What is A17: Linear equations?

Keep an equation balanced by applying the same operation to both sides until the unknown is isolated. Expand brackets first, then collect unknown terms on one side and constants on the other.

How many marks are A17 questions worth?

In verified past-paper appearances, A17 questions on Edexcel GCSE Maths 1MA1 have been worth 1–7 marks.

Which past exam papers has A17 appeared in?

A17 has appeared in 24 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

A17

Solve linear equations in one unknown algebraically (including those with the unknown on both sides of the equation); find approximate solutions using a graph

Linear equations

Worked answers, methods and verified real exam appearances for A17 on Edexcel GCSE Maths 1MA1.

Explanation

Keep an equation balanced by applying the same operation to both sides until the unknown is isolated. Expand brackets first, then collect unknown terms on one side and constants on the other.
With fractions, multiply every term by a common multiple of the denominators before simplifying.
A graphical solution is the $x$ -coordinate where graphs representing the two sides intersect, so its accuracy is limited by the graph scale.
The examiner awards method marks for valid algebraic steps; unexplained sign changes are not valid operations.
Check the solution by substituting it into the original equation and confirming both sides are equal.

Worked example

Solve $\dfrac{3(x-2)}{4}-\dfrac{x+1}{3}=\dfrac56$ .

1.Multiply every term by $12$ : $9(x-2)-4(x+1)=10$ .
2.Expand and collect terms: $9x-18-4x-4=10$ , so $5x=32$ .
3.Divide by $5$ : $x=\dfrac{32}{5}$ .

Answer: $x=\dfrac{32}{5}$ .

Common mistakes

Don't multiply only the fractional terms, not every term, when clearing denominators.
Don't change the sign of a term when moving it without performing the same addition or subtraction on both sides.

Exam tip

For a multi-mark 'solve' question, keep each balancing step visible so an arithmetic slip does not lose the method marks.

Worked practice

Tier 1 · Easy

Solve

7x+5=33

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$x=4$	2	Subtract $5$ from both sides to get $7x=28$ . Divide both sides by $7$ , giving $x=4$ .

Tier 2 · Standard

Draw

y=4.5x-2

and

y=9.2-x

on the same axes. Use the intersection to solve

4.5x-2=9.2-x

, giving an estimate to one decimal place.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$x\approx2.0$	4	The graphical solution is the $x$ -coordinate of the intersection. The lines meet at about $(2.04,7.16)$ , so the requested estimate is $x\approx2.0$ to one decimal place.

Tier 3 · Hard

Solve

\frac{3x+1}{5}+\frac{13-2x}{7}=\frac{5x+9}{8}

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$x=3$	4	Multiply every term by $280$ : $56(3x+1)+40(13-2x)=35(5x+9)$ . Expanding gives $168x+56+520-80x=175x+315$ , so $261=87x$ and $x=3$ .

Tier 1 · Easy

Solve

8x-5=3x+30

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$x=7$	2	Subtract $3x$ from both sides to get $5x-5=30$ . Add $5$ to both sides, so $5x=35$ and $x=7$ .

Tier 2 · Standard

Solve

0.6x-1.7=0.2x+4.3

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$x=15$	3	Subtract $0.2x$ to obtain $0.4x-1.7=4.3$ . Add $1.7$ , giving $0.4x=6$ . Dividing by $0.4$ gives $x=15$ .

Tier 3 · Hard

Solve

5-\dfrac{2x-3}{4}=\dfrac{x+7}{3}

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$x=\dfrac{41}{10}$	4	Multiply every term by $12$ to get $60-3(2x-3)=4(x+7)$ . Expanding gives $69-6x=4x+28$ , so $41=10x$ and $x=\dfrac{41}{10}$ .

Tier 2 · Standard

Solve

4(2x-3)+5=3(x+6)

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$x=5$	3	Expand both sides to get $8x-12+5=3x+18$ . Therefore $8x-7=3x+18$ , so $5x=25$ and $x=5$ .

Tier 3 · Hard

Solve

\dfrac{5x-4}{7}=\dfrac{3x+8}{11}

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$x=\dfrac{50}{17}$	4	Cross-multiply to get $11(5x-4)=7(3x+8)$ . Expanding gives $55x-44=21x+56$ , so $34x=100$ and $x=\dfrac{50}{17}$ . Substitution gives $26/17$ on each side.

Tier 3 · Hard

The value

x=-4

is the solution of

5(kx-3)+7=2(kx+8)-12

. Work out

k

and show the algebra leading to your answer.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$k=-1$	4	Substitute $x=-4$ : $5(-4k-3)+7=2(-4k+8)-12$ . This simplifies to $-20k-8=-8k+4$ , so $-12k=12$ and $k=-1$ . Checking gives $12$ on both sides of the original equation.

Q10

Tier 3 · Hard

A student says that

x=5

solves

4(3x-2)-5=7x+22

. Show by substitution that this is incorrect. Solve the equation to find the correct value of

x

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	At $x=5$ , the two sides are $47$ and $57$ , so $x=5$ is incorrect $x=7$	4	When $x=5$ , the left side is $4(15-2)-5=47$ and the right side is $35+22=57$ , so the claim fails. Expanding the equation gives $12x-13=7x+22$ . Hence $5x=35$ and $x=7$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2023-11	1H	Q7	5	Non-calculator	Higher	QP MS
2022-11	1F	Q16	4	Non-calculator	Foundation	QP MS
2022-06	3F	Q17	4	Allowed	Foundation	QP MS
2024-11	3F	Q6	3	Allowed	Foundation	QP MS
2022-11	1F	Q8	3	Non-calculator	Foundation	QP MS
2023-11	1F	Q24	5	Non-calculator	Foundation	QP MS
2021-11	3F	Q17	7	Allowed	Foundation	QP MS
2023-11	2H	Q4	3	Allowed	Higher	QP MS
2023-11	3F	Q16	2	Allowed	Foundation	QP MS
2023-06	3F	Q18	3	Allowed	Foundation	QP MS
2021-11	1F	Q16	4	Non-calculator	Foundation	QP MS
2022-11	2F	Q14	7	Allowed	Foundation	QP MS
2019-11	1F	Q17	3	Non-calculator	Foundation	QP MS
2023-06	1F	Q5	1	Non-calculator	Foundation	QP MS
2021-11	1F	Q15	4	Non-calculator	Foundation	QP MS
2024-06	2H	Q10	3	Allowed	Higher	QP MS
2024-06	1F	Q17	3	Non-calculator	Foundation	QP MS
2023-06	3F	Q11	3	Allowed	Foundation	QP MS
2019-06	1F	Q10	4	Non-calculator	Foundation	QP MS
2019-06	3F	Q15	4	Allowed	Foundation	QP MS
2019-11	3H	Q10	4	Allowed	Higher	QP MS
2024-11	1F	Q15	4	Non-calculator	Foundation	QP MS
2019-11	3F	Q19	2	Allowed	Foundation	QP MS
2023-11	2F	Q25	3	Allowed	Foundation	QP MS

Other points in A Algebra · graphs

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