Free GCSE Maths lesson: Algebra
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Free Lessons → GCSE / Key Stage 4 → Maths → Linear Equations

Lesson 30 · GCSE / Key Stage 4 · Maths · Algebra

Linear Equations

Learn how to solve equations by keeping both sides balanced and undoing operations in a clear order.

Qualification: GCSE Key Stage 4 Subject: Maths Strand: Algebra

GCSE specification fit

A core algebra method for every exam board.

Linear equations ask you to find the value of an unknown. They appear as direct algebra questions and inside worded GCSE problems.

QualificationGCSE Mathematics
Key stageKey Stage 4
StrandAlgebra
Tier guidanceFoundation and Higher

What you will learn

  • What it means to solve an equation.
  • How to use inverse operations.
  • How to solve two-step equations.
  • How to handle brackets and check by substitution.
  • How to solve equations with the unknown on both sides.
  • How to form and solve equations from worded contexts.

Why this matters

Equations turn unknown quantities into something you can work out. They are used in number problems, geometry, formulae, graphs and ratio questions.

Clear equation solving is also one of the easiest ways to protect method marks, because each balanced step shows what you did.

Prior knowledge

You should already be comfortable with:

  • using inverse operations such as add and subtract, multiply and divide,
  • substituting a number for a letter,
  • expanding brackets,
  • collecting like terms,
  • working with negative numbers.

Clear explanation

What solving means

An equation says two expressions are equal. To solve it, find the value of the unknown that makes the statement true.

x + 5 = 12 If x = 7, then 7 + 5 = 12 So x = 7 is the solution

Keep the equation balanced

Whatever you do to one side, do to the other side. This keeps the equality true while you isolate the unknown.

3x + 4 = 19 3x = 15 x = 5

Undo in the reverse order

If the unknown has been multiplied and then added to, undo the addition first, then undo the multiplication.

5y − 8 = 27 5y = 35 y = 7

Expand brackets first when needed

Brackets are easier to solve after you expand and simplify the equation.

2(x + 3) = 18 2x + 6 = 18 2x = 12 x = 6

Unknowns on both sides

Move the smaller unknown term first if it keeps the coefficient positive. Then solve the simpler equation.

7x − 5 = 3x + 11 4x − 5 = 11 4x = 16 x = 4

Form an equation from words

In problem questions, define the unknown first. Then translate each part of the sentence into algebra before solving.

Three tickets and a £2 booking fee cost £20 Let t be the ticket price: 3t + 2 = 20 3t = 18, so t = 6

Check at the end

Substitute your answer into the original equation. If both sides match, your solution works.

Check x = 4 in 7x − 5 = 3x + 11 Left side: 7 × 4 − 5 = 23 Right side: 3 × 4 + 11 = 23

Worked examples

Example 1: One-step equation

Solve x − 9 = 14.

Add 9 to both sides. x = 23
Answer: x = 23.

Example 2: Two-step equation

Solve 4a + 7 = 31.

4a = 24 a = 6
Answer: a = 6.

Example 3: Brackets

Solve 3(2m − 1) = 27.

6m − 3 = 27 6m = 30 m = 5
Answer: m = 5.

Example 4: Unknown on both sides

Solve 9p + 2 = 5p + 18.

4p + 2 = 18 4p = 16 p = 4
Answer: p = 4.

Quick checks

Choose an answer, then check your thinking.

1. Solve x + 6 = 15.

2. Solve 2x + 5 = 17.

3. Solve 6x − 4 = 2x + 20.

Practice questions

Question 1

Solve x + 8 = 21.

Reveal answer and marking guidance

Answer: x = 13.

Marking: Subtract 8 from both sides and write the final value of x.

Question 2

Solve 5y = 45.

Reveal answer and marking guidance

Answer: y = 9.

Marking: Divide both sides by 5.

Question 3

Solve 4a − 3 = 25.

Reveal answer and marking guidance

Answer: a = 7.

Marking: Add 3 first to get 4a = 28, then divide by 4.

Question 4

Solve 2(n + 5) = 28.

Reveal answer and marking guidance

Answer: n = 9.

Marking: Expand to 2n + 10 = 28, or divide by 2 first; both methods should reach n = 9.

Question 5

Solve 8p − 7 = 3p + 18.

Reveal answer and marking guidance

Answer: p = 5.

Marking: Subtract 3p to collect unknown terms, then add 7 and divide by 5.

Question 6

Solve x ÷ 3 + 4 = 10

Reveal answer and marking guidance

Answer: x = 18.

Marking: Subtract 4 to get x ÷ 3 = 6, then multiply by 3.

Question 7

Solve 3(2x − 1) = x + 22.

Reveal answer and marking guidance

Answer: x = 5.

Marking: Expand to 6x − 3 = x + 22, collect to get 5x = 25, then divide by 5.

Question 8

Three identical notebooks and a £2 pen cost £14.60 altogether. Form and solve an equation to find the cost of one notebook.

Reveal answer and marking guidance

Answer: one notebook costs £4.20.

Marking: Let n be the notebook cost. Solve 3n + 2 = 14.60, so 3n = 12.60 and n = 4.20. Include the money unit.

Question 9

Solve (x + 4) ÷ 3 = (x − 2) ÷ 2

Reveal answer and marking guidance

Answer: x = 14.

Marking: Multiply by 6 to clear the denominators: 2(x + 4) = 3(x − 2). Then 2x + 8 = 3x − 6, so x = 14.

Question 10

Solve (2x − 6) ÷ 4 = (x + 1) ÷ 3

Reveal answer and marking guidance

Answer: x = 11.

Marking: Multiply by 12 to clear the denominators: 3(2x − 6) = 4(x + 1). Then 6x − 18 = 4x + 4, so 2x = 22 and x = 11. Check in the original equation: (22 − 6) ÷ 4 = 4 and (11 + 1) ÷ 3 = 4.

Answers and marking guidance

The exact practice answers are hidden under each question so you can try first. For linear equations, marks usually come from defining the unknown when needed, showing balanced inverse operations, expanding brackets accurately, clearing fractions or decimals carefully, collecting unknown terms on one side, keeping signs attached to their terms, and checking the answer in the original equation or context.

Common mistakes

  • Changing only one side: every operation must keep the equation balanced.
  • Undoing steps in the wrong order: in 3x + 4 = 19, subtract 4 before dividing by 3.
  • Losing negative signs: −2x and +2x behave differently when moved or divided.
  • Forgetting to expand brackets: 2(x + 3) is 2x + 6, not 2x + 3.
  • Not defining the unknown: in worded questions, say what the letter represents before forming the equation.
  • Not checking: substituting the answer back catches many arithmetic slips.

Extension challenge

Solve 3(2x − 5) = 4x + 17.

Reveal answer

Answer: x = 16.

Expand to 6x − 15 = 4x + 17, then 2x = 32, so x = 16.

Exam-board guidance

Linear equations are assessed by every GCSE Maths board. They can be direct algebra questions or part of a worded problem where you must form the equation first.

AQA GCSE Maths

Write each inverse operation on a new line, keep both sides balanced, and substitute your answer into the original equation to check it.

OCR GCSE Maths

In worded problems, define the unknown first, form one balanced equation, then give the solved value in the original context.

Pearson Edexcel GCSE Maths

If brackets, fractions or decimals appear, simplify carefully before isolating the unknown and give contextual answers with units where needed.

Eduqas GCSE Maths

Keep equal signs lined up, collect like terms carefully, and avoid skipping the line where the equation becomes simpler.

WJEC Wales

Equations may model costs, measures or limits, so define the unknown, translate the words into algebra before solving, and finish with sensible units.

CCEA GCSE Maths

Be ready for equations in calculator and non-calculator units, with clear balancing steps and a quick substitution check.

Next lesson

Next, move into Inequalities.

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