Linear Equations · Study Notes · Edexcel IGCSE 4MA1 Higher

Solve any linear equation. Brackets, fractions, unknowns on both sides — Higher Tier expects fluency with all variants.

At a glance

Linear = highest power $1$ . ONE solution at most.
Brackets: expand first.
Both sides: gather $x$ terms one side.
Fractions: multiply through by LCM (or cross-multiply).
Word problems: define variable, then translate.
Always show every step.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

2.7 — Solve linear equations in one variable.
2.7 — Solve equations with brackets, fractions, unknowns on both sides.
2.7 — Set up linear equations from real-world contexts.

The standard solving method

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Inverse operations. Both sides. One step at a time.

Steps:

Expand any brackets.
Multiply through to clear fractions if convenient.
Gather variable terms on ONE side.
Gather constants on the OTHER side.
Divide by the coefficient of the variable.

Example. $3(2x - 5) + 4 = 25$ .

Expand: $6x - 15 + 4 = 25$ → $6x - 11 = 25$ .
Add $11$ : $6x = 36$ .
Divide: $x = 6$ .

An equation balances — every step must be applied to both sides.

With unknown on both sides.

Move the variable terms to the side with the larger coefficient (keeps positive coefficient).

Example. $5x + 4 = 2x + 19$ .

Subtract $2x$ from both sides: $3x + 4 = 19$ .
Subtract $4$ : $3x = 15$ .
Divide: $x = 5$ .

With fractions.

Multiply both sides by the LCM of the denominators to clear fractions.

Example. $\dfrac{x + 2}{3} = \dfrac{2x - 1}{4}$ .

LCM of $3, 4$ is $12$ . Or: cross-multiply.
$4(x + 2) = 3(2x - 1)$ .
Expand: $4x + 8 = 6x - 3$ .
Rearrange: $11 = 2x$ , so $x = 5.5$ .

Worked qualitative. Why move variable terms to the side with the larger coefficient?

Keeps the coefficient positive.
Less error-prone with sign mistakes.
$5x - 2x = 3x$ is cleaner than $-2x + 5x = 3x$ (but mathematically the same).

Edexcel tip. Show every step. Mark schemes credit each operation explicitly.

Expand brackets first.
Clear fractions by multiplying.
Gather $x$ on one side.
Divide by coefficient.

Equations with fractions

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Multiply through by LCM to clear. Or cross-multiply for two-fraction equations.

Method 1: Multiply by LCM.

For $\dfrac{x}{3} + \dfrac{x}{4} = 7$ :

LCM of $3, 4$ is $12$ .
Multiply through: $4x + 3x = 84$ .
Solve: $7x = 84$ , $x = 12$ .

Method 2: Cross-multiply (two fractions equal).

$\dfrac{a}{b} = \dfrac{c}{d}$ implies $ad = bc$ .

Example: $\dfrac{x + 1}{2} = \dfrac{x - 3}{5}$ .

Cross: $5(x + 1) = 2(x - 3)$ .
Expand: $5x + 5 = 2x - 6$ .
Solve: $3x = -11$ , $x = -11/3$ .

More complex.

$\dfrac{2x + 1}{3} - \dfrac{x - 1}{2} = 1$ .

Multiply through by LCM $= 6$ :

$2(2x + 1) - 3(x - 1) = 6$ .
$4x + 2 - 3x + 3 = 6$ .
$x + 5 = 6$ .
$x = 1$ .

Worked qualitative. Why does multiplying both sides by the LCM clear the fractions?

Each fraction is multiplied by the LCM, eliminating its denominator.
Result is a fraction-free equation.
Easier to solve.

Edexcel tip. Show the multiplication step: 'multiplying both sides by $12$ '. Mark schemes give M1 for clearing the fractions.

Multiply by LCM of denominators.
Or cross-multiply for two equal fractions.
Expand brackets after.
Gather, factor, divide.

Word problems — setting up linear equations

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Define variable. Translate words to equation. Solve.

Process:

Read carefully — what's the unknown?
Define the variable. 'Let $x$ be ...'.
Translate the relationships to an equation.
Solve the equation.
Check the answer makes sense.

Common phrases:

Phrase	Maths
'A number'	$x$
'Twice a number'	$2x$
'A number plus $5$ '	$x + 5$
'A number squared'	$x^{2}$
'Three times a number, plus $7$ '	$3x + 7$
'The product of $a$ and $b$ '	$ab$
'The sum of $a$ and $b$ '	$a + b$
'The difference of $a$ and $b$ '	$a - b$

Worked example. I think of a number. I double it then add $7$ . The result is $25$ .

Let $n$ = the number.
Double then add $7$ : $2n + 7$ .
Equals $25$ : $2n + 7 = 25$ .
Solve: $2n = 18$ , $n = 9$ .

Worked example. Three consecutive integers sum to $54$ . Find them.

Let smallest be $n$ . Other two: $n + 1, n + 2$ .
Sum: $n + (n + 1) + (n + 2) = 54$ .
$3n + 3 = 54$ , $3n = 51$ , $n = 17$ .
Three numbers: $17, 18, 19$ .

Translate each phrase in turn into algebra to build the equation.

Worked qualitative. Why is defining the variable EXPLICITLY the first step?

Makes the equation clear.
Reader (and you) can see what each letter means.
Prevents you from setting up the WRONG equation.
Edexcel mark schemes typically award the first M1 for this step.

Edexcel tip. Always state 'Let $x$ be ...' explicitly. The phrasing pays for itself in marks.

Define the variable first.
Translate words systematically.
Solve carefully.
Check the answer.

Quick recap

Linear: power 1, one solution.
Expand brackets, clear fractions.
Gather $x$ on one side.
Divide by coefficient.
Word problems: define, translate, solve, check.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Cross-multiply for two equal fractions.
Multiply by LCM for multiple fractions.
Define variables before setting up.
Both sides for every operation.

How it’s examined

Linear equations appear every Paper 1H + 2H. Standalone (3-4 marks) and embedded throughout. Examiner reports flag (1) sign errors when moving terms, (2) not expanding brackets, (3) not defining the variable in word problems.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H Jan 2024, 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on linear equations, written exactly the way a tutor would explain them at the board.

1Solve a simple linear equation
Foundation• linear, solve
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Question
Solve $4x - 7 = 13$ .
Step-by-step solution
1. Step 1
  Add $7$ to both sides: $4x = 20$ .
2. Step 2
  Divide by $4$ : $x = 5$ .
Answer
$x = 5$
2Solve with brackets
Foundation• linear, brackets
▼
Question
Solve $3(2x - 5) = 21$ .
Step-by-step solution
1. Step 1
  Expand: $6x - 15 = 21$ .
2. Step 2
  Add $15$ : $6x = 36$ .
3. Step 3
  Divide by $6$ : $x = 6$ .
Answer
$x = 6$
Examiner tip
Always expand brackets first, then solve. Some students try to divide by $3$ first; that works, but expanding is safer.
3Unknown on both sides
Foundation• Adapted from 4MA1/1H Jan 2024 Q3• linear, both sides
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Question
Solve $5x + 4 = 2x + 19$ .
Step-by-step solution
1. Step 1
  Subtract $2x$ from both sides: $3x + 4 = 19$ .
2. Step 2
  Subtract $4$ : $3x = 15$ .
3. Step 3
  Divide by $3$ : $x = 5$ .
Answer
$x = 5$
Examiner tip
Move smaller-coefficient $x$ to the side with bigger one to keep the coefficient positive. Helps avoid sign errors.
4Equation with fractions
Higher• Adapted from 4MA1/2H May/Jun 2024 Q7• linear, fractions
▼
Question
Solve $\dfrac{x + 2}{3} = \dfrac{2x - 1}{4}$ .
Step-by-step solution
1. Step 1
  Cross-multiply (multiply by $12$ , the LCM of $3$ and $4$ ):
  $4(x + 2) = 3(2x - 1)$
2. Step 2
  Expand both sides: $4x + 8 = 6x - 3$ .
3. Step 3
  Rearrange: $11 = 2x$ .
4. Step 4
  Divide: $x = 5.5$ (or $\dfrac{11}{2}$ ).
Answer
$x = 5.5$
Examiner tip
For two equal fractions, cross-multiply: $\dfrac{a}{b} = \dfrac{c}{d} \Leftrightarrow ad = bc$ . Mark schemes credit this method explicitly.
5Set up and solve from a word problem
Higher• word problem
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Question
I think of a number. I double it, then add $7$ . The result is $25$ . What was the number?
Step-by-step solution
1. Step 1
  Let the number be $n$ .
2. Step 2
  'Double, add 7' translates to: $2n + 7 = 25$ .
3. Step 3
  Subtract $7$ : $2n = 18$ . Divide: $n = 9$ .
Answer
$n = 9$
Examiner tip
Always START by defining the unknown. Mark schemes give M1 for setting up the correct equation, A1 for solving.
6An answer that is a fraction [1 mark]
Getting started• Paper 1H/2H style — 1 mark• linear equations, fractions
▼
Question
Solve $8x = 6$

Give your answer as a fraction in its simplest form. [1]
Step-by-step solution
1. Step 1
  B1 Divide both sides by $8$ , then cancel the common factor of $2$ .
  $x = \frac{6}{8} = \frac{3}{4}$
Answer
$x = \dfrac{3}{4}$
Examiner tip
Examiner tip: The number in front of $x$ goes on the bottom: $8x = 6$ means $x = 6 \div 8$ .

Common mistake: Writing $x = \dfrac{8}{6} = \dfrac{4}{3}$ , dividing the wrong way round. Check: $8 \times \dfrac{3}{4} = 6$ . Leaving $\dfrac{6}{8}$ also loses the mark, because the question asks for the simplest form.
7Undoing a division and an addition [2 marks]
Getting started• Paper 1H/2H style — 2 marks• linear equations, negative answer
▼
Question
Solve $\dfrac{x}{3} + 5 = 2$ [2]
Step-by-step solution
1. Step 1
  M1 Subtract $5$ from both sides first, because the $5$ is not divided by $3$ .
  $\frac{x}{3} = -3$
2. Step 2
  A1 Multiply both sides by $3$ .
  $x = -9$
Answer
$x = -9$
Examiner tip
Examiner tip: Undo the operations in reverse order. The $x$ was divided by $3$ and then $5$ was added, so take away $5$ first, then multiply by $3$ .

Common mistake: Multiplying only the $\dfrac{x}{3}$ by $3$ and writing $x + 5 = 6$ , so $x = 1$ . If you multiply first, every term must be multiplied: $x + 15 = 6$ , giving $x = -9$ .
8Forming an equation from costs in mixed units [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• forming equations, money
▼
Question
A pen costs $x$ pence.

A notebook costs $45$ pence more than a pen.

$4$ pens and $3$ notebooks cost £ $4.43$ in total.

Work out the cost of one pen. Show clear algebraic working. [3]
Step-by-step solution
1. Step 1
  M1 Work in pence throughout (£ $4.43 = 443$ pence). A notebook costs $(x + 45)$ pence.
  $4x + 3(x + 45) = 443$
2. Step 2
  M1 Expand and collect terms.
  $7x + 135 = 443 \;\Rightarrow\; 7x = 308$
3. Step 3
  A1 Divide by $7$ .
  $x = 44$
Answer
$44$ pence
Examiner tip
Examiner tip: Change everything into one unit before writing the equation. Check: $4 \times 44 + 3 \times 89 = 176 + 267 = 443$ .

Common mistake: Writing $4x + 3(x + 45) = 4.43$ , mixing pence and pounds, which gives a negative price. That should tell you something has gone wrong.
9Clearing fractions when there is a whole-number term [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• linear equations, fractions, brackets
▼
Question
Solve $\dfrac{2x - 1}{3} + 2 = \dfrac{x + 4}{2}$

Show clear algebraic working. [4]
Step-by-step solution
1. Step 1
  M1 Multiply every term by $6$ , the lowest common multiple of $3$ and $2$ . The $2$ on its own is a term too.
  $2(2x - 1) + 12 = 3(x + 4)$
2. Step 2
  M1 Expand the brackets.
  $4x - 2 + 12 = 3x + 12$
3. Step 3
  M1 Collect like terms.
  $4x + 10 = 3x + 12 \;\Rightarrow\; 4x - 3x = 12 - 10$
4. Step 4
  A1 Simplify.
  $x = 2$
Answer
$x = 2$
Examiner tip
Examiner tip: Before expanding, count the terms: there are three, and all three are multiplied by $6$ . Check: $\dfrac{3}{3} + 2 = 3$ and $\dfrac{6}{2} = 3$ .

Common mistake: Leaving the $+2$ as $+2$ instead of $+12$ . That gives $4x = 3x + 12$ and $x = 12$ , which fails the check.
10A trapezium and a rectangle with equal area [5 marks]
Stretch• Paper 1H/2H style — 5 marks• forming equations, area
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Question
A trapezium has parallel sides of length $(x + 3)$ cm and $(2x - 1)$ cm. The perpendicular distance between the parallel sides is $8$ cm.

A rectangle has length $(2x + 7)$ cm and width $4$ cm.

The area of the trapezium is equal to the area of the rectangle.

Work out the area of the rectangle. Show clear algebraic working. [5]
Step-by-step solution
1. Step 1
  M1 Use the area of a trapezium, $\tfrac{1}{2}(a + b)h$ , with the two parallel sides added first.
  $\tfrac{1}{2}(x + 3 + 2x - 1) \times 8 = 4(3x + 2) = 12x + 8$
2. Step 2
  M1 Set it equal to the area of the rectangle.
  $12x + 8 = 4(2x + 7)$
3. Step 3
  M1 Expand and collect terms.
  $12x + 8 = 8x + 28 \;\Rightarrow\; 4x = 20$
4. Step 4
  A1 Solve.
  $x = 5$
5. Step 5
  A1 The rectangle is $2(5) + 7 = 17$ cm by $4$ cm.
  $17 \times 4 = 68$
Answer
$68$ cm²
Examiner tip
Examiner tip: Check with the trapezium: its parallel sides are $8$ cm and $9$ cm, so its area is $\tfrac{1}{2} \times 17 \times 8 = 68$ cm².

Common mistake: Writing $\tfrac{1}{2} \times x + 3 + 2x - 1 \times 8$ with no brackets, so only part of the sum is halved and multiplied. Stopping at $x = 5$ also costs the final mark: the question asks for the area.

Question 1
Paper 1H/2H style — 1 mark1 mark
Solve $-3x = 27$ [1]
Model answer
$x = 27 \div (-3)$

Answer: $x = -9$
Why this scores
Mark scheme:
- B1 for $-9$ cao
Why this scores full marks: The candidate divides by the whole coefficient, $-3$ , sign included. The usual wrong answer is $9$ .
Question 2
Paper 1H/2H style — 2 marks2 marks
Solve $4(x + 3) = 10$ [2]
Model answer
$4x + 12 = 10$

$4x = -2$

Answer: $x = -0.5$
Why this scores
Mark scheme:
- M1 for $4x + 12 = 10$ or $x + 3 = 2.5$ oe
- A1 for $-0.5$ oe
Why this scores full marks: Both terms in the bracket are multiplied by $4$ , so the constant is $12$ , not $3$ . The answer is negative and not a whole number, and the candidate trusts it rather than forcing a tidier value. Check: $4(-0.5 + 3) = 4 \times 2.5 = 10$ .
Question 3
Paper 1H/2H style — 2 marks2 marks
Solve $3x + 2(x - 4) = 17$ [2]
Model answer
$3x + 2x - 8 = 17$

$5x = 25$

Answer: $x = 5$
Why this scores
Mark scheme:
- M1 for $3x + 2x - 8 = 17$ or $5x = 25$ oe
- A1 for $5$ cao
Why this scores full marks: The $2$ multiplies only the bracket; the $3x$ is left alone. A script that adds $3x + 2$ first and writes $5(x - 4)$ gets $x = 7.4$ and scores 0.
Question 4
Paper 1H/2H style — 3 marks3 marks
Solve $7(2 - x) = 3(x + 8)$ [3]
Model answer
$14 - 7x = 3x + 24$

$14 - 24 = 3x + 7x$

$-10 = 10x$

Answer: $x = -1$
Why this scores
Mark scheme:
- M1 for expanding at least one bracket correctly
- M1 for $10x = -10$ or $-10x = 10$ oe
- A1 for $-1$ cao
Why this scores full marks: Moving the $-7x$ to the right-hand side keeps the $x$ coefficient positive. Check: $7(3) = 21$ and $3(7) = 21$ .
Question 5
Paper 1H/2H style — 3 marks3 marks
Kofi has £ $x$ .

Ama has £ $15$ more than Kofi.

Ama gives Kofi £ $20$ . Kofi now has twice as much money as Ama.

Work out how much money Kofi had at the start. Show clear algebraic working. [3]
Model answer
After the gift: Kofi has $x + 20$ , Ama has $x + 15 - 20 = x - 5$

$x + 20 = 2(x - 5)$

$x + 20 = 2x - 10$

$x = 30$

Answer: £ $30$
Why this scores
Mark scheme:
- M1 for $x + 20$ and $x - 5$ (or $x + 15 - 20$ ) oe
- M1 for $x + 20 = 2(x - 5)$ oe
- A1 for £ $30$ cao
Why this scores full marks: Both amounts change: Kofi gains £ $20$ and Ama loses it. The "twice" goes on the smaller amount, inside a bracket. Check: Kofi £ $50$ , Ama £ $25$ .
Question 6
Paper 1H/2H style — 3 marks3 marks
Solve $0.2(3x - 5) = x - 3.4$

Show clear algebraic working. [3]
Model answer
$0.6x - 1 = x - 3.4$

$3.4 - 1 = x - 0.6x$

$2.4 = 0.4x$

Answer: $x = 6$
Why this scores
Mark scheme:
- M1 for $0.6x - 1$ oe, or for multiplying through by $5$ to get $3x - 5 = 5x - 17$
- M1 for $0.4x = 2.4$ or $2x = 12$ oe
- A1 for $6$ cao
Why this scores full marks: The decimal multiplies both terms in the bracket, and the $x$ terms go to the side where the coefficient stays positive. Multiplying every term by $5$ first is an equally good route that removes the decimals. Check: $0.2 \times 13 = 2.6$ and $6 - 3.4 = 2.6$ .
Question 7
Paper 1H/2H style — 4 marks4 marks
Triangle $ABC$ is isosceles with $AB = AC$ .

Angle $ABC = (3x + 10)^{\circ}$ and angle $ACB = (5x - 14)^{\circ}$ .

Work out the size of angle $BAC$ . Show clear algebraic working. [4]
Model answer
$AB = AC$ , so the base angles are equal:

$3x + 10 = 5x - 14$

$24 = 2x \Rightarrow x = 12$

Angle $ABC = 3(12) + 10 = 46^{\circ}$ , angle $ACB = 5(12) - 14 = 46^{\circ}$

Angle $BAC = 180 - 46 - 46 = 88$

Answer: $88^{\circ}$
Why this scores
Mark scheme:
- M1 for $3x + 10 = 5x - 14$ oe
- A1 for $x = 12$
- M1 for $180 - 2 \times 46$ oe (ft their $x$ )
- A1 for $88$ cao
Why this scores full marks: The candidate uses the fact that $AB = AC$ makes angles $ABC$ and $ACB$ equal, and says so. Working out both base angles checks the value of $x$ . Adding the two expressions to $180$ without using the isosceles fact cannot work: there are then two unknown quantities, $x$ and the third angle.
Question 8
Paper 1H/2H style — 4 marks4 marks
Solve $5 - \dfrac{2x + 3}{4} = \dfrac{x - 1}{3}$

Show clear algebraic working. [4]
Model answer
Multiply every term by $12$ :

$60 - 3(2x + 3) = 4(x - 1)$

$60 - 6x - 9 = 4x - 4$

$55 = 10x$

Answer: $x = 5.5$
Why this scores
Mark scheme:
- M1 for multiplying all terms by $12$ (or a common multiple), e.g. $60 - 3(2x + 3) = 4(x - 1)$
- M1 for correct expansion, $60 - 6x - 9 = 4x - 4$
- M1 for $10x = 55$ oe
- A1 for $5.5$ oe
Why this scores full marks: The whole number $5$ is multiplied by $12$ as well, and the bracket after the minus sign gives $-9$ , not $+9$ . Examiners regularly report both errors. Check: $5 - \dfrac{14}{4} = 1.5$ and $\dfrac{4.5}{3} = 1.5$ .
Question 9
Paper 1H/2H style — 5 marks5 marks
Mei cycles for $t$ hours at an average speed of $18$ km/h.

She then cycles for $(t + 0.5)$ hours at an average speed of $12$ km/h.

She cycles a total distance of $51$ km.

Work out the total time Mei spends cycling. Give your answer in hours and minutes. Show clear algebraic working. [5]
Model answer
Distances: $18t$ km and $12(t + 0.5)$ km

$18t + 12(t + 0.5) = 51$

$18t + 12t + 6 = 51$

$30t = 45 \Rightarrow t = 1.5$

Total time $= 1.5 + 2 = 3.5$ hours

Answer: $3$ hours $30$ minutes
Why this scores
Mark scheme:
- M1 for $18t$ or $12(t + 0.5)$ as a distance
- M1 for $18t + 12(t + 0.5) = 51$ oe
- M1 for $30t = 45$ oe
- A1 for $t = 1.5$
- A1 for $3$ hours $30$ minutes cao
Why this scores full marks: Distance is speed times time, so each part of the journey becomes an expression. The candidate then adds both times, not just $t$ , and converts $3.5$ hours correctly. Writing " $3$ hours $50$ minutes" is a common final slip. Check: $27 + 24 = 51$ km.
Question 10
Paper 1H/2H style — 5 marks5 marks
Tom spends $\dfrac{1}{3}$ of his money on a book.

He then spends $\dfrac{2}{5}$ of the money he has left on lunch.

He now has £ $18$ left.

Work out how much money Tom had at the start. Show clear algebraic working. [5]
Model answer
Let Tom start with £ $x$ .

Left after the book: $x - \dfrac{1}{3}x = \dfrac{2}{3}x$

Lunch: $\dfrac{2}{5} \times \dfrac{2}{3}x = \dfrac{4}{15}x$

$\dfrac{2}{3}x - \dfrac{4}{15}x = 18$

$\dfrac{10}{15}x - \dfrac{4}{15}x = \dfrac{6}{15}x = 18$

$6x = 270$

Answer: £ $45$
Why this scores
Mark scheme:
- M1 for $\dfrac{2}{3}x$ oe as the amount left after the book
- M1 for $\dfrac{2}{5} \times \dfrac{2}{3}x$ oe as the cost of lunch
- M1 for $\dfrac{2}{3}x - \dfrac{4}{15}x = 18$ oe
- M1 for $6x = 270$ or $\dfrac{2}{5}x = 18$ oe
- A1 for £ $45$ cao
Why this scores full marks: The $\dfrac{2}{5}$ is applied to the money left, not to the original amount. Taking $\dfrac{2}{5}$ of $x$ gives $x - \dfrac{1}{3}x - \dfrac{2}{5}x = 18$ and £ $67.50$ , the most common wrong answer. Because the question asks for clear algebraic working, the marks are for forming and solving the equation: £ $45$ found only by working backwards from £ $18$ , with no equation, does not earn them. Check: book £ $15$ , lunch £ $12$ , £ $18$ left.

4MA1 Linear Equations

At a glance

What you’ll learn

The standard solving method

Equations with fractions

Word problems — setting up linear equations

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Solve a simple linear equation

2Solve with brackets

3Unknown on both sides

4Equation with fractions

5Set up and solve from a word problem

6An answer that is a fraction [1 mark]

7Undoing a division and an addition [2 marks]

8Forming an equation from costs in mixed units [3 marks]

9Clearing fractions when there is a whole-number term [4 marks]

10A trapezium and a rectangle with equal area [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10