What are simultaneous equations in the context of Cambridge IGCSE Mathematics?

Simultaneous equations are a set of two or more equations with multiple variables that are solved together, as they share common solutions. In the Cambridge IGCSE Mathematics curriculum, students typically encounter simultaneous linear equations, which involve finding the values of variables that satisfy all equations in the set simultaneously.

How do you solve simultaneous equations using the substitution method?

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

What is the elimination method for solving simultaneous equations?

The elimination method involves adding or subtracting the equations to eliminate one of the variables, allowing you to solve for the other variable. This is done by aligning the equations and manipulating them so that adding or subtracting them cancels out one of the variables. Once one variable is found, substitute it back into one of the original equations to find the other variable.

Can simultaneous equations have no solution or infinite solutions?

Yes, simultaneous equations can have no solution or infinite solutions. If the equations represent parallel lines, they will never intersect, resulting in no solution. If the equations are equivalent, representing the same line, they have infinite solutions, as every point on the line satisfies both equations.

What are some real-life applications of simultaneous equations?

Simultaneous equations are used in various real-life applications, such as in economics to find equilibrium points, in engineering for circuit analysis, and in business for optimizing resource allocation. They help in modeling situations where multiple conditions must be satisfied simultaneously, making them a vital tool in problem-solving across different fields.

Why is "Reporting only one pair when one equation is quadratic" a common mistake in Simultaneous Equations?

Why it happens: Students solve one equation and stop, forgetting that the quadratic has two roots. How to avoid it: After finding both x-values, substitute each back to find its corresponding y. State BOTH pairs. Source: 0580/42 — recurring

Why is "Adding when you should subtract (or vice versa)" a common mistake in Simultaneous Equations?

Why it happens: If y-coefficients are both +2, you SUBTRACT to eliminate, not add. How to avoid it: Same sign → subtract. Opposite signs → add. Always check the variable disappears in the resulting equation.

Why is "Multiplying only one term when scaling an equation" a common mistake in Simultaneous Equations?

Why it happens: Students multiply a coefficient but forget to multiply the constant on the right. How to avoid it: Multiplying an equation by k means multiplying **every term** by k.

Why is "Substituting back into the equation you already used" a common mistake in Simultaneous Equations?

Why it happens: Doesn't catch arithmetic errors. How to avoid it: Substitute into the OTHER equation as a check; if it doesn't balance, you have an error to find.

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Short Notes - Simultaneous Equations

Detailed Study Notes

Detailed notes on Algebra for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Simultaneous Equations — Cambridge IGCSE 0580 Maths Extended (2026)

Find the values that satisfy two equations at once. Three reliable methods: elimination, substitution, and (graphically) intersection. Cambridge tests both linear-linear and linear-quadratic systems.

At a glance

A solution is a pair $(x, y)$ that satisfies BOTH equations.
Linear-linear: ELIMINATION (match coefficients) or SUBSTITUTION.
Linear-quadratic: SUBSTITUTION (rearrange linear, sub into quadratic).
Always state both values: $x = \ldots, y = \ldots$ .
Always check by substituting both values into BOTH equations.
A linear-quadratic system can have 0, 1, or 2 solution pairs.
Graphically: solutions are intersection points of the two graphs.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.11 — Solve simultaneous linear equations in two unknowns.
E2.12 — Solve simultaneous equations involving one linear and one quadratic equation.

Elimination method (linear-linear)

Match the coefficient of one variable, then add or subtract to eliminate it.

Method.

Multiply one or both equations to make the coefficients of one variable equal in magnitude.
Add (if signs are opposite) or subtract (if signs match) to eliminate that variable.
Solve the resulting one-variable equation.
Substitute back into either original equation to find the other variable.
State both values and check.

Worked. Solve $\begin{cases} 2x + 3y = 13 \\ 4x - y = 5 \end{cases}$ .

Aim: eliminate $y$ . Multiply equation 2 by $3$ to get $12x - 3y = 15$ . Now add to equation 1:

$(2x + 3y) + (12x - 3y) = 13 + 15$ .
$14x = 28 \Rightarrow x = 2$ .

Substitute back: $4(2) - y = 5 \Rightarrow y = 3$ .

Check in equation 1: $2(2) + 3(3) = 4 + 9 = 13$ ✓.

Solution: $x = 2$ , $y = 3$ .

Scaling equation ② by 3 makes the y-coefficients opposite (+3y and −3y), so adding removes y.

Sign rule.

Signs SAME on the variable to eliminate → SUBTRACT.
Signs OPPOSITE → ADD.

Match the coefficient of one variable in both equations.
Same sign → subtract. Opposite sign → add.
Solve the resulting single-variable equation.
Always state both values and check.

Substitution method

Rearrange one equation for one variable, then substitute that expression into the other.

When to choose. Substitution is best when one equation already has a coefficient of $\pm 1$ on a variable, OR when one of the equations is quadratic.

Method.

Rearrange one equation: $y = \ldots$ or $x = \ldots$ .
Substitute that expression into the other equation.
Solve.
Substitute back to find the other variable.

Worked (linear-linear). Solve $\begin{cases} y = 2x - 3 \\ 3x + y = 12 \end{cases}$ .

Equation 1 already gives $y$ . Substitute into equation 2:

$3x + (2x - 3) = 12$ .
$5x - 3 = 12 \Rightarrow x = 3$ .

Then $y = 2(3) - 3 = 3$ .

Solution: $x = 3$ , $y = 3$ . Check: $3(3) + 3 = 12$ ✓.

Because equation ① already gives y, drop it straight into equation ② and solve for x.

Rearrange one equation, substitute into the other.
Best when one variable already has coefficient $\pm 1$ .
Always good for linear-quadratic systems.

Linear with quadratic

Substitution is the only practical method. Expect 0, 1, or 2 solution pairs.

Method.

Rearrange the LINEAR equation for one variable.
Substitute into the QUADRATIC.
Solve the resulting quadratic.
Substitute each solution back into the linear equation to find the other variable.

Worked. Solve $\begin{cases} y = x + 1 \\ x^{2} + y^{2} = 25 \end{cases}$ .

Substitute $y = x + 1$ into the quadratic:

$x^{2} + (x + 1)^{2} = 25$ .
Expand: $x^{2} + x^{2} + 2x + 1 = 25$ .
Combine: $2x^{2} + 2x - 24 = 0$ .
Divide by $2$ : $x^{2} + x - 12 = 0$ .
Factorise: $(x + 4)(x - 3) = 0$ .
$x = -4$ or $x = 3$ .

For each $x$ , find $y$ :

$x = -4 \Rightarrow y = -3$ .
$x = 3 \Rightarrow y = 4$ .

Solutions: $(-4, -3)$ and $(3, 4)$ . Check both pairs in both equations.

The two solution pairs are exactly where the line crosses the circle.

Geometric meaning. A line and a circle (or other conic) can meet at 0, 1 (tangent), or 2 points.

The number of solution pairs equals the number of crossing points.

Rearrange linear, substitute into quadratic.
Solve the resulting quadratic.
Each $x$ gives a corresponding $y$ via the linear equation.
Always state pairs $(x, y)$ together.

Graphical interpretation

Solutions are the points where the two graphs cross.

Each equation can be drawn as a graph:

A linear equation → a straight line.
A quadratic in $x, y$ → a parabola, circle, ellipse, etc.

The solutions to the simultaneous system are the intersection points.

A solution exists wherever the two graphs meet.

Two parallel lines → no solution.
Two identical lines → infinite solutions.
One line crossing a parabola → 0, 1 (tangent), or 2 points.
Two non-parallel lines → 1 solution.

A graphical sketch is sometimes asked for in addition to (or instead of) algebraic solving.

Each equation is a curve.
Solutions = intersection points.
Parallel lines → no solution.
Identical lines → infinitely many solutions.

Quick recap

Linear-linear: elimination (match coefficients, add/subtract) OR substitution.
Linear-quadratic: substitution — rearrange linear, sub into quadratic, solve.
Always state pairs $(x, y)$ , never $x$ and $y$ as separate sets.
Always check both values in BOTH original equations.
Graphically: solutions = intersections.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Simultaneous equations — two or more equations satisfied at the same time by the same values.
Elimination method — match coefficients of one variable and add/subtract to remove it.
Substitution method — rearrange one equation, substitute into the other.

How it’s examined

Linear-linear systems appear on every Paper 2 (3-4 marks). Linear-quadratic show up on Paper 4 most years (5-6 marks). Examiner reports flag pairing slip-ups (matching the wrong $y$ to each $x$ ) and arithmetic errors when adding/subtracting equations.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.11-2.12); 0580/22 May/Jun 2024 — Q11 (linear-linear); 0580/42 Oct/Nov 2024 — Q13 (linear-quadratic); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Simultaneous Equations

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

Question type:

Question patterns to master — Simultaneous Equations

Almost every simultaneous equations exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

Solve simultaneously with two linear equations already written down — your job is purely the algebra.

How to approach it

Choose elimination (scale so one variable's coefficients match, then add or subtract) or substitution (isolate one variable and replace it). Solve for one unknown, then back-substitute for the other.

Common trap

Adding when you should subtract — same-sign coefficients require subtraction. Examiner reports also flag scaling only one term of an equation; multiply every term.

Multi-step problem

Recognise it by

Solve simultaneously where one equation is non-linear — a quadratic or a circle $x^2+y^2=r^2$ paired with a line.

How to approach it

Substitute the linear equation into the non-linear one, expand to a single quadratic, solve it, then back-substitute each $x$ to get its paired $y$ .

Common trap

Reporting only one solution pair. A one-linear-one-quadratic system has two pairs — examiner reports stress stating both $(x,y)$ coordinates.

Word problem

Recognise it by

A real-world context — tickets, pens, a rectangle's perimeter and area — with no equations stated.

How to approach it

Define a letter for each unknown, turn each piece of information into an equation, then solve the resulting system by elimination or substitution and interpret the answer.

Common trap

Mis-pairing quantities when forming the equations, or stating swapped length/width as two different rectangles. Examiner reports note candidates should give one dimension pair.

1Solve by elimination
CoreDirect calculation• elimination
Question
Solve simultaneously: $3x + 2y = 13$ and $5x - 2y = 11$ .
Step-by-step solution
1. Step 1
  Coefficients of $y$ are $+2$ and $-2$ — add the equations to eliminate $y$ .
  $(3x + 2y) + (5x - 2y) = 13 + 11$
2. Step 2
  Simplify.
  $8x = 24 \implies x = 3$
3. Step 3
  Substitute back into $3x + 2y = 13$ .
  $9 + 2y = 13 \implies y = 2$
Answer
$x = 3,\ y = 2$
Examiner tip
Always verify in the OTHER equation: $5(3) - 2(2) = 15 - 4 = 11$ . ✓
2Elimination requiring multiplication
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q9• elimination
Question
Solve simultaneously: $2x + 3y = 12$ and $5x - 2y = 11$ .
Step-by-step solution
1. Step 1
  Multiply the first by $2$ and the second by $3$ to make $y$ -coefficients equal in size.
  $4x + 6y = 24,\quad 15x - 6y = 33$
2. Step 2
  Add to eliminate $y$ .
  $19x = 57 \implies x = 3$
3. Step 3
  Substitute back: $2(3) + 3y = 12$ .
  $y = 2$
Answer
$x = 3,\ y = 2$
3Solve by substitution
CoreDirect calculation• substitution
Question
Solve simultaneously: $y = 2x + 1$ and $3x + y = 16$ .
Step-by-step solution
1. Step 1
  Substitute the first into the second.
  $3x + (2x + 1) = 16$
2. Step 2
  Solve.
  $5x = 15 \implies x = 3$
3. Step 3
  Substitute back.
  $y = 2(3) + 1 = 7$
Answer
$x = 3,\ y = 7$
4One linear and one quadratic
ExtendedMulti-step problem• Adapted from 0580/42 Oct/Nov 2023 Q12• quadratic, substitution
Question
Solve simultaneously: $y = x + 2$ and $x^{2} + y^{2} = 20$ .
Step-by-step solution
1. Step 1
  Substitute the linear into the quadratic.
  $x^{2} + (x + 2)^{2} = 20$
2. Step 2
  Expand.
  $x^{2} + x^{2} + 4x + 4 = 20$
3. Step 3
  Simplify and rearrange.
  $2x^{2} + 4x - 16 = 0 \implies x^{2} + 2x - 8 = 0$
4. Step 4
  Factorise.
  $(x + 4)(x - 2) = 0 \implies x = -4\ \text{or}\ x = 2$
5. Step 5
  Find $y$ for each.
  $x = -4: y = -2.\quad x = 2: y = 4.$
Answer
$(x, y) = (-4, -2)$ or $(2, 4)$
Examiner tip
When you solve a one-linear-one-quadratic system, you always have two pairs of solutions. State both.
5Word problem to simultaneous equations
ExtendedWord problem• word problem
Question
Two adult tickets and three child tickets cost $\$ 32 $. Three adult tickets and one child ticket cost$ $27$. Find the cost of an adult ticket and a child ticket.
Step-by-step solution
1. Step 1
  Let $a$ = adult cost, $c$ = child cost.
  $2a + 3c = 32,\quad 3a + c = 27$
2. Step 2
  Multiply the second by $3$ .
  $9a + 3c = 81$
3. Step 3
  Subtract first from this.
  $7a = 49 \implies a = 7$
4. Step 4
  Substitute back.
  $3(7) + c = 27 \implies c = 6$
Answer
Adult $\$ 7 $, child$ $6$
6Substitution requiring rearrangement first
CoreDirect calculation• Adapted from 0580/22 Feb/Mar 2024 Q8• substitution, rearranging
Question
Solve simultaneously: $x + 2y = 11$ and $4x - y = 8$ .
Step-by-step solution
1. Step 1
  Rearrange the second equation to make $y$ the subject.
  $y = 4x - 8$
2. Step 2
  Substitute into the first equation.
  $x + 2(4x - 8) = 11$
3. Step 3
  Expand and solve.
  $x + 8x - 16 = 11 \implies 9x = 27 \implies x = 3$
4. Step 4
  Back-substitute.
  $y = 4(3) - 8 = 4$
Answer
$x = 3,\ y = 4$
Examiner tip
The examiner report flags candidates often forget to put brackets around the substituted expression, writing $x + 2 \cdot 4x - 8 = 11$ — which gives the wrong answer. Always bracket the substitution.
7Simultaneous equations with fractional coefficients
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2023 Q14• elimination, fractions
Question
Solve simultaneously: $\dfrac{x}{2} + \dfrac{y}{3} = 5$ and $\dfrac{x}{4} - \dfrac{y}{6} = 1$ .
Step-by-step solution
1. Step 1
  Multiply the first equation by $6$ (LCM of $2$ and $3$ ).
  $3x + 2y = 30$
2. Step 2
  Multiply the second equation by $12$ (LCM of $4$ and $6$ ).
  $3x - 2y = 12$
3. Step 3
  Add the two cleared equations to eliminate $y$ .
  $6x = 42 \implies x = 7$
4. Step 4
  Substitute back into $3x + 2y = 30$ .
  $21 + 2y = 30 \implies y = \dfrac{9}{2}$
Answer
$x = 7,\ y = \dfrac{9}{2}$
Examiner tip
Examiners reward candidates who clear fractions in one step by multiplying by the LCM of all denominators. Adding the equations as fractions almost always introduces arithmetic errors.
8Line and circle intersection (one-linear-one-quadratic)
ExtendedMulti-step problem• Adapted from 0580/42 May/Jun 2024 Q21• quadratic, geometry, A*
Question
Solve simultaneously: $y = 3x - 1$ and $x^{2} + y^{2} = 25$ .
Step-by-step solution
1. Step 1
  Substitute $y = 3x - 1$ into the quadratic.
  $x^{2} + (3x - 1)^{2} = 25$
2. Step 2
  Expand the square.
  $x^{2} + 9x^{2} - 6x + 1 = 25$
3. Step 3
  Collect and rearrange.
  $10x^{2} - 6x - 24 = 0 \implies 5x^{2} - 3x - 12 = 0$
4. Step 4
  Apply the quadratic formula with $a = 5$ , $b = -3$ , $c = -12$ .
  $x = \dfrac{3 \pm \sqrt{9 + 240}}{10} = \dfrac{3 \pm \sqrt{249}}{10}$
5. Step 5
  Compute to 2 d.p.: $x \approx 1.88$ or $x \approx -1.28$ . Then $y = 3x - 1$ for each.
Answer
$(x, y) \approx (1.88, 4.64)$ or $(-1.28, -4.84)$ (to 2 d.p.)
Examiner tip
A* candidates secure full marks by stating BOTH coordinate pairs. The 2024 mark scheme also accepts answers given in exact surd form.
9Mixed-coin word problem with three constraints
ChallengeWord problem• Adapted from 0580/42 Oct/Nov 2024 Q12• word problem, elimination
Question
Six identical pens and four identical pencils cost $\$ 11.20 $. Five pens and three pencils cost$ $9.05$. Find the cost of one pen and one pencil.
Step-by-step solution
1. Step 1
  Let $p$ = cost of a pen, $q$ = cost of a pencil.
  $6p + 4q = 11.20,\quad 5p + 3q = 9.05$
2. Step 2
  Multiply the first by $3$ and the second by $4$ to match $q$ -coefficients.
  $18p + 12q = 33.60,\quad 20p + 12q = 36.20$
3. Step 3
  Subtract the first from the second.
  $2p = 2.60 \implies p = 1.30$
4. Step 4
  Substitute back into $6p + 4q = 11.20$ .
  $7.80 + 4q = 11.20 \implies 4q = 3.40 \implies q = 0.85$
Answer
Pen $\$ 1.30 $, pencil$ $0.85$
Examiner tip
Examiners reward candidates who scale BOTH equations to match coefficients. The 2024 mark scheme also accepts substitution as an alternative method.
10Word problem leading to a non-linear system
ChallengeWord problem• Adapted from 0580/42 Oct/Nov 2024 Q22• word problem, quadratic
Question
A rectangle has perimeter $26$ cm and area $40$ cm $^{2}$ . Find the lengths of its sides.
Step-by-step solution
1. Step 1
  Let the sides be $x$ and $y$ cm. The perimeter gives $2x + 2y = 26 \Rightarrow x + y = 13$ . The area gives $xy = 40$ .
2. Step 2
  Express $y = 13 - x$ and substitute.
  $x(13 - x) = 40$
3. Step 3
  Expand and rearrange.
  $13x - x^{2} = 40 \implies x^{2} - 13x + 40 = 0$
4. Step 4
  Factorise: $(x - 5)(x - 8) = 0$ , so $x = 5$ or $x = 8$ .
5. Step 5
  Both correspond to the same rectangle (length $8$ , width $5$ ).
Answer
Sides are $5$ cm and $8$ cm.
Examiner tip
The examiner report flags candidates often state two separate rectangles rather than recognising that swapping length and width is the same rectangle. State the dimensions as a single pair.

Key Formulae — Simultaneous Equations

The formulae you need to memorise for simultaneous equations on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Elimination strategy
$\text{Make matching coefficients} \to \text{add or subtract} \to \text{solve} \to \text{back-substitute}$
When to use
Standard for two linear equations in two variables.
Substitution strategy
$\text{Solve one equation for one variable} \to \text{substitute into the other} \to \text{solve}$
When to use
Best when one equation already has a variable isolated, or when one of the equations is non-linear (quadratic).

Key Definitions and Keywords — Simultaneous Equations

Definitions to memorise and the exact keywords mark schemes credit for simultaneous equations answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Simultaneous equations
Examiner keyword
Two or more equations whose solution must satisfy all of them at the same time.
Elimination method
Examiner keyword
Add or subtract scaled equations so that one variable cancels, leaving a single-variable equation.
Substitution method
Examiner keyword
Express one variable from one equation and substitute into the other.
Back-substitute
After finding one variable, substitute its value into one of the original equations to find the other.

Common Mistakes and Misconceptions — Simultaneous Equations

The traps other students keep falling into on simultaneous equations questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Reporting only one pair when one equation is quadratic
0580/42 — recurring
Why it happens
Students solve one equation and stop, forgetting that the quadratic has two roots.
How to avoid it
After finding both $x$ -values, substitute each back to find its corresponding $y$ . State BOTH pairs.
✕Adding when you should subtract (or vice versa)
Why it happens
If $y$ -coefficients are both $+2$ , you SUBTRACT to eliminate, not add.
How to avoid it
Same sign → subtract. Opposite signs → add. Always check the variable disappears in the resulting equation.
✕Multiplying only one term when scaling an equation
Why it happens
Students multiply a coefficient but forget to multiply the constant on the right.
How to avoid it
Multiplying an equation by $k$ means multiplying every term by $k$ .
✕Substituting back into the equation you already used
Why it happens
Doesn't catch arithmetic errors.
How to avoid it
Substitute into the OTHER equation as a check; if it doesn't balance, you have an error to find.

Simultaneous Equations — frequently asked questions

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

Cambridge IGCSE Mathematics (0580)

Simultaneous Equations

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Short Notes - Simultaneous Equations

Detailed Study Notes

Detailed notes on Algebra for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Simultaneous Equations — Cambridge IGCSE 0580 Maths Extended (2026)

Find the values that satisfy two equations at once. Three reliable methods: elimination, substitution, and (graphically) intersection. Cambridge tests both linear-linear and linear-quadratic systems.

At a glance

A solution is a pair $(x, y)$ that satisfies BOTH equations.
Linear-linear: ELIMINATION (match coefficients) or SUBSTITUTION.
Linear-quadratic: SUBSTITUTION (rearrange linear, sub into quadratic).
Always state both values: $x = \ldots, y = \ldots$ .
Always check by substituting both values into BOTH equations.
A linear-quadratic system can have 0, 1, or 2 solution pairs.
Graphically: solutions are intersection points of the two graphs.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.11 — Solve simultaneous linear equations in two unknowns.
E2.12 — Solve simultaneous equations involving one linear and one quadratic equation.

Elimination method (linear-linear)

Match the coefficient of one variable, then add or subtract to eliminate it.

Method.

Multiply one or both equations to make the coefficients of one variable equal in magnitude.
Add (if signs are opposite) or subtract (if signs match) to eliminate that variable.
Solve the resulting one-variable equation.
Substitute back into either original equation to find the other variable.
State both values and check.

Worked. Solve $\begin{cases} 2x + 3y = 13 \\ 4x - y = 5 \end{cases}$ .

Aim: eliminate $y$ . Multiply equation 2 by $3$ to get $12x - 3y = 15$ . Now add to equation 1:

$(2x + 3y) + (12x - 3y) = 13 + 15$ .
$14x = 28 \Rightarrow x = 2$ .

Substitute back: $4(2) - y = 5 \Rightarrow y = 3$ .

Check in equation 1: $2(2) + 3(3) = 4 + 9 = 13$ ✓.

Solution: $x = 2$ , $y = 3$ .

Scaling equation ② by 3 makes the y-coefficients opposite (+3y and −3y), so adding removes y.

Sign rule.

Signs SAME on the variable to eliminate → SUBTRACT.
Signs OPPOSITE → ADD.

Match the coefficient of one variable in both equations.
Same sign → subtract. Opposite sign → add.
Solve the resulting single-variable equation.
Always state both values and check.

Substitution method

Rearrange one equation for one variable, then substitute that expression into the other.

When to choose. Substitution is best when one equation already has a coefficient of $\pm 1$ on a variable, OR when one of the equations is quadratic.

Method.

Rearrange one equation: $y = \ldots$ or $x = \ldots$ .
Substitute that expression into the other equation.
Solve.
Substitute back to find the other variable.

Worked (linear-linear). Solve $\begin{cases} y = 2x - 3 \\ 3x + y = 12 \end{cases}$ .

Equation 1 already gives $y$ . Substitute into equation 2:

$3x + (2x - 3) = 12$ .
$5x - 3 = 12 \Rightarrow x = 3$ .

Then $y = 2(3) - 3 = 3$ .

Solution: $x = 3$ , $y = 3$ . Check: $3(3) + 3 = 12$ ✓.

Because equation ① already gives y, drop it straight into equation ② and solve for x.

Rearrange one equation, substitute into the other.
Best when one variable already has coefficient $\pm 1$ .
Always good for linear-quadratic systems.

Linear with quadratic

Substitution is the only practical method. Expect 0, 1, or 2 solution pairs.

Method.

Rearrange the LINEAR equation for one variable.
Substitute into the QUADRATIC.
Solve the resulting quadratic.
Substitute each solution back into the linear equation to find the other variable.

Worked. Solve $\begin{cases} y = x + 1 \\ x^{2} + y^{2} = 25 \end{cases}$ .

Substitute $y = x + 1$ into the quadratic:

$x^{2} + (x + 1)^{2} = 25$ .
Expand: $x^{2} + x^{2} + 2x + 1 = 25$ .
Combine: $2x^{2} + 2x - 24 = 0$ .
Divide by $2$ : $x^{2} + x - 12 = 0$ .
Factorise: $(x + 4)(x - 3) = 0$ .
$x = -4$ or $x = 3$ .

For each $x$ , find $y$ :

$x = -4 \Rightarrow y = -3$ .
$x = 3 \Rightarrow y = 4$ .

Solutions: $(-4, -3)$ and $(3, 4)$ . Check both pairs in both equations.

The two solution pairs are exactly where the line crosses the circle.

Geometric meaning. A line and a circle (or other conic) can meet at 0, 1 (tangent), or 2 points.

The number of solution pairs equals the number of crossing points.

Rearrange linear, substitute into quadratic.
Solve the resulting quadratic.
Each $x$ gives a corresponding $y$ via the linear equation.
Always state pairs $(x, y)$ together.

Graphical interpretation

Solutions are the points where the two graphs cross.

Each equation can be drawn as a graph:

A linear equation → a straight line.
A quadratic in $x, y$ → a parabola, circle, ellipse, etc.

The solutions to the simultaneous system are the intersection points.

A solution exists wherever the two graphs meet.

Two parallel lines → no solution.
Two identical lines → infinite solutions.
One line crossing a parabola → 0, 1 (tangent), or 2 points.
Two non-parallel lines → 1 solution.

A graphical sketch is sometimes asked for in addition to (or instead of) algebraic solving.

Each equation is a curve.
Solutions = intersection points.
Parallel lines → no solution.
Identical lines → infinitely many solutions.

Quick recap

Linear-linear: elimination (match coefficients, add/subtract) OR substitution.
Linear-quadratic: substitution — rearrange linear, sub into quadratic, solve.
Always state pairs $(x, y)$ , never $x$ and $y$ as separate sets.
Always check both values in BOTH original equations.
Graphically: solutions = intersections.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Simultaneous equations — two or more equations satisfied at the same time by the same values.
Elimination method — match coefficients of one variable and add/subtract to remove it.
Substitution method — rearrange one equation, substitute into the other.

How it’s examined

Take this whole topic with you

Step-by-step worked examples — Simultaneous Equations

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

Question type:

Question patterns to master — Simultaneous Equations

Almost every simultaneous equations exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

Solve simultaneously with two linear equations already written down — your job is purely the algebra.

How to approach it

Choose elimination (scale so one variable's coefficients match, then add or subtract) or substitution (isolate one variable and replace it). Solve for one unknown, then back-substitute for the other.

Common trap

Adding when you should subtract — same-sign coefficients require subtraction. Examiner reports also flag scaling only one term of an equation; multiply every term.

Multi-step problem

Recognise it by

Solve simultaneously where one equation is non-linear — a quadratic or a circle $x^2+y^2=r^2$ paired with a line.

How to approach it

Substitute the linear equation into the non-linear one, expand to a single quadratic, solve it, then back-substitute each $x$ to get its paired $y$ .

Common trap

Reporting only one solution pair. A one-linear-one-quadratic system has two pairs — examiner reports stress stating both $(x,y)$ coordinates.

Word problem

Recognise it by

A real-world context — tickets, pens, a rectangle's perimeter and area — with no equations stated.

How to approach it

Define a letter for each unknown, turn each piece of information into an equation, then solve the resulting system by elimination or substitution and interpret the answer.

Common trap

Mis-pairing quantities when forming the equations, or stating swapped length/width as two different rectangles. Examiner reports note candidates should give one dimension pair.

1Solve by elimination
CoreDirect calculation• elimination
Question
Solve simultaneously: $3x + 2y = 13$ and $5x - 2y = 11$ .
Step-by-step solution
1. Step 1
  Coefficients of $y$ are $+2$ and $-2$ — add the equations to eliminate $y$ .
  $(3x + 2y) + (5x - 2y) = 13 + 11$
2. Step 2
  Simplify.
  $8x = 24 \implies x = 3$
3. Step 3
  Substitute back into $3x + 2y = 13$ .
  $9 + 2y = 13 \implies y = 2$
Answer
$x = 3,\ y = 2$
Examiner tip
Always verify in the OTHER equation: $5(3) - 2(2) = 15 - 4 = 11$ . ✓
2Elimination requiring multiplication
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q9• elimination
Question
Solve simultaneously: $2x + 3y = 12$ and $5x - 2y = 11$ .
Step-by-step solution
1. Step 1
  Multiply the first by $2$ and the second by $3$ to make $y$ -coefficients equal in size.
  $4x + 6y = 24,\quad 15x - 6y = 33$
2. Step 2
  Add to eliminate $y$ .
  $19x = 57 \implies x = 3$
3. Step 3
  Substitute back: $2(3) + 3y = 12$ .
  $y = 2$
Answer
$x = 3,\ y = 2$
3Solve by substitution
CoreDirect calculation• substitution
Question
Solve simultaneously: $y = 2x + 1$ and $3x + y = 16$ .
Step-by-step solution
1. Step 1
  Substitute the first into the second.
  $3x + (2x + 1) = 16$
2. Step 2
  Solve.
  $5x = 15 \implies x = 3$
3. Step 3
  Substitute back.
  $y = 2(3) + 1 = 7$
Answer
$x = 3,\ y = 7$
4One linear and one quadratic
ExtendedMulti-step problem• Adapted from 0580/42 Oct/Nov 2023 Q12• quadratic, substitution
Question
Solve simultaneously: $y = x + 2$ and $x^{2} + y^{2} = 20$ .
Step-by-step solution
1. Step 1
  Substitute the linear into the quadratic.
  $x^{2} + (x + 2)^{2} = 20$
2. Step 2
  Expand.
  $x^{2} + x^{2} + 4x + 4 = 20$
3. Step 3
  Simplify and rearrange.
  $2x^{2} + 4x - 16 = 0 \implies x^{2} + 2x - 8 = 0$
4. Step 4
  Factorise.
  $(x + 4)(x - 2) = 0 \implies x = -4\ \text{or}\ x = 2$
5. Step 5
  Find $y$ for each.
  $x = -4: y = -2.\quad x = 2: y = 4.$
Answer
$(x, y) = (-4, -2)$ or $(2, 4)$
Examiner tip
When you solve a one-linear-one-quadratic system, you always have two pairs of solutions. State both.
5Word problem to simultaneous equations
ExtendedWord problem• word problem
Question
Two adult tickets and three child tickets cost $\$ 32 $. Three adult tickets and one child ticket cost$ $27$. Find the cost of an adult ticket and a child ticket.
Step-by-step solution
1. Step 1
  Let $a$ = adult cost, $c$ = child cost.
  $2a + 3c = 32,\quad 3a + c = 27$
2. Step 2
  Multiply the second by $3$ .
  $9a + 3c = 81$
3. Step 3
  Subtract first from this.
  $7a = 49 \implies a = 7$
4. Step 4
  Substitute back.
  $3(7) + c = 27 \implies c = 6$
Answer
Adult $\$ 7 $, child$ $6$
6Substitution requiring rearrangement first
CoreDirect calculation• Adapted from 0580/22 Feb/Mar 2024 Q8• substitution, rearranging
Question
Solve simultaneously: $x + 2y = 11$ and $4x - y = 8$ .
Step-by-step solution
1. Step 1
  Rearrange the second equation to make $y$ the subject.
  $y = 4x - 8$
2. Step 2
  Substitute into the first equation.
  $x + 2(4x - 8) = 11$
3. Step 3
  Expand and solve.
  $x + 8x - 16 = 11 \implies 9x = 27 \implies x = 3$
4. Step 4
  Back-substitute.
  $y = 4(3) - 8 = 4$
Answer
$x = 3,\ y = 4$
Examiner tip
The examiner report flags candidates often forget to put brackets around the substituted expression, writing $x + 2 \cdot 4x - 8 = 11$ — which gives the wrong answer. Always bracket the substitution.
7Simultaneous equations with fractional coefficients
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2023 Q14• elimination, fractions
Question
Solve simultaneously: $\dfrac{x}{2} + \dfrac{y}{3} = 5$ and $\dfrac{x}{4} - \dfrac{y}{6} = 1$ .
Step-by-step solution
1. Step 1
  Multiply the first equation by $6$ (LCM of $2$ and $3$ ).
  $3x + 2y = 30$
2. Step 2
  Multiply the second equation by $12$ (LCM of $4$ and $6$ ).
  $3x - 2y = 12$
3. Step 3
  Add the two cleared equations to eliminate $y$ .
  $6x = 42 \implies x = 7$
4. Step 4
  Substitute back into $3x + 2y = 30$ .
  $21 + 2y = 30 \implies y = \dfrac{9}{2}$
Answer
$x = 7,\ y = \dfrac{9}{2}$
Examiner tip
Examiners reward candidates who clear fractions in one step by multiplying by the LCM of all denominators. Adding the equations as fractions almost always introduces arithmetic errors.
8Line and circle intersection (one-linear-one-quadratic)
ExtendedMulti-step problem• Adapted from 0580/42 May/Jun 2024 Q21• quadratic, geometry, A*
Question
Solve simultaneously: $y = 3x - 1$ and $x^{2} + y^{2} = 25$ .
Step-by-step solution
1. Step 1
  Substitute $y = 3x - 1$ into the quadratic.
  $x^{2} + (3x - 1)^{2} = 25$
2. Step 2
  Expand the square.
  $x^{2} + 9x^{2} - 6x + 1 = 25$
3. Step 3
  Collect and rearrange.
  $10x^{2} - 6x - 24 = 0 \implies 5x^{2} - 3x - 12 = 0$
4. Step 4
  Apply the quadratic formula with $a = 5$ , $b = -3$ , $c = -12$ .
  $x = \dfrac{3 \pm \sqrt{9 + 240}}{10} = \dfrac{3 \pm \sqrt{249}}{10}$
5. Step 5
  Compute to 2 d.p.: $x \approx 1.88$ or $x \approx -1.28$ . Then $y = 3x - 1$ for each.
Answer
$(x, y) \approx (1.88, 4.64)$ or $(-1.28, -4.84)$ (to 2 d.p.)
Examiner tip
A* candidates secure full marks by stating BOTH coordinate pairs. The 2024 mark scheme also accepts answers given in exact surd form.
9Mixed-coin word problem with three constraints
ChallengeWord problem• Adapted from 0580/42 Oct/Nov 2024 Q12• word problem, elimination
Question
Six identical pens and four identical pencils cost $\$ 11.20 $. Five pens and three pencils cost$ $9.05$. Find the cost of one pen and one pencil.
Step-by-step solution
1. Step 1
  Let $p$ = cost of a pen, $q$ = cost of a pencil.
  $6p + 4q = 11.20,\quad 5p + 3q = 9.05$
2. Step 2
  Multiply the first by $3$ and the second by $4$ to match $q$ -coefficients.
  $18p + 12q = 33.60,\quad 20p + 12q = 36.20$
3. Step 3
  Subtract the first from the second.
  $2p = 2.60 \implies p = 1.30$
4. Step 4
  Substitute back into $6p + 4q = 11.20$ .
  $7.80 + 4q = 11.20 \implies 4q = 3.40 \implies q = 0.85$
Answer
Pen $\$ 1.30 $, pencil$ $0.85$
Examiner tip
Examiners reward candidates who scale BOTH equations to match coefficients. The 2024 mark scheme also accepts substitution as an alternative method.
10Word problem leading to a non-linear system
ChallengeWord problem• Adapted from 0580/42 Oct/Nov 2024 Q22• word problem, quadratic
Question
A rectangle has perimeter $26$ cm and area $40$ cm $^{2}$ . Find the lengths of its sides.
Step-by-step solution
1. Step 1
  Let the sides be $x$ and $y$ cm. The perimeter gives $2x + 2y = 26 \Rightarrow x + y = 13$ . The area gives $xy = 40$ .
2. Step 2
  Express $y = 13 - x$ and substitute.
  $x(13 - x) = 40$
3. Step 3
  Expand and rearrange.
  $13x - x^{2} = 40 \implies x^{2} - 13x + 40 = 0$
4. Step 4
  Factorise: $(x - 5)(x - 8) = 0$ , so $x = 5$ or $x = 8$ .
5. Step 5
  Both correspond to the same rectangle (length $8$ , width $5$ ).
Answer
Sides are $5$ cm and $8$ cm.
Examiner tip
The examiner report flags candidates often state two separate rectangles rather than recognising that swapping length and width is the same rectangle. State the dimensions as a single pair.

Key Formulae — Simultaneous Equations

The formulae you need to memorise for simultaneous equations on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Elimination strategy
$\text{Make matching coefficients} \to \text{add or subtract} \to \text{solve} \to \text{back-substitute}$
When to use
Standard for two linear equations in two variables.
Substitution strategy
$\text{Solve one equation for one variable} \to \text{substitute into the other} \to \text{solve}$
When to use
Best when one equation already has a variable isolated, or when one of the equations is non-linear (quadratic).

Key Definitions and Keywords — Simultaneous Equations

Definitions to memorise and the exact keywords mark schemes credit for simultaneous equations answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Simultaneous equations
Examiner keyword
Two or more equations whose solution must satisfy all of them at the same time.
Elimination method
Examiner keyword
Add or subtract scaled equations so that one variable cancels, leaving a single-variable equation.
Substitution method
Examiner keyword
Express one variable from one equation and substitute into the other.
Back-substitute
After finding one variable, substitute its value into one of the original equations to find the other.

Common Mistakes and Misconceptions — Simultaneous Equations

The traps other students keep falling into on simultaneous equations questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Reporting only one pair when one equation is quadratic
0580/42 — recurring
Why it happens
Students solve one equation and stop, forgetting that the quadratic has two roots.
How to avoid it
After finding both $x$ -values, substitute each back to find its corresponding $y$ . State BOTH pairs.
✕Adding when you should subtract (or vice versa)
Why it happens
If $y$ -coefficients are both $+2$ , you SUBTRACT to eliminate, not add.
How to avoid it
Same sign → subtract. Opposite signs → add. Always check the variable disappears in the resulting equation.
✕Multiplying only one term when scaling an equation
Why it happens
Students multiply a coefficient but forget to multiply the constant on the right.
How to avoid it
Multiplying an equation by $k$ means multiplying every term by $k$ .
✕Substituting back into the equation you already used
Why it happens
Doesn't catch arithmetic errors.
How to avoid it
Substitute into the OTHER equation as a check; if it doesn't balance, you have an error to find.

Simultaneous Equations — frequently asked questions

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

Simultaneous Equations

Short Notes - Simultaneous Equations

Detailed Study Notes

At a glance

What you’ll learn

Elimination method (linear-linear)

Substitution method

Linear with quadratic

Graphical interpretation

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Simultaneous Equations

1Solve by elimination

2Elimination requiring multiplication

3Solve by substitution

4One linear and one quadratic

5Word problem to simultaneous equations

6Substitution requiring rearrangement first

7Simultaneous equations with fractional coefficients

8Line and circle intersection (one-linear-one-quadratic)

9Mixed-coin word problem with three constraints

10Word problem leading to a non-linear system

Elimination strategy

Substitution strategy

Simultaneous equations

Elimination method

Substitution method

Back-substitute

✕Reporting only one pair when one equation is quadratic

✕Adding when you should subtract (or vice versa)

✕Multiplying only one term when scaling an equation

✕Substituting back into the equation you already used

Simultaneous Equations — frequently asked questions

Simultaneous Equations

Short Notes - Simultaneous Equations

Detailed Study Notes

At a glance

What you’ll learn

Elimination method (linear-linear)

Substitution method

Linear with quadratic

Graphical interpretation

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Simultaneous Equations

1Solve by elimination

2Elimination requiring multiplication

3Solve by substitution

4One linear and one quadratic

5Word problem to simultaneous equations

6Substitution requiring rearrangement first

7Simultaneous equations with fractional coefficients

8Line and circle intersection (one-linear-one-quadratic)

9Mixed-coin word problem with three constraints

10Word problem leading to a non-linear system

Elimination strategy

Substitution strategy

Simultaneous equations

Elimination method

Substitution method

Back-substitute

✕Reporting only one pair when one equation is quadratic

✕Adding when you should subtract (or vice versa)

✕Multiplying only one term when scaling an equation

✕Substituting back into the equation you already used

Simultaneous Equations — frequently asked questions