Linear equations (incl. unknowns both sides, fractions) – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator4 marks

(a) Solve $4x - 7 = 21$ [2]

(b) Solve $\dfrac{y}{3} + 5 = 9$ [2]

(a) for $4x = 28$ or $(21 + 7) \div 4$M1
$x = 7$A1
(b) for $\frac{y}{3} = 4$ or $y + 15 = 27$M1
$y = 12$A1
Answer: (a) $x = 7$ (b) $y = 12$
Examiner tip: In (b) take away the 5 first, THEN multiply by 3 – undoing in the wrong order ($9 \times 3 - 5$) gives the wrong answer $y = 22$.

2Grade 4–5Non-calculator3 marks

Solve $7x - 4 = 3x + 18$ [3]

for a correct first step to collect the $x$ terms on one side or the number terms on the other, e.g. $4x - 4 = 18$ or $7x = 3x + 22$ or $-4 = -4x + 18$M1
for $4x = 22$ (or $-4x = -22$)M1
$x = 5.5$ oe, e.g. $\frac{11}{2}$, $5\frac{1}{2}$ or $\frac{22}{4}$A1
Answer: $x = 5.5$
Examiner tip: The answer does not have to be a whole number – give $x = 5.5$ (or $\frac{11}{2}$) exactly; do not round it to 6.

3Grade 4–5Non-calculator3 marks

Solve $5(2x - 3) = 3(x + 4)$ [3]

for expanding at least one bracket correctly, e.g. $10x - 15$ or $3x + 12$M1
for collecting terms correctly from a correct equation, to reach $ax = b$, e.g. $7x = 27$M1
$x = \frac{27}{7}$ oe, e.g. $3\frac{6}{7}$ (accept 3.857… or 3.86)A1
Answer: $x = \dfrac{27}{7}$ ($= 3\frac{6}{7}$)
Examiner tip: On a non-calculator paper leave the answer as an exact fraction – a rounded decimal like 3.9 loses the A mark.

4Grade 6–7HigherNon-calculator4 marks

Solve $\dfrac{2x + 1}{3} - \dfrac{x - 2}{4} = 2$ [4]

for multiplying every term by 12 (or a common multiple of 3 and 4), e.g. $4(2x + 1) - 3(x - 2) = 24$M1
for expanding correctly, including the sign: $8x + 4 - 3x + 6 = 24$M1
for collecting terms to reach $5x = 14$M1
$x = 2.8$ oe, e.g. $\frac{14}{5}$A1
Answer: $x = 2.8$
Examiner tip: The minus sign belongs to the whole of $3(x - 2)$, so it becomes $-3x + 6$ – writing $-6$ is the most common lost mark.

5Grade 4–5Non-calculator3 marks

The diagram shows a triangle. The angles are given in degrees.

Work out the size of the largest angle in the triangle. You must show your working. [3]

for forming an equation using the angle sum of a triangle: $(2x + 10) + (3x - 5) + (x + 25) = 180$ or $6x + 30 = 180$P1
for solving to find $x = 25$P1
$70^\circ$A1
Answer: $70^\circ$
Examiner tip: Finding $x = 25$ is not the answer – substitute back into all three expressions and give the largest angle.

6Grade 6–7HigherNon-calculator4 marks

A rectangle has length $(3x - 2)$ cm and width $(x + 4)$ cm.
A square has sides of length $(x + 5)$ cm.

The perimeter of the rectangle is equal to the perimeter of the square.

Work out the area of the square. [4]

for a correct expression for the perimeter of the rectangle, e.g. $2(3x - 2) + 2(x + 4)$ or $8x + 4$P1
for forming a correct equation, e.g. $8x + 4 = 4(x + 5)$P1
for solving to find $x = 4$ (so side of square $= 9$ cm)P1
$81$ cm²A1
Answer: $81\text{ cm}^2$
Examiner tip: Use the perimeter for the equation but the AREA for the answer – stopping at $x = 4$ or giving the perimeter (36 cm) loses the final mark.

7Grade 6–7HigherNon-calculator4 marks

A bag contains only red counters and blue counters.
The ratio of red counters to blue counters is $3 : 5$

Asha puts 12 more red counters into the bag. She does not add or remove any blue counters.
Now the ratio of red counters to blue counters is $3 : 4$

Work out how many blue counters are in the bag. [4]

for representing the counters algebraically, e.g. red $= 3k$, blue $= 5k$, or for writing both ratios with the same number of blue parts, e.g. $12 : 20$ and $15 : 20$P1
for forming a correct equation, e.g. $\frac{3k + 12}{5k} = \frac{3}{4}$ or $4(3k + 12) = 15k$, or for $3$ parts $= 12$ (from $15 - 12 = 3$ parts)P1
for $k = 16$ (or 1 part $= 4$), or red $= 48$ before / $60$ afterP1
$80$A1
Answer: $80$ blue counters
Examiner tip: The number of blue counters does not change – use the same blue amount ($5k$) in both ratios, then cross-multiply: $4(3k + 12) = 3 \times 5k$.

8Grade 1–3FoundationNon-calculator4 marks

(a) Solve $15 - 2x = 9$ [2]

(b) Solve $2(y + 7) = 8$ [2]

(a) for $-2x = -6$ or $2x = 6$ or $15 - 9 = 2x$ oeM1
$x = 3$A1
(b) for $y + 7 = 4$ or $2y + 14 = 8$ or $2y = -6$M1
$y = -3$A1
Answer: (a) $x = 3$ (b) $y = -3$
Examiner tip: In (a) the $x$ term is being subtracted, so $x = -3$ is a common sign slip; in (b) a negative answer is fine, so do not 'correct' it to 3.

9Grade 4–5Non-calculator3 marks

Priya thinks of a number.

She multiplies her number by 4 and then subtracts 9.
Her answer is the same as when she adds 15 to her number.

By forming and solving an equation, find Priya's number. [3]

for forming a correct equation, e.g. $4n - 9 = n + 15$ (any letter)M1
for a correct collection of terms, e.g. $3n = 24$ or $4n - n = 15 + 9$M1
$8$A1
Answer: $8$
Examiner tip: Write the equation before doing anything else; trial and improvement giving 8 with no equation may lose the method marks when the question says 'by forming and solving an equation'.

10Grade 4–5Non-calculator3 marks

Ben is solving the equation $5x + 3 = 2x - 9$

Here is his working.

$$\begin{aligned} 5x - 2x &= -9 + 3 \\ 3x &= -6 \\ x &= -2 \end{aligned}$$

(a) Explain the mistake Ben has made. [1]

(b) Solve $5x + 3 = 2x - 9$ correctly. [2]

(a) for a correct explanation, e.g. 'he added 3 to the right-hand side instead of subtracting 3' or 'it should be $-9 - 3$'; accept 'he did not change the sign of the 3 when moving it'. Ignore incorrect working in (b) for this markC1
(b) for $3x = -12$ or $5x - 2x = -9 - 3$ oeM1
$x = -4$A1
Answer: (a) He should have subtracted 3 from both sides (giving $-9 - 3$), not added 3. (b) $x = -4$
Examiner tip: In 'explain' questions say exactly which step is wrong and what it should be; 'he got the wrong answer' scores nothing.

11Grade 4–5Non-calculator4 marks

The graphs of $y = 2x + 1$ and $y = 6 - x$ are drawn on the grid.

(a) Use the graph to solve $2x + 1 = 4$ [1]

(b) Use the graph to find an estimate for the solution of $2x + 1 = 6 - x$ [1]

(a) $x = 1.5$ (accept 1.4 to 1.6 from the graph)B1
(b) any value from 1.6 to 1.8 inclusive (the $x$-coordinate of the point of intersection); ignore a $y$-value given as well, but do not accept a $y$-value aloneB1
(c) for $3x = 5$ or $2x + x = 6 - 1$ oeM1
$x = \frac{5}{3}$ (accept $1\frac{2}{3}$)A1
Answer: (a) $x = 1.5$ (b) $x \approx 1.7$ (c) $x = \frac{5}{3}$
Examiner tip: The solution of an equation is the $x$-coordinate where the graphs meet; students often give the $y$-coordinate (about 4.3) instead.

12Grade 6–7HigherNon-calculator3 marks

Solve $\dfrac{4 - 3x}{5} = \dfrac{x + 2}{3}$ [3]

for removing both fractions correctly, e.g. $3(4 - 3x) = 5(x + 2)$ or $\frac{15(4 - 3x)}{5} = \frac{15(x + 2)}{3}$M1
for expanding and collecting terms correctly to reach $ax = b$, e.g. $14x = 2$ or $-14x = -2$ (from $12 - 9x = 5x + 10$)M1
$x = \frac{1}{7}$ oe, e.g. $\frac{2}{14}$ (accept 0.142… or 0.14 only if a correct fraction is seen first)A1
Answer: $x = \dfrac{1}{7}$
Examiner tip: Multiply the whole numerator by the other denominator, using brackets: writing $3 \times 4 - 3x$ instead of $3(4 - 3x)$ loses the first method mark and the marks that follow.

13Grade 4–5Calculator5 marks

Two taxi companies charge for a journey of $d$ miles as follows.

Speedy Cabs: £3.20 plus £1.85 per mile
City Rides: £5.60 plus £1.45 per mile

(a) For one particular journey both companies charge the same amount.
Form and solve an equation to find the length of this journey. [3]

(b) Lena makes a journey of 9 miles.
Which company is cheaper for Lena, and by how much? [2]

(a) for forming a correct equation, e.g. $3.20 + 1.85d = 5.60 + 1.45d$ (allow in pence: $320 + 185d = 560 + 145d$)M1
for $0.4d = 2.4$ oe, e.g. $40d = 240$M1
$6$ (miles)A1
(b) for working out the cost of 9 miles for at least one company: £19.85 (Speedy) or £18.65 (City), or for $0.40 \times (9 - 6)$M1
City Rides, cheaper by £1.20 (both company and amount needed)A1
Answer: (a) 6 miles (b) City Rides, by £1.20
Examiner tip: In (b) a correct method with no clear statement of which company is cheaper loses the final mark – always answer the question in words.

14Grade 6–7HigherNon-calculator4 marks

(a) Show that the equation $3(2x - 4) = 6x + 5$ has no solution. [2]

(b) $k$ is a constant.
The equation $4(x + k) - x = 3x + 10$ is true for all values of $x$.
Find the value of $k$. [2]

(a) for expanding to $6x - 12 = 6x + 5$M1
for a correct conclusion, e.g. 'the $x$ terms cancel to leave $-12 = 5$, which is false, so there is no solution' or '$6x - 12$ is always 17 less than $6x + 5$'C1
(b) for simplifying the left side to $3x + 4k$ (or for $4k = 10$)M1
$k = 2.5$ oe, e.g. $\frac{5}{2}$A1
Answer: (a) $6x - 12 = 6x + 5 \Rightarrow -12 = 5$, which is impossible, so no solution. (b) $k = \frac{5}{2}$
Examiner tip: Reaching $0x = 17$ is not enough on its own for (a) – you must state that this is impossible for any $x$.

15Grade 8–9HigherCalculator5 marks

Sam drives $x$ km at an average speed of 60 km/h.
He then drives a further $(x + 30)$ km at an average speed of 80 km/h.

His average speed for the whole journey is 70 km/h.

Work out the total distance Sam drives. You must show your working. [5]

for an expression for the time of either part, e.g. $\frac{x}{60}$ or $\frac{x + 30}{80}$ (hours)P1
for forming a correct equation for total time, e.g. $\frac{x}{60} + \frac{x + 30}{80} = \frac{2x + 30}{70}$P1
for clearing the fractions correctly, e.g. $28x + 21(x + 30) = 24(2x + 30)$ (multiplying by 1680) or $49x + 630 = 48x + 720$P1
for $x = 90$P1
$210$ (km)A1
Answer: $210$ km
Examiner tip: Average speed is total distance ÷ total time – it is not $\frac{60 + 80}{2}$, and $x = 90$ is not the final answer.

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