Simultaneous equations (linear & linear/quadratic) – GCSE Maths exam questions with mark scheme

1Grade 4–5Non-calculator2 marks

The graphs of $y = x + 1$ and $y = 7 - 2x$ are drawn on the grid.

Use the graphs to solve the simultaneous equations

$$y = x + 1 \qquad y = 7 - 2x$$ [2]

$x = 2$ (accept the coordinate pair $(2, 3)$ for both marks)B1
$y = 3$B1
Answer: $x = 2$, $y = 3$
Examiner tip: The solution is the point where the lines cross – give BOTH the $x$ and the $y$ value, not just one coordinate.

2Grade 4–5Non-calculator3 marks

Solve the simultaneous equations

$$4x + 3y = 18$$ $$2x - y = 4$$ [3]

for a correct method to eliminate one variable, e.g. multiplying the second equation by 3 and adding: $10x = 30$ (condone one arithmetic error), or rearranging $y = 2x - 4$ and substitutingM1
(dep) for substituting their found value into one of the equations, e.g. $2 \times 3 - y = 4$M1
$x = 3$, $y = 2$A1
Answer: $x = 3$, $y = 2$
Examiner tip: Check your answer in the equation you did NOT use for substitution – $4 \times 3 + 3 \times 2 = 18$ ✓.

3Grade 4–5Non-calculator4 marks

Solve the simultaneous equations

$$5x + 3y = 7$$ $$3x - 2y = 8$$ [4]

for a correct method to eliminate one variable, e.g. $10x + 6y = 14$ and $9x - 6y = 24$ added to give $19x = 38$ (condone one arithmetic error), or eliminating $x$: $15x+9y=21$ and $15x-10y=40$ subtracted to give $19y=-19$M1
$x = 2$ (or $y = -1$ if $x$ eliminated)A1
(dep on M1) for substituting their found value into one of the equations, e.g. $5 \times 2 + 3y = 7$M1
$x = 2$ and $y = -1$A1
Answer: $x = 2$, $y = -1$
Examiner tip: Same signs subtract, different signs add: $+6y$ and $-6y$ must be ADDED to eliminate $y$.

4Grade 4–5Non-calculator4 marks

At a cinema, 3 adult tickets and 2 child tickets cost £47 in total.

2 adult tickets and 5 child tickets cost £57 in total.

Work out the cost of one adult ticket and the cost of one child ticket. [4]

for forming two equations, e.g. $3a + 2c = 47$ and $2a + 5c = 57$P1
for a correct process to eliminate one variable, e.g. $6a + 4c = 94$ and $6a + 15c = 171$ leading to $11c = 77$P1
for substituting their found value to find the other, e.g. $3a + 2 \times 7 = 47$P1
adult £11, child £7A1
Answer: Adult £11, child £7
Examiner tip: Define your letters clearly ($a$ = cost of adult ticket) and give the answer back in context with £ signs.

5Grade 6–7HigherNon-calculator4 marks

The line $y = 2x - 4$ meets the curve $y = x^2 - 3x + 2$ at the points $A$ and $B$.

Find the coordinates of $A$ and $B$. [4]

for equating, $x^2 - 3x + 2 = 2x - 4$M1
for $x^2 - 5x + 6 = 0$A1
for solving their quadratic, e.g. $(x - 2)(x - 3) = 0$M1
$(2, 0)$ and $(3, 2)$A1
Answer: $(2, 0)$ and $(3, 2)$
Examiner tip: After finding the $x$-values, substitute into the LINEAR equation to get the $y$-values – it's quicker and safer.

6Grade 8–9HigherNon-calculator5 marks

Solve algebraically the simultaneous equations

$$x^2 + y^2 = 13$$ $$y = x + 1$$ [5]

for substituting $y = x + 1$ into $x^2 + y^2 = 13$: $x^2 + (x + 1)^2 = 13$M1
for $2x^2 + 2x - 12 = 0$ oe, e.g. $x^2 + x - 6 = 0$A1
for solving their 3-term quadratic, e.g. $(x + 3)(x - 2) = 0$M1
$x = 2$ and $x = -3$A1
$x = 2, y = 3$ and $x = -3, y = -2$, correctly pairedA1
Answer: $x = 2, y = 3$ and $x = -3, y = -2$
Examiner tip: $(x + 1)^2 = x^2 + 2x + 1$, not $x^2 + 1$ – expand it with a grid or FOIL.

7Grade 8–9HigherCalculator6 marks

The line $2x + y = 7$ intersects the circle $x^2 + y^2 = 25$ at the points $P$ and $Q$.

Work out the length of $PQ$. Give your answer correct to 3 significant figures. [6]

for rearranging and substituting, e.g. $x^2 + (7 - 2x)^2 = 25$M1
for $5x^2 - 28x + 24 = 0$A1
for correct substitution into the quadratic formula: $\dfrac{28 \pm \sqrt{(-28)^2 - 4 \times 5 \times 24}}{2 \times 5}$M1
$x = 4.543\ldots$ and $x = 1.056\ldots$A1
for finding the $y$-values ($-2.087\ldots$, $4.887\ldots$) and using Pythagoras, e.g. $\sqrt{(4.543 - 1.056)^2 + (-2.087 - 4.887)^2}$M1
$7.80$ (accept $7.79$–$7.80$)A1
Answer: $PQ = 7.80$ units
Examiner tip: Keep full calculator values until the end – rounding the coordinates to 1 dp first can push your length out of the accepted range.

8Grade 1–3FoundationNon-calculator2 marks

Priya thinks of two numbers.

The two numbers add up to 23.

The difference between the two numbers is 7.

Work out the two numbers. [2]

for a correct method, e.g. $(23 + 7) \div 2$ or $(23 - 7) \div 2$, or $x + y = 23$ and $x - y = 7$ added or subtracted to give $2x = 30$ or $2y = 16$; or a pair of numbers with sum 23 or difference 7 found by systematic trial (e.g. 14 and 9 tested)M1
15 and 8 (both required, in either order)A1
Answer: 15 and 8
Examiner tip: Check BOTH conditions: 15 + 8 = 23 and 15 − 8 = 7 – a pair that only adds to 23 (e.g. 16 and 7) scores at most the method mark.

9Grade 4–5Non-calculator3 marks

Ben is solving the simultaneous equations

$$3x + 2y = 19$$ $$x - 2y = 1$$

Here is his working.

(1) − (2):   2x = 18
              x = 9
9 − 2y = 1   so   y = 4

(a) Explain what Ben has done wrong. [1]

(b) Solve the simultaneous equations correctly. [2]

(a) explanation, e.g. he should have added the equations because the $y$ terms have opposite signs, or $2y - (-2y) = 4y$ so subtracting does not eliminate $y$, or $3x - x$ is $2x$ but $2y - (-2y)$ is $4y$ not $0$. Do not accept only showing that $x = 9, y = 4$ does not satisfy $3x + 2y = 19$ (e.g. $27 + 8 = 35$) – that shows his answer is wrong but not what he did wrongC1
(b) for a correct method to eliminate one variable, e.g. adding to get $4x = 20$, or substitution $x = 1 + 2y$ into the first equation giving $3(1 + 2y) + 2y = 19$M1
(b) $x = 5$ and $y = 2$A1
Answer: (a) He subtracted; he should have added (subtracting $-2y$ gives $4y$, not $0$). (b) $x = 5$, $y = 2$
Examiner tip: Same signs subtract, different signs add – and always check your pair in the equation you did NOT use to find the second value.

10Grade 4–5Calculator4 marks

A taxi company charges a fixed amount plus a cost for each mile travelled.

A 4-mile journey costs £9.60.

A 7-mile journey costs £14.40.

Lena has £25.

Work out the greatest whole number of miles Lena can travel in one journey with this taxi company. [4]

for setting up the two equations, e.g. $c + 4m = 9.60$ and $c + 7m = 14.40$, or for $14.40 - 9.60 = 4.80$ for 3 milesP1
for finding the cost per mile, $m = 1.60$ (£1.60 or 160p), or the fixed charge $c = 3.20$P1
for a complete process to find the number of miles, e.g. $(25 - 3.20) \div 1.60\ (= 13.625)$, or showing 13 miles costs £24 and 14 miles costs £25.60P1
13 (miles) cao – do not accept 13.6 or 14A1
Answer: 13 miles
Examiner tip: The answer must be rounded DOWN – 13.625 miles means 14 miles would cost more than £25 (£25.60).

11Grade 6–7HigherNon-calculator4 marks

The graph of $y = x^2 - 3x + 1$ is drawn on the grid for values of $x$ from $-1$ to $4$.

(a) By drawing a suitable straight line on the grid, find estimates for the solutions of the simultaneous equations

$$y = x^2 - 3x + 1 \qquad y = x - 1$$ [3]

(b) Explain why your answers to part (a) are estimates. [1]

(a) the line $y = x - 1$ drawn correctly across at least $0 \le x \le 4$ (e.g. through $(0, -1)$ and $(4, 3)$); may be implied by correct answersB1
(a) for reading off the $x$-coordinates of both points of intersection of the curve with their straight line (ft their line if it is a straight line)M1
(a) $x = 0.6, y = -0.4$ and $x = 3.4, y = 2.4$, correctly paired (accept $x$ in the range 0.5 to 0.7 with $y$ in −0.5 to −0.3; $x$ in 3.3 to 3.5 with $y$ in 2.3 to 2.5)A1
(b) explanation, e.g. the values are read from a graph and the intersections are not on exact grid points / the exact solutions are not whole numbers (they are $2 \pm \sqrt{2}$), so readings can only be approximateC1
Answer: (a) $x \approx 0.6, y \approx -0.4$ and $x \approx 3.4, y \approx 2.4$ (b) read from a graph; intersections are not at exact grid values
Examiner tip: Give BOTH coordinates for BOTH intersection points and keep each $y$ with its own $x$ – an $x$ value on its own only answers half the question.

12Grade 6–7HigherCalculator5 marks

Solve the simultaneous equations

$$y = 2x^2 - 3$$ $$x + y = 4$$

Give your answers correct to 2 decimal places. [5]

for substituting to eliminate one variable, e.g. $x + 2x^2 - 3 = 4$ or $4 - x = 2x^2 - 3$M1
for $2x^2 + x - 7 = 0$ oe (a correct 3-term quadratic $= 0$)A1
for correct substitution into the quadratic formula, e.g. $x = \dfrac{-1 \pm \sqrt{1^2 - 4(2)(-7)}}{2(2)}$ ft their 3-term quadratic (or completing the square to $\left(x + \tfrac{1}{4}\right)^2 = \tfrac{57}{16}$)M1
$x = 1.64$ and $x = -2.14$ (allow $x = 1.637\ldots$ and $-2.137\ldots$)A1
$y = 2.36$ and $y = 6.14$, correctly paired: $x = 1.64, y = 2.36$ and $x = -2.14, y = 6.14$A1
Answer: $x = 1.64, y = 2.36$ and $x = -2.14, y = 6.14$
Examiner tip: The quadratic doesn't factorise – if you can't find factors, go straight to the formula, and find $y$ from the LINEAR equation using the unrounded $x$.

13Grade 8–9HigherNon-calculator5 marks

A rectangle has a perimeter of 46 cm.

The length of a diagonal of the rectangle is 17 cm.

Work out the area of the rectangle. [5]

You must show your working.

for using the perimeter to form a linear equation, e.g. $2x + 2y = 46$ or $x + y = 23$P1
for using Pythagoras to form $x^2 + y^2 = 17^2\ (= 289)$P1
for substituting to form a quadratic in one variable, e.g. $x^2 + (23 - x)^2 = 289$ leading to $x^2 - 23x + 120 = 0$ oe; OR for using $(x + y)^2 = x^2 + 2xy + y^2$: $529 = 289 + 2xy$P1
for solving to find both side lengths, 8 and 15, e.g. $(x - 8)(x - 15) = 0$; OR for $2xy = 240$P1
120 (cm²)A1
Answer: 120 cm²
Examiner tip: Two unknown sides need two equations – the diagonal gives you Pythagoras, so this is a linear/quadratic pair, not a guess-and-check.

14Grade 8–9HigherNon-calculator4 marks

The line $L$ has equation $y = 2x - 2$.

The curve $C$ has equation $y = x^2 - 4x + 7$.

Show that $L$ meets $C$ at exactly one point, and find the coordinates of this point. [4]

for equating to eliminate $y$, e.g. $x^2 - 4x + 7 = 2x - 2$M1
for $x^2 - 6x + 9 = 0$A1
for $(x - 3)^2 = 0$ (or $(x - 3)(x - 3) = 0$) with a statement that there is only one (repeated) solution, $x = 3$, so the line meets the curve at one point onlyC1
$(3, 4)$A1
Answer: $x^2 - 6x + 9 = (x - 3)^2 = 0$ has the single repeated root $x = 3$, so they meet only at $(3, 4)$
Examiner tip: A ‘show that’ needs a conclusion: say that $(x - 3)^2 = 0$ gives only ONE value of $x$, so there is only one point – finding $(3, 4)$ alone is not enough.

15Grade 4–5Non-calculator4 marks

Solve the simultaneous equations

$$3y + 7 = -4x$$ $$6x = y + 6$$ [4]

for rearranging both equations into a consistent form, e.g. $4x + 3y = -7$ and $6x - y = 6$ (or making $y$ the subject of the second: $y = 6x - 6$)M1
for a correct method to eliminate one variable (condone one arithmetic error), e.g. multiplying the second by 3 and adding: $22x = 11$, or substituting $y = 6x - 6$: $4x + 3(6x - 6) = -7$M1
$x = \frac{1}{2}$ oe (accept 0.5)A1
$y = -3$A1
Answer: $x = \frac{1}{2}$, $y = -3$
Examiner tip: Line the equations up as $ax + by = c$ BEFORE eliminating – and don't panic when $x$ comes out as a fraction; $22x = 11$ gives $x = \frac{1}{2}$.

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