Simultaneous Equations · Study Notes · Edexcel IGCSE 4MA1 Higher

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

At a glance

Linear-linear: elimination or substitution.
Elimination: align coefficients, add or subtract.
Substitution: solve one equation for one variable, sub into other.
Linear-quadratic: substitute the linear into the quadratic.
Two solution pairs for linear-quadratic.
Always check by substituting back.

What you’ll learn

Mapped to the Pearson Edexcel 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

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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

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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.

The two intersection points are the two solution pairs of the linear-quadratic system.

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

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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.

Quick recap

Elimination: align, add or subtract.
Substitution: solve, sub in.
Linear-quadratic: sub linear into quadratic.
Two solution pairs for linear-quadratic.
Word: define, write, solve.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Elimination: same signs subtract, opposite add.
Substitution: useful when one var isolated.
Linear-quadratic: ALWAYS substitute.
Two solution pairs for linear-quadratic.

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.

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 = £ $2$ .
6Equating two expressions for y [2 marks]
Getting started• Paper 1H/2H style — 2 marks• simultaneous equations, intersection of lines
▼
Question
Solve the simultaneous equations

$y = 4x - 9$

$y = x + 3$

Show clear algebraic working. [2]
Step-by-step solution
1. Step 1
  M1 Both equations give $y$ , so the two right-hand sides are equal.
  $4x - 9 = x + 3 \;\Rightarrow\; 3x = 12 \;\Rightarrow\; x = 4$
2. Step 2
  A1 Substitute $x = 4$ into either equation. Check in the other: $4(4) - 9 = 7$ .
  $y = 4 + 3 = 7$
Answer
$x = 4$ , $y = 7$
Examiner tip
Examiner tip: Setting the two expressions equal is a form of substitution. Graphically, $(4, 7)$ is where the two lines cross.

Common mistake: Moving terms across and changing only one sign, for example $4x - x = 3 - 9$ , giving $x = -2$ . Collect $x$ terms on one side and numbers on the other, changing the sign of each term that moves.
7Finding m and c from two points on a line [3 marks]
Getting started• Paper 1H/2H style — 3 marks• simultaneous equations, straight line, forming equations
▼
Question
The line with equation $y = mx + c$ passes through the points $(2, 7)$ and $(-3, -8)$ .

Find the value of $m$ and the value of $c$ . Show clear algebraic working. [3]
Step-by-step solution
1. Step 1
  M1 Each point satisfies $y = mx + c$ . Substitute both to form a pair of equations in $m$ and $c$ .
  $7 = 2m + c \qquad -8 = -3m + c$
2. Step 2
  M1 Subtract the second equation from the first to eliminate $c$ .
  $7 - (-8) = 2m - (-3m) \;\Rightarrow\; 15 = 5m$
3. Step 3
  A1 Solve, then substitute $m = 3$ into $7 = 2m + c$ .
  $m = 3, \qquad c = 7 - 6 = 1$
Answer
$m = 3$ , $c = 1$ (so the line is $y = 3x + 1$ )
Examiner tip
Examiner tip: Here the unknowns are $m$ and $c$ , not $x$ and $y$ . The coordinates are the numbers you substitute.

Common mistake: Subtracting negatives wrongly: $7 - (-8) = 15$ and $2m - (-3m) = 5m$ . Writing the brackets in prevents answers such as $-1 = -m$ . Check with the second point: $3(-3) + 1 = -8$ .
8Simultaneous equations with decimal coefficients [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• simultaneous equations, elimination
▼
Question
Solve the simultaneous equations

$0.5x + 0.2y = 3$

$0.3x - 0.4y = -0.8$

Show clear algebraic working. [3]
Step-by-step solution
1. Step 1
  Multiply every term of both equations by $10$ to remove the decimals.
  $5x + 2y = 30 \qquad 3x - 4y = -8$
2. Step 2
  M1 Multiply the first by $2$ to give $+4y$ , then add to eliminate $y$ .
  $10x + 4y = 60 \qquad 13x = 52 \;\Rightarrow\; x = 4$
3. Step 3
  M1 Substitute $x = 4$ into $5x + 2y = 30$ .
  $20 + 2y = 30 \;\Rightarrow\; y = 5$
4. Step 4
  A1 Both values correct. Check in an original equation: $0.3(4) - 0.4(5) = 1.2 - 2 = -0.8$ .
Answer
$x = 4$ , $y = 5$
Examiner tip
Examiner tip: Scaling by $10$ is allowed because it is the same operation on every term, so the pair has the same solution.

Common mistake: Multiplying the left-hand side by $10$ but leaving $-0.8$ unchanged, which gives $3x - 4y = -0.8$ and a wrong answer. Always check in one of the original decimal equations.
9Opposite sides of a rectangle: forming the equations [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• simultaneous equations, forming equations, geometry
▼
Question
$ABCD$ is a rectangle. The lengths of its sides, in centimetres, are

$AB = 3x + 2y$ , $\quad DC = x + 4y + 2$ , $\quad AD = 2x + y$ , $\quad BC = 5y - x + 1$

Work out the area of the rectangle. [4]
Step-by-step solution
1. Step 1
  M1 Opposite sides of a rectangle are equal, so $AB = DC$ and $AD = BC$ . Write both equations and simplify each.
  $3x + 2y = x + 4y + 2 \;\Rightarrow\; x - y = 1 \qquad 2x + y = 5y - x + 1 \;\Rightarrow\; 3x - 4y = 1$
2. Step 2
  M1 Solve the pair. From $x - y = 1$ , $x = y + 1$ ; substitute into $3x - 4y = 1$ .
  $3(y + 1) - 4y = 1 \;\Rightarrow\; -y = -2$
3. Step 3
  A1 Both values.
  $y = 2, \quad x = 3$
4. Step 4
  A1 Find the side lengths and the area. Check: $DC = 3 + 8 + 2 = 13$ and $BC = 10 - 3 + 1 = 8$ .
  $AB = 9 + 4 = 13 \qquad AD = 6 + 2 = 8 \qquad 13 \times 8 = 104$
Answer
$104$ cm²
Examiner tip
Examiner tip: The equations come from a property of the shape, not from given numbers. Say which sides are equal, then collect terms so each equation is in the form $ax + by = c$ .

Common mistake: Stopping at $x = 3$ , $y = 2$ . Those values earn three marks, but the question asks for the area, so the final mark needs $104$ cm². Another slip is multiplying $x$ by $y$ ( $6$ ) instead of the two side lengths.
10A line that meets a circle at only one point [5 marks]
Stretch• Paper 1H/2H style — 5 marks• linear and quadratic, circle, intersection
▼
Question
(a) Solve the simultaneous equations

$x^{2} + y^{2} = 20$

$x + 2y = 10$

Show clear algebraic working. [4]

(b) What does your answer to part (a) show about the line $x + 2y = 10$ and the circle $x^{2} + y^{2} = 20$ ? [1]
Step-by-step solution
1. Step 1
  M1 Make $x$ the subject of the linear equation and substitute into the circle, keeping the bracket.
  $x = 10 - 2y \qquad (10 - 2y)^{2} + y^{2} = 20$
2. Step 2
  A1 Expand and collect into a three-term quadratic equal to zero, then divide by $5$ .
  $100 - 40y + 4y^{2} + y^{2} = 20 \;\Rightarrow\; 5y^{2} - 40y + 80 = 0 \;\Rightarrow\; y^{2} - 8y + 16 = 0$
3. Step 3
  M1 Factorise. The quadratic is a perfect square, so there is only one value of $y$ .
  $(y - 4)^{2} = 0 \;\Rightarrow\; y = 4$
4. Step 4
  A1 Find $x$ from $x = 10 - 2y$ .
  $x = 10 - 8 = 2$
5. Step 5
  B1 The quadratic has one repeated root, $y = 4$ , so there is only one solution and only one point where the line and the circle meet. The line touches the circle at $(2, 4)$ without crossing it: it is a tangent to the circle.
Answer
(a) $x = 2$ , $y = 4$ (b) The quadratic has one repeated root, so the line meets the circle at only one point, $(2, 4)$ : the line is a tangent to the circle.
Examiner tip
Examiner tip: The number of different roots of the quadratic is the number of meeting points. Two different roots: the line crosses the circle twice. One repeated root: it touches once (a tangent). No real roots: it misses the circle. Check: $2^{2} + 4^{2} = 20$ .

Common mistake: Expanding $(10 - 2y)^{2}$ as $100 + 4y^{2}$ and losing the $-40y$ term, or assuming there must be two answers and inventing a second pair.

Question 1
Paper 1H/2H style — 1 mark1 mark
Kai says that $x = 2$ , $y = -1$ is the solution of the simultaneous equations

$3x + 4y = 2$

$x - 2y = 3$

Is Kai correct? Give a reason for your answer. [1]
Model answer
$3(2) + 4(-1) = 6 - 4 = 2$ , which is correct.

$2 - 2(-1) = 2 + 2 = 4$ , which is not $3$ .

No, Kai is not correct, because $x = 2$ , $y = -1$ does not satisfy the second equation.
Why this scores
Mark scheme:
- B1 for No, with a reason that shows the second equation gives $4$ , not $3$
Why this scores full marks: A solution of a pair must satisfy both equations. The script tests both, and the reason quotes the failed check. "No" with no working scores 0.
Question 2
Paper 1H/2H style — 2 marks2 marks
$3x + 2y = 17$

$2x + 3y = 13$

Work out the value of $x + y$ . Show clear algebraic working. [2]
Model answer
Add the equations:

$5x + 5y = 30$

$x + y = 6$

Answer: $6$
Why this scores
Mark scheme:
- M1 for adding the equations to get $5x + 5y = 30$ , or for a correct method to find $x$ or $y$
- A1 for $6$ cao
Why this scores full marks: The coefficients of $x$ and $y$ swap between the equations, so adding them gives $5x + 5y$ , a multiple of $x + y$ . The script spots this and never needs $x$ and $y$ separately. Solving fully also works: $x = 5$ , $y = 1$ , and $5 + 1 = 6$ .
Question 3
Paper 1H/2H style — 2 marks2 marks
Solve the simultaneous equations

$6x + y = 17$

$2x + y = 5$

Show clear algebraic working. [2]
Model answer
Subtract: $(6x + y) - (2x + y) = 17 - 5$

$4x = 12 \Rightarrow x = 3$

$2(3) + y = 5 \Rightarrow y = -1$

Answer: $x = 3$ , $y = -1$
Why this scores
Mark scheme:
- M1 for a correct method to eliminate $y$ , e.g. $4x = 12$
- A1 for $x = 3$ and $y = -1$
Why this scores full marks: The $y$ terms have the same sign, so the script subtracts. Adding would give $8x + 2y = 22$ and eliminate nothing. Check: $6(3) - 1 = 17$ .
Question 4
Paper 1H/2H style — 3 marks3 marks
The sum of two numbers is $31$ .

Three times the larger number minus twice the smaller number is $38$ .

Find the two numbers. Show clear algebraic working. [3]
Model answer
Let the larger number be $x$ and the smaller be $y$ .

$x + y = 31$

$3x - 2y = 38$

$\times 2$ : $2x + 2y = 62$

Add: $5x = 100 \Rightarrow x = 20$

$y = 31 - 20 = 11$

Answer: $20$ and $11$
Why this scores
Mark scheme:
- M1 for $x + y = 31$ and $3x - 2y = 38$ oe
- M1 for a correct method to eliminate one variable
- A1 for $20$ and $11$
Why this scores full marks: The script defines both letters, so it is clear which number is tripled. Check: $3(20) - 2(11) = 60 - 22 = 38$ .
Question 5
Paper 1H/2H style — 3 marks3 marks
Solve the simultaneous equations

$3(x + 2y) = 2x - 4$

$4x - y = 2y + 11$

Show clear algebraic working. [3]
Model answer
$3x + 6y = 2x - 4 \Rightarrow x + 6y = -4$

$4x - 3y = 11$

$\times 2$ : $8x - 6y = 22$

Add: $9x = 18 \Rightarrow x = 2$

$2 + 6y = -4 \Rightarrow y = -1$

Answer: $x = 2$ , $y = -1$
Why this scores
Mark scheme:
- M1 for rearranging both equations to the form $ax + by = c$ , e.g. $x + 6y = -4$ and $4x - 3y = 11$
- M1 for a correct method to eliminate one variable from their equations
- A1 for $x = 2$ and $y = -1$
Why this scores full marks: Tidying each equation first turns an unfamiliar pair into a standard one. The script expands $3(x + 2y)$ fully, avoiding $3x + 2y$ , and collects $-y - 2y$ as $-3y$ . Check in the original: $3(2 - 2) = 0$ and $2(2) - 4 = 0$ .
Question 6
Paper 1H/2H style — 4 marks4 marks
A taxi fare is made up of a fixed charge of £ $f$ plus £ $r$ for each kilometre travelled.

A journey of $6$ km costs £ $13.10$

A journey of $11$ km costs £ $21.60$

Work out the cost of a journey of $15$ km. [4]
Model answer
$f + 6r = 13.10$

$f + 11r = 21.60$

Subtract: $5r = 8.50 \Rightarrow r = 1.70$

$f = 13.10 - 6 \times 1.70 = 2.90$

$15$ km: $2.90 + 15 \times 1.70 = 2.90 + 25.50 = 28.40$

Answer: £ $28.40$
Why this scores
Mark scheme:
- M1 for $f + 6r = 13.10$ and $f + 11r = 21.60$ oe
- M1 for a correct method to find $r$ or $f$ , e.g. $5r = 8.50$
- A1 for $r = 1.70$ and $f = 2.90$
- A1 for £ $28.40$ cao
Why this scores full marks: Subtracting the equations removes the fixed charge, so the extra $5$ km costs £ $8.50$ . The script then answers the question, a $15$ km fare, rather than stopping at the rate.
Question 7
Paper 1H/2H style — 4 marks4 marks
Show that the three lines

$2x + y = 7$

$x - 3y = 14$

$3x - 2y = 21$

all pass through the same point. [4]
Model answer
Solve the first two:

$\times 3$ : $6x + 3y = 21$

Add $x - 3y = 14$ : $7x = 35 \Rightarrow x = 5$

$2(5) + y = 7 \Rightarrow y = -3$

The first two lines meet at $(5, -3)$ .

Third line: $3(5) - 2(-3) = 15 + 6 = 21$ ✓

So $(5, -3)$ lies on all three lines: they pass through the same point.
Why this scores
Mark scheme:
- M1 for a correct method to eliminate one variable from two of the equations
- M1 (dep) for substituting their value to find the other variable
- A1 for $(5, -3)$
- B1 for substituting $(5, -3)$ into the third equation to show it gives $21$ , with a conclusion
Why this scores full marks: Two lines meet at one point; the script finds it and then shows the third line also passes through it. The final mark needs both the check and the concluding sentence, as in any "Show that".
Question 8
Paper 1H/2H style — 5 marks5 marks
Solve the simultaneous equations

$x^{2} + y^{2} = 25$

$3x + y = 5$

Show clear algebraic working. [5]
Model answer
$y = 5 - 3x$

$x^{2} + (5 - 3x)^{2} = 25$

$x^{2} + 25 - 30x + 9x^{2} = 25$

$10x^{2} - 30x = 0$

$10x(x - 3) = 0$

$x = 0$ or $x = 3$

$x = 0$ : $y = 5 - 0 = 5$

$x = 3$ : $y = 5 - 9 = -4$

Answer: $x = 0$ , $y = 5$ and $x = 3$ , $y = -4$
Why this scores
Mark scheme:
- M1 for substituting a rearranged linear equation into the circle, e.g. $x^{2} + (5 - 3x)^{2} = 25$
- A1 for $10x^{2} - 30x = 0$ oe
- M1 for a correct method to solve their quadratic, e.g. factorising
- A1 for $x = 0$ and $x = 3$
- A1 for $y = 5$ and $y = -4$ , correctly paired
Why this scores full marks: The constant terms cancel, so the quadratic has no number term. The script factorises $10x^{2} - 30x$ instead of dividing both sides by $x$ , which would throw away the solution $x = 0$ . Both pairs check in the circle: $0 + 25 = 25$ and $9 + 16 = 25$ .
Question 9
Paper 1H/2H style — 5 marks5 marks
Solve the simultaneous equations

$y = x^{2} - 2x - 1$

$y = 5 - 2x$

Give your answers in exact form. Show clear algebraic working. [5]
Model answer
$x^{2} - 2x - 1 = 5 - 2x$

$x^{2} = 6$

$x = \sqrt{6}$ or $x = -\sqrt{6}$

$x = \sqrt{6}$ : $y = 5 - 2\sqrt{6}$

$x = -\sqrt{6}$ : $y = 5 + 2\sqrt{6}$

Answer: $x = \sqrt{6}$ , $y = 5 - 2\sqrt{6}$ and $x = -\sqrt{6}$ , $y = 5 + 2\sqrt{6}$
Why this scores
Mark scheme:
- M1 for eliminating $y$ , e.g. $x^{2} - 2x - 1 = 5 - 2x$
- A1 for $x^{2} = 6$ oe
- A1 for $x = \pm\sqrt{6}$
- M1 for substituting at least one of their $x$ values into an equation to find $y$
- A1 for $y = 5 - 2\sqrt{6}$ and $y = 5 + 2\sqrt{6}$ , correctly paired
Why this scores full marks: The $-2x$ terms cancel, leaving $x^{2} = 6$ . The script takes both square roots; giving only $\sqrt{6}$ is the error the June 2024 report describes as taking only the positive root. The answers stay as surds, because decimals such as $2.449$ are not exact.
Question 10
Paper 1H/2H style — 6 marks6 marks
The line $y = 2x - 1$ intersects the curve $y = x^{2} - 3x + 2$ at the points $A$ and $B$ .

Work out the length of $AB$ . Give your answer correct to 3 significant figures. Show clear algebraic working. [6]
Model answer
$x^{2} - 3x + 2 = 2x - 1$

$x^{2} - 5x + 3 = 0$

$x = \dfrac{5 \pm \sqrt{25 - 12}}{2} = \dfrac{5 \pm \sqrt{13}}{2}$

$x = 4.3027\ldots$ or $x = 0.6972\ldots$

$y = 2x - 1$ : $y = 7.6055\ldots$ or $y = 0.3944\ldots$

Difference in $x$ : $\sqrt{13}$ . Difference in $y$ : $2\sqrt{13}$

$AB = \sqrt{(\sqrt{13})^{2} + (2\sqrt{13})^{2}} = \sqrt{13 + 52} = \sqrt{65}$

Answer: $8.06$
Why this scores
Mark scheme:
- M1 for eliminating $y$ , e.g. $x^{2} - 3x + 2 = 2x - 1$
- A1 for $x^{2} - 5x + 3 = 0$
- M1 for correct substitution into the quadratic formula for their three-term quadratic
- A1 for $x = \dfrac{5 \pm \sqrt{13}}{2}$ or awrt $4.30$ and $0.697$
- M1 for finding both $y$ values and using Pythagoras with their coordinates
- A1 for awrt $8.06$
Why this scores full marks: The script finds both points before using Pythagoras, and works with exact values, so no rounding error enters. Rounding the coordinates to $4.3$ , $0.7$ , $7.6$ and $0.4$ gives $AB = 8.05$ , which loses the final mark.

4MA1 Simultaneous Equations

At a glance

What you’ll learn

Elimination method

Substitution method

Linear and quadratic together

Word problems with simultaneous equations

Quick recap

Memorise this

How it’s examined

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

1Elimination method

2Multiply to eliminate

3Substitution method

4Linear with quadratic

5Setting up from a word problem

6Equating two expressions for y [2 marks]

7Finding m and c from two points on a line [3 marks]

8Simultaneous equations with decimal coefficients [3 marks]

9Opposite sides of a rectangle: forming the equations [4 marks]

10A line that meets a circle at only one point [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