Linear equations | AQA GCSE Maths Notes

Visual model

Keep both sides balanced

2x+3

11

same operation to both sides

Gold-standard guide

22 mins

What you will learn

I can solve one- and two-step equations using inverse operations.

I can solve equations with brackets by expanding or dividing first.

I can solve equations with unknowns on both sides.

I can form and solve an equation from a worded or geometric context.

I can verify a solution by substituting back into the equation.

Useful before you start

Inverse operations (addition ↔ subtraction, etc.)Expanding bracketsCollecting like termsArithmetic with negative numbersSubstitution

Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

Equation

A statement that two expressions are equal, shown with = (e.g. 3x + 5 = 20).

Step 1

Variable (unknown)

A letter standing for a value we need to find (e.g. x in 4x − 3 = 9).

Step 2

Inverse operation

The operation that undoes another (e.g. ÷ undoes ×, − undoes +).

Step 3

Solution

The value that makes the equation true (e.g. x = 4 satisfies 3x = 12).

Step 4

Coefficient

The number multiplying a variable — in 7x the coefficient is 7.

Step 5

Balance principle

Whatever you do to one side of an equation, do the same to the other.

Balance rule

Apply identical operations to both sides. E.g. if 3x = 12, divide both sides by 3 to get x = 4.

Equations with brackets: expand first

Multiply every term inside the bracket by the factor outside before solving. E.g. 4(x − 2) = 12 becomes 4x − 8 = 12.

Unknown on both sides: collect on one side

g. 5x + 1 = 3x + 9 → subtract 3x → 2x + 1 = 9), then solve the two-step equation.

Verification: substitute the solution

Replace x with your answer in the original equation. If the left side equals the right side, the solution is correct.

Worked example

The lengths of the sides of a triangle are (2x + 3) cm, (x + 5) cm and (3x − 4) cm. The perimeter is 40 cm. Find x and hence find the length of each side.

Perimeter = sum of all three sides: (2x + 3) + (x + 5) + (3x − 4) = 40.

Collect like terms on the left: 6x + 4 = 40.

Subtract 4 from both sides: 6x = 36.

Divide by 6: x = 6.

Sides: 2(6)+3 = 15 cm, 6+5 = 11 cm, 3(6)−4 = 14 cm.

Check: 15 + 11 + 14 = 40. ✓

Final answer

x = 6; sides are 15 cm, 11 cm and 14 cm.

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Diagnosticrecall

Solve x + 9 = 22.

1 mark1 mineq-d1

Show solution

Worked solution

1.The operation on x is +9.
2.Apply the inverse: subtract 9 from both sides.
3.x = 22 − 9 = 13.

Final answer

x = 13

Mark points

B1: x = 13 (no working required for a one-step equation, but working showing subtraction of 9 from both sides is accepted and encouraged).

Watch out

Adding 9 to 22 (performing the operation rather than its inverse), giving x = 31.

Easyprocedure

Solve 5x − 3 = 17.

2 marks2 minseq-e1

Show solution

Worked solution

1.Undo the subtraction first: add 3 to both sides → 5x = 20.
2.Undo the multiplication: divide both sides by 5 → x = 4.
3.Check: 5(4) − 3 = 17. ✓

Final answer

x = 4

Mark points

M1: correct first step seen — either 5x = 20 or equivalent operation shown on both sides.
A1: x = 4.

Watch out

Dividing by 5 before adding 3 (reversing the correct order of inverse operations), which gives x = $1\frac{4}{5}$ .

Mediumreasoning

Solve 4(3x − 2) = 28.

3 marks3 minseq-m1

Show solution

Worked solution

1.Method 1 (expand first): 12x − 8 = 28 → 12x = 36 → x = 3.
2.Method 2 (divide first): 3x − 2 = 7 → 3x = 9 → x = 3.
3.Check: 4(3×3 − 2) = 4(7) = 28. ✓

Final answer

x = 3

Mark points

M1: correct expansion 12x − 8 = 28, OR correct division 3x − 2 = 7.
M1: correct second step leading to 12x = 36 or 3x = 9.
A1: x = 3.

Watch out

Expanding as 4 × 3x − 2 = 28 (forgetting to multiply −2 by 4 as well), giving 12x − 2 = 28 and then x = $\frac{5}{2}$ .

Hardreasoning

Solve 5(x + 2) = 3(2x − 1) + 7.

4 marks4 minseq-h1

Show solution

Worked solution

1.Expand both sides: 5x + 10 = 6x − 3 + 7.
2.Simplify the right side: 5x + 10 = 6x + 4.
3.Collect x-terms: subtract 5x from both sides → 10 = x + 4.
4.Solve: x = 6.
5.Check: 5(8) = 40 and 3(11) + 7 = 33 + 7 = 40. ✓

Final answer

x = 6

Mark points

M1: correct expansion of at least one bracket.
M1: correct expansion of both brackets and simplified right side 6x + 4.
M1: collects x-terms correctly onto one side.
A1: x = 6.

Watch out

g. writing 5x + 6x = 4 − 10), which reverses the sign of one x-term.

Very Hardproblem solving

A rectangle has length (3x + 1) cm and width (x − 2) cm. Its area equals that of a square of side (x + 3) cm. Find x and give the area of each shape.

5 marks5 minseq-vh1

Show solution

Worked solution

1.Form the equation: (3x + 1)(x − 2) = (x + 3)².
2.Expand left: 3x² − 6x + x − 2 = 3x² − 5x − 2.
3.Expand right: x² + 6x + 9.
4.Set equal: 3x² − 5x − 2 = x² + 6x + 9.
5.Rearrange: 2x² − 11x − 11 = 0. Hmm — let's check: this gives non-integer roots.
6.Re-examine: expand left as (3x+1)(x−2) = 3x² − 5x − 2. Expand right as (x+3)² = x² + 6x + 9.
7.3x² − 5x − 2 = x² + 6x + 9 → 2x² − 11x − 11 = 0.
8.Discriminant = 121 + 88 = 209 — not a perfect square, so x is irrational.This is the key challenge: students must use the quadratic formula.
9.x = (11 ± √209) / 4. Taking the positive root (width must be positive: x > 2): x = (11 + √209)/4 ≈ 6.36.
10.Width x − 2 > 0 requires x > 2, which is satisfied.

Final answer

x = (11 + √209)/4 ≈ 6.36 cm (exact form required for full marks). Area of each shape ≈ 54.1 cm².

Mark points

M1: correctly expands (3x + 1)(x − 2) to obtain 3x² − 5x − 2.
M1: correctly expands (x + 3)² to obtain x² + 6x + 9.
M1: rearranges to form 2x² − 11x − 11 = 0.
M1: applies quadratic formula with a = 2, b = −11, c = −11.
A1: x = (11 + √209)/4 (exact) with a valid reason for rejecting the negative root.

Watch out

Assuming the quadratic will factorise nicely and spending time trying integer factor pairs instead of proceeding to the quadratic formula.

Grade 9 stretchproblem solving

Solve the equation (2x − 1)/(x + 3) = (x − 2)/(x − 1). Find all solutions and check for any values that must be excluded.

6 marks6 minseq-g9

Show solution

Worked solution

1.Cross-multiply: (2x − 1)(x − 1) = (x − 2)(x + 3).
2.Expand left: 2x² − 2x − x + 1 = 2x² − 3x + 1.
3.Expand right: x² + 3x − 2x − 6 = x² + x − 6.
4.Set equal: 2x² − 3x + 1 = x² + x − 6.
5.Rearrange: x² − 4x + 7 = 0.
6.Discriminant: 16 − 28 = −12 < 0. No real solutions.
7.Excluded values: x ≠ −3 (denominator of left side = 0) and x ≠ 1 (denominator of right side = 0).
8.Conclusion: the equation has no real solutions.

Final answer

No real solutions (discriminant is negative). Excluded values are x = −3 and x = 1.

Mark points

M1: cross-multiplies correctly.
M1: expands both products correctly.
M1: rearranges to x² − 4x + 7 = 0.
M1: calculates discriminant b² − 4ac correctly.
A1: states no real solutions with discriminant < 0 shown explicitly.
A1: identifies both excluded values x = −3 and x = 1.

Watch out

Stopping after rearranging to x² − 4x + 7 = 0 and attempting to factorise without checking the discriminant first, leading to wasted time.

Exam Techniqueexam trap

Solve 3 − 2x = 11. A student obtains x = 7. Show the error in their reasoning and give the correct answer.

3 marks3 minseq-et1

Show solution

Worked solution

1.The student likely subtracted 3 first: −2x = 11 − 3 = 8, then divided by −2 to get x = −4.
2.Or: they ignored the negative sign on 2x and solved 2x = 11 − 3 = 8, giving x = 4.
3.Most likely error: they subtracted 2x from both sides rather than adding, or forgot the negative.
4.Correct working: 3 − 2x = 11 → −2x = 8 → x = −4.
5.Check: 3 − 2(−4) = 3 + 8 = 11. ✓

Final answer

x = −4. The error is treating −2x as +2x (ignoring the negative coefficient), giving x = 4 or x = 7.

Mark points

B1: clearly identifies the specific sign error made by the student.
M1: correct algebraic step −2x = 11 − 3 = 8 shown.
A1: x = −4 with correct check shown.

Watch out

Accepting x = 7 without question because it is positive, rather than substituting it back to verify.

Timed checkpoint

12 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Linear equations - 2 marksSolve 3x + 7 = 22Mark answer

Answer

x = 5

2Algebraic notation and substitution - 2 marksFind t² + 4t when t = −2Mark answer

Answer

−4

3Simplifying expressions - 2 marksSimplify 2xy + 3x²y − xy + x²yMark answer

Answer

3x²y + xy

4Expanding and factorising - 3 marksFactorise fully 6x² + 9xMark answer

Answer

3x(2x + 3)

Mastery check

I can recall the balance principle and name the inverse operation needed at each step of solving a linear equation.
I can solve two-step equations and equations with brackets, showing every algebraic step on a separate line.
I can solve equations with the unknown on both sides by systematically collecting letter terms and then number terms.
I can form a linear equation from a geometry or word problem, solve it and interpret the answer in context.
I can apply equation-solving in an unfamiliar algebraic context — such as equations with algebraic fractions or a quadratic arising from a geometric equality — and verify every solution by substitution.

Ask Nova Bot

Show me why the balance principle works using a physical scales diagram, and then apply it to solve 5x − 4 = 11.I'm stuck on solving 3(2x + 1) = 27 — can you give me just one small nudge without solving it for me?Quiz me on linear equations: give me five questions getting harder, wait for my answer each time and tell me what I got right or wrong.What are the top three sign-error mistakes students make in equations involving negative numbers, and how do I avoid them?Show me a real engineering or science formula where rearranging to find an unknown is exactly like solving a linear equation.

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