5 — Quadratic formula
Solve: 2x² − 3x − 2 = 0
1
Identify a = 2, b = −3, c = −2
2
Substitute into x = (−b ± √(b² − 4ac)) / 2a: x = (3 ± √(9 + 16)) / 4 = (3 ± 5) / 4
3
Two solutions: x = 8/4 = 2 or x = −2/4 = −0.5
Answerx = 2 or x = −0.5
Inequalities
Inequalities are solved in the same way as equations — with one important exception: when you multiply or divide both sides by a negative number, the inequality sign flips.
Number line notation: an open circle ○ means the value is not included (< or >); a filled circle ● means it is included (≤ or ≥).
Worked Example 6 — Linear inequality
Solve: 3x − 4 > 11 and represent it on a number line
1
Add 4 to both sides: 3x > 15
3
On a number line: open circle at 5, arrow pointing right
Answerx > 5
Worked Example 7 — Double inequality
Solve: −1 ≤ 2x + 3 < 11
1
Subtract 3 from all three parts: −4 ≤ 2x < 8
2
Divide all three parts by 2: −2 ≤ x < 4
Answer−2 ≤ x < 4
To change the subject of a formula, apply inverse operations to isolate the desired variable — exactly like solving an equation. Work step by step: undo addition/subtraction first, then multiplication/division, then powers/roots.
Worked Example 8 — Simple rearrangement
Make u the subject of: v = u + at
1
Subtract at from both sides: u = v − at
Answeru = v − at
Worked Example 9 — Rearrangement with a power
Make r the subject of: V = (4/3)πr³
1
Multiply both sides by 3: 3V = 4πr³
2
Divide both sides by 4π: r³ = 3V / 4π
3
Take the cube root: r = ∛(3V / 4π)
Answerr = ∛(3V / 4π)
Practice questions
Work through each question before checking the answers. Difficulty is shown for each question.
Foundation (Grade 3–5)
Q1Solve: 3x + 7 = 22Foundation
Q2Solve: 4x − 5 = 2x + 9Foundation
Q3Solve: 2(x + 3) = 14Foundation
Q4Solve the inequality: 2x + 1 ≤ 9Foundation
Q5Make t the subject of: s = d / tFoundation
Higher (Grade 5–7)
Q6Solve: x² − 7x + 12 = 0 by factorisingHigher
Q7Solve: 5 − 2x > 1, and state the largest integer that satisfies the inequalityHigher
Q8Solve: −3 < 2x − 1 ≤ 7Higher
Q9Make x the subject of: y = (x + 3) / 5Higher
Q10Make a the subject of: v² = u² + 2asHigher
Higher — Hard (Grade 8–9)
Q11Solve using the quadratic formula (give answers to 2 d.p.): 3x² + 5x − 2 = 0Grade 8–9
Q12Solve: x² − 5x ≤ 0 and represent the solution on a number lineGrade 8–9
Q13Make r the subject of: A = 2πr² + 2πrhGrade 8–9
Q14Solve: (x + 1)² = 25Grade 8–9
Q15Make x the subject of: p = √((x + q) / r)Grade 8–9
Answers
Foundation (Q1–Q5)
Q1x = 5(subtract 7, divide by 3)
Q2x = 7(subtract 2x: 2x − 5 = 9; add 5; divide by 2)
Q3x = 4(expand: 2x + 6 = 14; subtract 6; divide by 2)
Q4x ≤ 4(subtract 1: 2x ≤ 8; divide by 2)
Q5t = d / s(multiply both sides by t, then divide by s)
Higher (Q6–Q10)
Q6x = 3 or x = 4((x − 3)(x − 4) = 0)
Q7x < 2; largest integer = 1(subtract 5: −2x > −4; divide by −2, flip sign)
Q8−1 < x ≤ 4(add 1 throughout: −2 < 2x ≤ 8; divide by 2)
Q9x = 5y − 3(multiply by 5: x + 3 = 5y; subtract 3)
Q10a = (v² − u²) / 2s(subtract u²; divide by 2s)
Higher — Hard (Q11–Q15)
Q11x = 0.33 or x = −2.00(a=3, b=5, c=−2; discriminant = 25+24 = 49; x = (−5 ± 7)/6)
Q120 ≤ x ≤ 5(factorise: x(x − 5) ≤ 0; roots at x=0 and x=5; parabola ≤ 0 between roots)
Q13r = (−2πh ± √(4π²h² + 8πA)) / 4π(treat as quadratic in r: 2πr² + 2πrh − A = 0)
Q14x = 4 or x = −6(square root both sides: x + 1 = ±5)
Q15x = p²r − q(square both sides: p² = (x+q)/r; multiply by r; subtract q)
Common mistakes
These are the errors Alamin sees most frequently with equations, inequalities and formulae at GCSE.
Common Mistake 1
Not applying the operation to both sides
When solving 3x + 7 = 22, some students subtract 7 from only the left side to get 3x = 22. Every operation must be applied to both sides — the correct step is 3x = 22 − 7 = 15.
Common Mistake 2
Forgetting ± when square rooting
Solving x² = 25, students often write only x = 5. Square roots have two solutions: x = 5 and x = −5. Always write ±.
Common Mistake 3
Not flipping the inequality sign
Dividing or multiplying an inequality by a negative number flips the sign. For −2x > 6, dividing by −2 gives x < −3, not x > −3. This is the most commonly lost mark on inequality questions.
Common Mistake 4
Errors expanding brackets before solving
When solving 3(2x − 1) = 15, a common error is writing 6x − 1 = 15 (forgetting to multiply the −1 by 3). Always expand every term: 6x − 3 = 15.
Common Mistake 5
Rearranging formulae — only moving part of a term
Making v the subject of E = ½mv²: dividing by m gives E/m = ½v² — but students often skip the ½ and write v² = E/m. Divide by the entire coefficient: v² = 2E/m.
Common Mistake 6
Using open/closed circles incorrectly on number lines
For x > 3, use an open circle (3 is not included). For x ≥ 3, use a closed (filled) circle (3 is included). Mixing these up loses a mark even when the inequality itself is correct.
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