Algebra

Simultaneous equations

21 past-paper questions, easiest first. Try each one, then reveal the answer.

Foundation & Higher7 of 21 from Foundation papers

Revision guide by The Full Marks

Simultaneous equations: what you need to know

Simultaneous equations are two equations with two unknowns that are both true at once. You find the pair of values that works for both, either algebraically or from where two graphs cross.

Foundation needs two linear equations; a linear equation with a quadratic is Higher only.

Key facts

Elimination: make the coefficients of one letter match, then add (different signs) or subtract (same signs).
Substitution: rearrange one equation for a letter and put it into the other.
On a graph, the solution is the point where the lines cross.
Higher: a line and a curve can meet at 2 points, 1 point (a tangent) or not at all.

Linear pair by elimination

Multiply one or both equations so one letter has the same number in front.
Same signs: subtract. Different signs: add.
Solve for the remaining letter, substitute back, then check in the other equation.

Word problems

Choose a letter for each unknown, e.g. $t$ for the price of a tea and $c$ for a coffee.
Write two equations from the information, solve them, then answer the question actually asked.

Higher: linear and quadratic by substitution

Rearrange the linear equation to $y = \ldots$ or $x = \ldots$ .
Substitute into the quadratic, expand and rearrange to $= 0$ .
Solve the quadratic, then find the matching value of the other letter for each answer.

Worked examples

Example 1 Foundation · grade 5

Solve $3x + 2y = 4$ and $4x + 5y = 17$ .

Show the solution

Make the $y$ terms match: multiply the first equation by 5 and the second by 2.
$15x + 10y = 20$
$8x + 10y = 34$
Same signs, so subtract: $7x = -14$ , so $x = -2$
$3(-2) + 2y = 4$ , so $2y = 10$ and $y = 5$
Check: $4(-2) + 5(5) = -8 + 25 = 17$

Example 2 Higher · grade 8

Solve $y = x + 1$ and $x^2 + y^2 = 13$ .

Show the solution

$x^2 + (x + 1)^2 = 13$
$x^2 + x^2 + 2x + 1 = 13$
$2x^2 + 2x - 12 = 0$
$x^2 + x - 6 = 0$
$(x + 3)(x - 2) = 0$
$x = 2$ , $y = 3$ or $x = -3$ , $y = -2$

Common mistakes

Subtracting when the signs are different (or adding when they are the same).
Multiplying only some terms of an equation, e.g. the $x$ term but not the number.
Squaring a bracket wrongly: $(x + 1)^2 = x^2 + 2x + 1$ , not $x^2 + 1$ .
Giving the $x$ answers without pairing each with its $y$ value.
Reading a solution off a graph the wrong way round: give the $x$ value (across) first, then $y$ (up).

Exam tips

Answers with no working often get no marks: show the elimination or substitution step clearly.
For "show that the line and curve meet at only one point", end with a sentence: the quadratic has only one (repeated) solution, so there is only one point of intersection.

Questions people ask

How do you solve simultaneous equations?

Make the number in front of one letter the same in both equations, add or subtract to get rid of it, solve for the other letter, then substitute back.

What do simultaneous equations look like on a graph?

Each equation is a line or curve. The solutions are the coordinates where they cross.

Exam questions: Simultaneous equations (21, easiest first)

Simultaneous equations5 marksGrade 4No calculatorFoundation

November 2024 · Paper 1 Foundation · Q16

Use two combined prices of tea and coffee to find the cost of a different order.

Show answer

Answer	$£15.50$	5 marks

Worked solution Checked against the mark scheme

The difference is 2 teas: $8.50 - 4.50 = £4$ , so 1 tea costs $£2$ .

1 coffee $= 4.50 - 2 = £2.50$

$4 \times 2 + 3 \times 2.50 = 8 + 7.50 = £15.50$

Watch out: Dividing £4.50 equally between tea and coffee.

Simultaneous equations4 marksGrade 5No calculatorHigher

June 2024 · Paper 1 Higher · Q10

Solve a pair of linear simultaneous equations.

Show answer

Answer	$x = 3,\ y = -4$	4 marks

Worked solution Checked against the mark scheme

Multiply the first equation by 3 and the second by 2:

$15x - 6y = 69$ and $4x - 6y = 36$

Subtract: $11x = 33$ , so $x = 3$

Substitute: $5(3) - 2y = 23 \Rightarrow -2y = 8 \Rightarrow y = -4$

Watch out: Adding instead of subtracting when the $y$ terms have the same sign.

Simultaneous equations4 marksGrade 5CalculatorHigher

November 2024 · Paper 2 Higher · Q10

Solve a pair of linear simultaneous equations by elimination.

Show answer

Answer	$x = -2.5$ , $y = -6$	4 marks

Worked solution Checked against the mark scheme

Multiply the first by 7 and the second by 5:

$28x - 35y = 140$ and $30x + 35y = -285$

Add: $58x = -145$ , so $x = -2.5$

$4(-2.5) - 5y = 20 \Rightarrow -5y = 30 \Rightarrow y = -6$

Watch out: Adding when the signs mean you should subtract, or a sign slip with the negatives.

Simultaneous equations1 markGrade 5No calculatorOn both tiers

June 2023 · Paper 1 Foundation · Q28

Read the solution of a pair of simultaneous equations from where their graphs cross.

Show answer

Answer	$x = 6,\ y = -2$	1 mark

Worked solution Checked against the mark scheme

The solution is where the two lines meet, at $(6, -2)$ .

Check: $2 - 2(-2) = 6$ ✓ and $2(-2) = -4 = 3(6) - 22$ ✓

So $x = 6$ , $y = -2$ .

Watch out: Swapping x and y.

Simultaneous equations3 marksGrade 5CalculatorFoundation

November 2023 · Paper 2 Foundation · Q30

Solve a pair of linear simultaneous equations with decimal constants.

Show answer

Answer	$x = -2,\ y = 1.5$	3 marks

Worked solution Checked against the mark scheme

Multiply the first equation by 3: $9x + 3y = -13.5$

Subtract the second: $5x = -10$ , so $x = -2$

Substitute: $3(-2) + y = -4.5$ , so $y = 1.5$

Watch out: Getting the sign wrong when subtracting negative numbers.

Simultaneous equations4 marksGrade 5No calculatorHigher

November 2022 · Paper 1 Higher · Q12

Show answer

Answer	$x = 3$ , $y = -2$	4 marks

Worked solution Checked against the mark scheme

Multiply the first by 3 and the second by 2:

$15x + 6y = 33$ and $8x + 6y = 12$

Subtract: $7x = 21$ , so $x = 3$

$5(3) + 2y = 11 \Rightarrow 2y = -4 \Rightarrow y = -2$

Watch out: Adding the equations instead of subtracting after matching the y terms.

Simultaneous equations3 marksGrade 5CalculatorFoundation

June 2022 · Paper 2 Foundation · Q28

Show answer

Answer	$x = 6.5,\ y = -2.75$	3 marks

Worked solution Checked against the mark scheme

Double the first equation: $10x + 4y = 54$

Subtract the second: $(10x + 4y) - (6x + 4y) = 54 - 28$ , so $4x = 26$ and $x = 6.5$

Substitute: $5(6.5) + 2y = 27$ , so $2y = -5.5$ and $y = -2.75$

Watch out: Adding the equations instead of subtracting when the signs are the same.

Simultaneous equations2 marksGrade 5No calculatorHigher

June 2019 · Paper 1 Higher · Q10

Read the solution of a pair of simultaneous linear equations from where their graphs cross.

Show answer

Answer	$x \approx 2.25,\ y \approx -1.35$ (x from 2.2 to 2.3, y from −1.3 to −1.4)	2 marks

Worked solution Checked against the mark scheme

The solution is where the two lines cross.

Reading the grid, they cross at about $x = 2.25$ and $y = -1.35$ .

(Solving exactly gives $x = \frac{9}{4}$ and $y = -\frac{4}{3}$ .)

Watch out: Swapping the x and y values, or reading where a line meets an axis.

Simultaneous equations3 marksGrade 5CalculatorFoundation

June 2019 · Paper 3 Foundation · Q30

Show answer

Answer	$x = 1$ , $y = -2$	3 marks

Worked solution Checked against the mark scheme

$3x - 4y = 11$ and $9x + 2y = 5$

Double the second: $18x + 4y = 10$

Add to the first: $21x = 21$ , so $x = 1$

$9(1) + 2y = 5$ , so $2y = -4$ and $y = -2$

Watch out: Sign errors when eliminating, e.g. subtracting instead of adding.

Simultaneous equations3 marksGrade 5No calculatorOn both tiers

November 2018 · Paper 1 Foundation · Q25

Show answer

Answer	$x = 4.5$ , $y = -1.5$	3 marks

Worked solution Checked against the mark scheme

Multiply the first equation by 3: $15x + 3y = 63$

Add $x - 3y = 9$ : $16x = 72$ , so $x = 4.5$

$5(4.5) + y = 21 \Rightarrow y = 21 - 22.5 = -1.5$

Check: $4.5 - 3(-1.5) = 9$ ✓

Watch out: Sign error when eliminating y, e.g. subtracting instead of adding.

Simultaneous equations3 marksGrade 5No calculatorFoundation

Specimen · Paper 1 Foundation · Q29

Show answer

Answer	$x = 7$ , $y = -3$	3 marks

Worked solution Checked against the mark scheme

Multiply the first equation by 3: $12x + 3y = 75$

Add the second: $12x + 3y + x - 3y = 75 + 16$ , so $13x = 91$ and $x = 7$ .

Substitute: $4(7) + y = 25$ , so $y = -3$ .

Check: $7 - 3(-3) = 16$ ✓

Watch out: Subtracting the equations when the y terms have opposite signs.

Simultaneous equations4 marksGrade 6CalculatorHigher

June 2023 · Paper 2 Higher · Q11

Solve a pair of linear simultaneous equations algebraically.

Show answer

Answer	$x = -2$ , $y = 1.5$	4 marks

Worked solution Checked against the mark scheme

Multiply the first by 2: $4x + 12y = 10$

Multiply the second by 3: $9x - 12y = -36$

Add: $13x = -26$ , so $x = -2$

$2(-2) + 6y = 5 \Rightarrow 6y = 9 \Rightarrow y = 1.5$

Watch out: Subtracting the equations when the signs mean they should be added, giving a wrong x.

Simultaneous equations3 marksGrade 7No calculatorHigher

November 2021 · Paper 1 Higher · Q20

Draw a straight line on a circle graph to estimate where the line and circle meet.

Show answer

Answer	$x \approx 2.1, y \approx 5.1$ and $x \approx -2.9, y \approx -4.7$	3 marks

Worked solution Checked against the mark scheme

Rearrange: $y = 2x + 1$ . Draw it through (0, 1), (2, 5) and (−3, −5).

Read off where it crosses the circle:

$x \approx 2.1, y \approx 5.1$

$x \approx -2.9, y \approx -4.7$

Watch out: Drawing $y = x + 1$ or giving the coordinates of only one crossing point.

Simultaneous equations5 marksGrade 7CalculatorHigher

November 2021 · Paper 3 Higher · Q13

Use the coordinates of two corners of a pattern of identical rectangles to form and solve equations for the rectangle's sides, then find a third point.

Show answer

Answer	$(9, 8)$	5 marks

Worked solution Checked against the mark scheme

Let each rectangle be $l$ long and $w$ wide.

Across from A to B: $11 - 3 = 8$ , so $2l - w = 8$

Up from A to B: $20 - 4 = 16$ , so $2l + w = 16$

Adding: $4l = 24$ , $l = 6$ , then $w = 4$ .

C is one length right and one width up from A: $(3 + 6, 4 + 4) = (9, 8)$

Watch out: Assuming the rectangles are twice as long as they are wide without checking.

Simultaneous equations4 marksGrade 7CalculatorHigher

November 2021 · Paper 3 Higher · Q16

Show algebraically that a quadratic curve and a straight line meet at exactly one point.

Show answer

Answer	$x^2 - 2x + 1 = 0$ gives $(x - 1)^2 = 0$ , a single root, so one point in common	4 marks

Worked solution Checked against the mark scheme

From the line: $y = 5x - 4$

Set equal: $x^2 + 3x - 3 = 5x - 4$

$x^2 - 2x + 1 = 0$

$(x - 1)^2 = 0$ , so $x = 1$ is the only solution.

One value of $x$ means only one point in common, $(1, 1)$ .

Watch out: Solving correctly but not writing a statement explaining why this means one point.

Simultaneous equations4 marksGrade 8CalculatorHigher

November 2024 · Paper 3 Higher · Q20

Solve a pair of simultaneous equations, one linear and one quadratic, giving answers to 3 significant figures.

Show answer

Answer	$x = 26.3,\ y = -45.6$ and $x = 1.71,\ y = 3.58$ (3.57 also accepted)	4 marks

Worked solution Checked against the mark scheme

$y = 7 - 2x$ , so $(7 - 2x)^2 = 3x^2 + 4$

$49 - 28x + 4x^2 = 3x^2 + 4$

$x^2 - 28x + 45 = 0$

$x = 14 \pm \sqrt{151}$ , so $x = 26.3$ or $x = 1.71$

Then $y = 7 - 2x$ : $y = -45.6$ or $y = 3.58$

Watch out: Expanding (7 − 2x)² as 49 + 4x² without the middle term.

Simultaneous equations5 marksGrade 8CalculatorHigher

June 2022 · Paper 3 Higher · Q22

Find the intersection points of a straight line and a curve by substituting and solving a quadratic.

Show answer

Answer	$(-3, -11)$ and $(5.5, 6)$	5 marks

Worked solution Checked against the mark scheme

Substitute $y = 2x - 5$ : $(2x - 5)^2 = 6x^2 - 25x - 8$

$4x^2 - 20x + 25 = 6x^2 - 25x - 8$

$2x^2 - 5x - 33 = 0 \Rightarrow (2x - 11)(x + 3) = 0$

$x = 5.5 \Rightarrow y = 6$ ; $x = -3 \Rightarrow y = -11$

Watch out: Expanding $(2x - 5)^2$ as $4x^2 + 25$ .

Simultaneous equations5 marksGrade 8No calculatorHigher

June 2019 · Paper 1 Higher · Q22

Use two probabilities, before and after adding counters to a bag, to find how many red and green counters there were at the start.

Show answer

Answer	$12$ red and $9$ green	5 marks

Worked solution Checked against the mark scheme

Green is $\frac{3}{7}$ , so $g = 3x$ and $r = 4x$ for some $x$ .

After adding: $\frac{3x + 3}{7x + 5} = \frac{6}{13}$

$39x + 39 = 42x + 30$ , so $3x = 9$ and $x = 3$ .

Red $= 4 \times 3 = 12$ , green $= 3 \times 3 = 9$ .

Watch out: Adding 2 to the denominator only, forgetting the 3 extra green counters also increase the total.

Simultaneous equations5 marksGrade 8CalculatorHigher

June 2019 · Paper 3 Higher · Q20

Solve a linear and a quadratic equation simultaneously by substitution.

Show answer

Answer	$x = 5, y = -2$ and $x = \frac{17}{5}, y = -\frac{4}{5}$	5 marks

Worked solution Checked against the mark scheme

From $3x + 4y = 7$ : $4y = 7 - 3x$ , so $4y^2 = \frac{(7-3x)^2}{4}$

$x^2 - \frac{49 - 42x + 9x^2}{4} = 9 \Rightarrow 4x^2 - 49 + 42x - 9x^2 = 36$

$5x^2 - 42x + 85 = 0 \Rightarrow (5x - 17)(x - 5) = 0$

$x = 5 \Rightarrow y = \frac{7 - 15}{4} = -2$

$x = \frac{17}{5} \Rightarrow y = \frac{7 - 10.2}{4} = -\frac{4}{5}$

Watch out: Squaring 4y as 4y² instead of 16y² when substituting.

Simultaneous equations5 marksGrade 8CalculatorHigher

June 2018 · Paper 2 Higher · Q16

Draw a circle from its equation and use it with a straight line to estimate solutions of simultaneous equations.

Show answer

(a)	Circle centre $(0, 0)$ , radius $3.5$	2 marks
(b)	$x = 2.0, y = -2.9$ and $x = -1.2, y = 3.3$	3 marks

Worked solution Checked against the mark scheme

(a) $x^2 + y^2 = 12.25$ is a circle centre $O$ with radius $\sqrt{12.25} = 3.5$

(b) Draw $y = 1 - 2x$ (through $(0, 1)$ and $(2, -3)$ ).

Read off the crossing points: about $(2.0, -2.9)$ and $(-1.2, 3.3)$

(Check: $5x^2 - 4x - 11.25 = 0$ gives $x = 1.95$ or $-1.15$ .)

Watch out: Using 12.25 as the radius instead of its square root.

Simultaneous equations5 marksGrade 8CalculatorHigher

November 2018 · Paper 3 Higher · Q19

Solve a pair of simultaneous equations where one is quadratic.

Show answer

Answer	$x = 3, y = -1$ and $x = -\frac{23}{7}, y = \frac{15}{7}$	5 marks

Worked solution Checked against the mark scheme

From the linear equation: $x = 1 - 2y$

$2(1 - 2y)^2 - y^2 = 17$

$2 - 8y + 8y^2 - y^2 = 17 \Rightarrow 7y^2 - 8y - 15 = 0$

$(7y - 15)(y + 1) = 0$ , so $y = \frac{15}{7}$ or $y = -1$

$y = -1$ : $x = 3$ . $\quad y = \frac{15}{7}$ : $x = 1 - \frac{30}{7} = -\frac{23}{7}$

Watch out: Expanding $(1 - 2y)^2$ as $1 + 4y^2$ .

Also tested in these questions

Their main topic is different, but part of each one tests this topic.

November 2023 · Paper 1 Higher · Q21: Draw the circle centred at the origin with radius 13 and use a straight line to estimate where they cross.
November 2020 · Paper 3 Foundation · Q28: Read the solution of simultaneous equations from the intersection of two lines, and estimate the roots of a quadratic from its graph.
November 2018 · Paper 2 Higher · Q17: Form and solve simultaneous equations from two ratio conditions to find the ratio of two numbers.
November 2018 · Paper 2 Higher · Q18: Find where a line through two points meets a line through the origin with a given gradient.
June 2018 · Paper 3 Higher · Q16: Form and solve simultaneous equations to find the coefficients of a quadratic nth term, then find the nth term of another quadratic sequence.