What is a linear equation in the context of Edexcel IGCSE Mathematics?

In Edexcel IGCSE Mathematics, a linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. Linear equations are typically in the form of ax + b = 0, where 'a' and 'b' are constants, and 'x' is the variable. These equations graph as straight lines on a coordinate plane.

How do you solve a linear equation with one variable?

To solve a linear equation with one variable, you need to isolate the variable on one side of the equation. This can be done by performing operations such as addition, subtraction, multiplication, or division on both sides of the equation. For example, to solve 2x + 3 = 7, you would subtract 3 from both sides to get 2x = 4, and then divide both sides by 2 to find x = 2.

What are the steps to graph a linear equation on a coordinate plane?

To graph a linear equation on a coordinate plane, follow these steps: 1) Rewrite the equation in the form y = mx + c, where 'm' is the slope and 'c' is the y-intercept. 2) Plot the y-intercept on the y-axis. 3) Use the slope to determine another point on the line by rising and running from the y-intercept. 4) Draw a straight line through the points. This line represents the solutions to the equation.

What is the significance of the slope and y-intercept in a linear equation?

In a linear equation of the form y = mx + c, the slope 'm' represents the rate of change or the steepness of the line. It indicates how much 'y' changes for a unit change in 'x'. The y-intercept 'c' is the point where the line crosses the y-axis, representing the value of 'y' when 'x' is zero. Understanding these components helps in interpreting and graphing linear relationships.

How can linear equations be used to solve real-world problems?

Linear equations are used to model and solve real-world problems involving relationships between two quantities. For example, they can be used to calculate distances, predict financial outcomes, or determine rates of change. By setting up an equation based on the problem's conditions, you can solve for unknown variables and make informed decisions based on the results.

Why is "Solving without expanding brackets" a common mistake in Linear Equations?

Why it happens: Looks shorter. How to avoid it: Expand brackets first unless dividing by the constant outside is exact and easy. 3(2x - 5) = 21 6x - 15 = 21.

Why is "Sign error when moving terms across the equals" a common mistake in Linear Equations?

Why it happens: Forgetting that subtracting from one side adds to the other. How to avoid it: 5x + 4 = 2x + 19. Subtracting 2x: 5x - 2x = 3x. Apply to BOTH sides.

Why is "Cross-multiplying when fractions are equal but not isolated" a common mistake in Linear Equations?

Why it happens: Confusion about applicability. How to avoid it: Cross-multiplication ONLY works when ONE fraction equals ANOTHER fraction. a/b = c → a = bc, not ad.

Equations, Formulae and IdentitiesEdexcel IGCSE Mathematics (4MA1)

Linear Equations

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Linear Equations Study Notes — Edexcel IGCSE 4MA1 Higher Tier (2026 onwards)

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

What you’ll learn

Mapped to the Cambridge 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

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: $x = 6$ .

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

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

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: $17, 18, 19$ .

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.

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.

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Step-by-step worked examples — Linear Equations

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
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
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
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.

Key Definitions and Keywords — Linear Equations

Definitions to memorise and the exact keywords mark schemes credit for linear equations answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4MA1 sitting.

Linear equation
Examiner keyword
An equation where the highest power of the variable is $1$ . Has at most ONE solution.
Example
$3x + 5 = 11$ is linear; $x^{2} = 9$ is not.
Solution
The value of the variable that makes the equation TRUE.
Example
Solution to $3x = 12$ is $x = 4$ .

Common Mistakes and Misconceptions — Linear Equations

The traps other students keep falling into on linear equations questions — taken from recent Pearson Edexcel IGCSE 4MA1 examiner reports and mark schemes — and how to avoid them.

✕Solving without expanding brackets
Why it happens
Looks shorter.
How to avoid it
Expand brackets first unless dividing by the constant outside is exact and easy. $3(2x - 5) = 21 \to 6x - 15 = 21$ .
✕Sign error when moving terms across the equals
Why it happens
Forgetting that subtracting from one side adds to the other.
How to avoid it
$5x + 4 = 2x + 19$ . Subtracting $2x$ : $5x - 2x = 3x$ . Apply to BOTH sides.
✕Cross-multiplying when fractions are equal but not isolated
Why it happens
Confusion about applicability.
How to avoid it
Cross-multiplication ONLY works when ONE fraction equals ANOTHER fraction. $\dfrac{a}{b} = c$ → $a = bc$ , not $ad$ .

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Linear Equations — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Linear Equations

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Solve a simple linear equation

2Solve with brackets

3Unknown on both sides

4Equation with fractions

5Set up and solve from a word problem

Linear equation

Solution

✕Solving without expanding brackets

✕Sign error when moving terms across the equals

✕Cross-multiplying when fractions are equal but not isolated

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Linear Equations — frequently asked questions