What are simultaneous linear equations in the context of Edexcel IGCSE Mathematics?

Simultaneous linear equations are a set of two or more linear equations with multiple variables that are solved together. In the Edexcel IGCSE Mathematics curriculum, students learn to find the values of variables that satisfy all equations in the set simultaneously.

How do you solve simultaneous linear equations using the substitution method?

To solve simultaneous linear equations using the substitution method, first solve one of the equations for one variable in terms of the other. Then, substitute this expression into the other equation. This will result in a single equation with one variable, which can be solved. Finally, substitute back to find the other variable.

What is the elimination method for solving simultaneous linear equations?

The elimination method involves adding or subtracting the equations to eliminate one of the variables, making it possible to solve for the remaining variable. Once one variable is found, substitute it back into one of the original equations to find the other variable. This method is commonly taught in the Edexcel IGCSE Mathematics curriculum.

Can simultaneous linear equations have no solution or infinitely many solutions?

Yes, simultaneous linear equations can have no solution if the lines represented by the equations are parallel. They can have infinitely many solutions if the lines coincide, meaning they are the same line. In such cases, the equations are dependent and consistent.

What are some real-world applications of simultaneous linear equations?

Simultaneous linear equations are used in various real-world applications, such as calculating the point of intersection between two lines, optimizing resource allocation, and solving problems involving mixtures or financial planning. Understanding these applications is part of the Edexcel IGCSE Mathematics curriculum.

Why is "Subtracting when should add (or vice versa)" a common mistake in Simultaneous Linear Equations ?

Why it happens: Coefficient signs. How to avoid it: If coefficients are EQUAL → SUBTRACT. If OPPOSITE signs → ADD. Always identify before combining.

Why is "Multiplying just one side of an equation" a common mistake in Simultaneous Linear Equations ?

Why it happens: Distractedly thinking about elimination. How to avoid it: When multiplying through an equation by a constant, multiply EVERY term — including the constant on the right. Source: 4MA1/1H Jan 2024 — examiner report Q14

Why is "Only stating one pair of solutions for linear-quadratic" a common mistake in Simultaneous Linear Equations ?

Why it happens: Familiar with linear (one pair). How to avoid it: Linear-quadratic systems usually have TWO solution PAIRS. Find both x's, find both y's.

Why is "Stating only one variable and not finding the other" a common mistake in Simultaneous Linear Equations ?

Why it happens: Forgetting to substitute back. How to avoid it: After finding x, ALWAYS substitute back to find y. State both.

Equations, Formulae and IdentitiesEdexcel IGCSE Mathematics (4MA1)

Simultaneous Linear Equations

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next.

Previous Next

Learn this Topic

Start with the slides for the quick version, then go deeper with the full study notes.

Short Study Notes in the form of Slides

Read the notes first. If the method in a worked example clicks, you're ready for the questions.

Page 1 / 0

Detailed Study Notes

Full prose, callouts and a recap — built for A* mastery, not just a quick scan.

Take these study notes with you

Download a branded PDF — full prose, callouts, recap and memorise list for Simultaneous Linear Equations , ready to print or save offline.

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

Solve two equations in two unknowns. Three methods (elimination, substitution, graphical), plus the linear-quadratic case.

What you’ll learn

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

2.7 — Solve two simultaneous linear equations.
2.7 — Solve simultaneous equations algebraically and graphically.
2.7 — Solve linear-quadratic simultaneous equations.

Elimination method

Align coefficients of one variable. Add or subtract to eliminate.

The procedure:

Make the coefficients of one variable EQUAL in magnitude.
ADD if signs are opposite, SUBTRACT if equal.
Solve the resulting single-variable equation.
Substitute back to find the other variable.

Example. $3x + 2y = 12$ , $5x - 2y = 4$ .

Coefficients of $y$ : $+2$ and $-2$ → opposite. Add:

$8x = 16$ → $x = 2$ .
Substitute: $3(2) + 2y = 12 \to y = 3$ .

With multiplication. $2x + 3y = 13$ , $4x + y = 11$ .

Multiply 2nd by $3$ to align $y$ : $12x + 3y = 33$ . Subtract 1st from this: $10x = 20$ → $x = 2$ . Substitute back: $4(2) + y = 11 \to y = 3$ .

Sign rules:

SAME sign coefficients → SUBTRACT.
OPPOSITE sign coefficients → ADD.

Edexcel tip. Always state your strategy: 'multiplying equation 2 by 3 to align $y$ -coefficients'. Mark schemes credit this clarity.

Align ONE variable's coefficient.
Same signs: subtract. Opposite: add.
Solve, then substitute back.
Multiply equations as needed.

Substitution method

Solve one equation for one variable; substitute into the other.

Useful when: one variable is already isolated (e.g. $y = ...$ ).

Method:

Solve one equation for one variable.
Substitute into the other equation.
Solve.
Substitute back to find the other variable.

Example. $y = 2x + 1$ , $3x + y = 11$ .

Sub $y = 2x + 1$ into 2nd: $3x + (2x + 1) = 11 \to 5x + 1 = 11 \to x = 2$ .

Sub back: $y = 2(2) + 1 = 5$ .

When to use substitution vs elimination.

Situation	Method
One variable already isolated	Substitution
Coefficients align cleanly	Elimination
Linear-quadratic	Substitution (must)

Worked qualitative. Why does substitution work for linear-quadratic?

The linear gives $y$ in terms of $x$ (or vice versa).
Substituting into the quadratic gives a single-variable quadratic.
Solve for $x$ , get the corresponding $y$ values.

Edexcel tip. Show the substitution explicitly. 'Substituting $y = 2x + 1$ into ...' earns the M1.

Isolate one variable.
Substitute into the other.
Solve the resulting equation.
Substitute back.

Linear and quadratic together

Substitute linear into quadratic. Get a quadratic in one variable. Solve.

Form: one linear equation + one quadratic equation.

Linear: $y = mx + c$ (or rearrange to this form). Quadratic: $x^{2} + y^{2} = r^{2}$ (circle), $y = x^{2} + ...$ , etc.

Method:

From the linear equation, express $y$ in terms of $x$ .
Substitute into the quadratic.
Solve the resulting quadratic in $x$ .
For each $x$ , find the corresponding $y$ .

Example. $y = x + 2$ , $x^{2} + y^{2} = 10$ .

Sub: $x^{2} + (x + 2)^{2} = 10$ . Expand: $2x^{2} + 4x + 4 = 10 \to 2x^{2} + 4x - 6 = 0$ . Divide by 2: $x^{2} + 2x - 3 = 0$ . Factor: $(x + 3)(x - 1) = 0$ . $x = -3$ or $x = 1$ .

For each, find $y$ :

$x = -3 \to y = -3 + 2 = -1$ . Pair: $(-3, -1)$ .
$x = 1 \to y = 1 + 2 = 3$ . Pair: $(1, 3)$ .

TWO solution pairs. Always state both as pairs.

Geometric interpretation. A line through a circle: 0, 1, or 2 intersection points. Two solutions = two intersections.

Worked qualitative. Why two solution pairs?

A linear equation has one degree of freedom.
A quadratic in two variables has many.
Their intersection is finite: at most degree-of-quadratic many points.
For a quadratic and a line: typically 2 intersections.

Edexcel tip. State BOTH pairs as $(x, y)$ . Mark schemes deduct for stating only one.

Sub linear into quadratic.
Solve the quadratic in one variable.
Find corresponding values.
Two solution pairs.
State as $(x, y)$ pairs.

Word problems with simultaneous equations

Define variables, write equations, solve.

Process:

Identify the unknowns. Define them with letters.
Set up TWO equations from the problem.
Solve simultaneously.
Check both unknowns make sense.

Example. Three pens + two books cost £ $13$ . Five pens + four books cost £ $23$ .

Let pen = £ $p$ , book = £ $b$ .
$3p + 2b = 13$ .
$5p + 4b = 23$ .
Multiply first by 2: $6p + 4b = 26$ .
Subtract second from this: $p = 3$ .
Substitute: $b = 2$ .

So pen = £ $3$ , book = £ $2$ .

Common contexts:

Cost / price problems (different items).
Mixture / mixing problems.
Speed / distance / time (current vs still water).
Ages.

Worked qualitative. Two trains leave at the same time. Train A travels at $60$ km/h, train B at $80$ km/h. After how long are they $560$ km apart (going in opposite directions)?

Let time = $t$ hours.
Distance A: $60t$ . Distance B: $80t$ .
Total apart: $60t + 80t = 140t = 560$ .
$t = 4$ hours.

(Single equation, but linear simultaneous in distance + time pattern.)

Edexcel tip. Always define variables explicitly. 'Let $p$ = price of a pen' earns the first M1.

Define variables.
Two equations from problem.
Solve simultaneously.
Check the answer makes sense.

How it’s examined

Simultaneous equations on every Higher Tier paper (4-7 marks). Linear-linear most common; linear-quadratic for A* candidates. Examiner reports flag (1) wrong sign in elimination, (2) only stating one pair for linear-quadratic, (3) skipping substitution-back step.

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.

Master this Topic

Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

Take this whole topic with you

Download a branded revision sheet — worked examples, formulae, definitions and common mistakes for Simultaneous Linear Equations , ready to print or save as PDF.

Step-by-step worked examples — Simultaneous Linear Equations

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

1Elimination method
Higher• simultaneous, elimination
Question
Solve $3x + 2y = 12$ and $5x - 2y = 4$ by elimination.
Step-by-step solution
1. Step 1
  Coefficients of $y$ are equal and opposite ( $+2$ and $-2$ ). Add the equations:
  $(3x + 2y) + (5x - 2y) = 12 + 4 \implies 8x = 16$
2. Step 2
  $x = 2$ . Substitute into first equation: $3(2) + 2y = 12 \implies 2y = 6 \implies y = 3$ .
Answer
$x = 2$ , $y = 3$
2Multiply to eliminate
Higher• Adapted from 4MA1/1H Jan 2024 Q14• simultaneous
Question
Solve $2x + 3y = 13$ and $4x + y = 11$ .
Step-by-step solution
1. Step 1
  Multiply second equation by $3$ to match $y$ -coefficients:
  $12x + 3y = 33$
2. Step 2
  Subtract original first from this:
  $(12x + 3y) - (2x + 3y) = 33 - 13 \implies 10x = 20$
3. Step 3
  $x = 2$ . Substitute back: $4(2) + y = 11 \implies y = 3$ .
Answer
$x = 2$ , $y = 3$
Examiner tip
When coefficients don't match, multiply ONE or BOTH equations to align. Mark schemes credit the multiplication step.
3Substitution method
Higher• substitution
Question
Solve $y = 2x + 1$ and $3x + y = 11$ .
Step-by-step solution
1. Step 1
  Substitute $y = 2x + 1$ into the second equation:
  $3x + (2x + 1) = 11$
2. Step 2
  Simplify: $5x + 1 = 11 \implies 5x = 10 \implies x = 2$ .
3. Step 3
  Substitute back: $y = 2(2) + 1 = 5$ .
Answer
$x = 2$ , $y = 5$
4Linear with quadratic
Higher• Adapted from 4MA1/2H May/Jun 2024 Q19• linear-quadratic
Question
Solve $y = x + 2$ and $x^{2} + y^{2} = 10$ .
Step-by-step solution
1. Step 1
  Substitute $y = x + 2$ into the second:
  $x^{2} + (x + 2)^{2} = 10$
2. Step 2
  Expand: $x^{2} + x^{2} + 4x + 4 = 10$ → $2x^{2} + 4x - 6 = 0$ .
3. Step 3
  Divide by $2$ : $x^{2} + 2x - 3 = 0$ . Factor: $(x + 3)(x - 1) = 0$ .
4. Step 4
  $x = -3$ or $x = 1$ . Find $y$ : $x = -3 \to y = -1$ ; $x = 1 \to y = 3$ .
Answer
$(x, y) = (-3, -1)$ or $(1, 3)$
Examiner tip
TWO solution PAIRS for linear-quadratic systems. Always state both pairs explicitly.
5Setting up from a word problem
Higher• simultaneous, word problem
Question
Three pens and two books cost £ $13$ . Five pens and four books cost £ $23$ . Find the cost of one pen and one book.
Step-by-step solution
1. Step 1
  Let pen = £ $p$ , book = £ $b$ .
2. Step 2
  Equations: $3p + 2b = 13$ and $5p + 4b = 23$ .
3. Step 3
  Multiply first by $2$ : $6p + 4b = 26$ . Subtract from second: $5p + 4b - 6p - 4b = 23 - 26 \implies -p = -3 \implies p = 3$ .
4. Step 4
  Substitute: $3(3) + 2b = 13 \implies b = 2$ .
Answer
Pen = £ $3$ , book = £ $2$ .

Key Definitions and Keywords — Simultaneous Linear Equations

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

Simultaneous equations
Examiner keyword
Two or more equations sharing variables. Solving means finding values that satisfy ALL equations at once.

Common Mistakes and Misconceptions — Simultaneous Linear Equations

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

✕Subtracting when should add (or vice versa)
Why it happens
Coefficient signs.
How to avoid it
If coefficients are EQUAL → SUBTRACT. If OPPOSITE signs → ADD. Always identify before combining.
✕Multiplying just one side of an equation
4MA1/1H Jan 2024 — examiner report Q14
Why it happens
Distractedly thinking about elimination.
How to avoid it
When multiplying through an equation by a constant, multiply EVERY term — including the constant on the right.
✕Only stating one pair of solutions for linear-quadratic
Why it happens
Familiar with linear (one pair).
How to avoid it
Linear-quadratic systems usually have TWO solution PAIRS. Find both x's, find both y's.
✕Stating only one variable and not finding the other
Why it happens
Forgetting to substitute back.
How to avoid it
After finding $x$ , ALWAYS substitute back to find $y$ . State both.

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.