What is A17: Linear equations?

Do the same thing to both sides until the unknown stands alone. With a fraction, multiply every term on both sides by the denominator, writing the line 5-x=2(2x-7) before expanding; the method mark needs the whole of each side multiplied, not just one term.

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.

Definition

Specification 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.

Explanation

Do the same thing to both sides until the unknown stands alone. With a fraction, multiply every term on both sides by the denominator, writing the line $5-x=2(2x-7)$ before expanding; the method mark needs the whole of each side multiplied, not just one term.
Expand brackets, then collect the $x$ terms on one side and the numbers on the other in a single visible step such as $4x+x=14+5$ .
Divide last, and leave the answer as an exact fraction or decimal; $\dfrac{19}{5}$ and $3.8$ both score, but $\dfrac{-19}{-5}$ does not.
Check by substituting into the original equation.
A graphical solution is the $x$ -coordinate where the two graphs meet, read to the accuracy of the grid.

Worked example

Solve

\dfrac{3(x-2)}{4}-\dfrac{x+1}{3}=\dfrac{5}{6}

(3)

(Total for Question 1 is 3 marks)

Working

1.Multiply every term by $12$ , the lowest common multiple of $4$ , $3$ and $6$ : $9(x-2)-4(x+1)=10$ .
2.Expand the brackets: $9x-18-4x-4=10$ .
3.Collect the $x$ terms and the numbers: $5x=10+18+4=32$ .
4.Divide by $5$ : $x=\dfrac{32}{5}$ .

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

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for a correct first step, eg $9(x-2)-4(x+1)=10$ or $9x-18-4x-4=10$	M1	Method must show every term multiplied by 12 (or a common multiple)
	(dep) for isolating terms in $x$ , eg $9x-4x=10+18+4$ or $5x=32$	M1
	for $\dfrac{32}{5}$	A1	oe; accept 6.4 or $6\dfrac{2}{5}$

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.

Mark-scheme wording to use

“for a correct first step, eg 5 – x = 2(2x − 7) or 5 – x = 4x − 14 or 5/2 – x/2 = 2x – 7” — Pearson 1MA1, June 2018 Paper 3H, Q7
“(dep) for isolating terms in x … Accept 19/5 , 3 4/5 oe but not −19/−5 oe” — Pearson 1MA1, June 2018 Paper 3H, Q7
“for correct expansion of the bracket or dividing all terms by 3 as a first step … for isolating terms in x on one side of an equation eg 5x – 6 – 3x = −3” — Pearson 1MA1, November 2017 Paper 2H, Q1

What the examiner wants

Every term treated the same. The June 2018 Paper 3H scheme states that the method must show the left side and both terms on the right multiplied by $2$ ; the November 2024 Paper 3H report says many knew to multiply both sides by the denominator but too often did not carry it out correctly, and then lost the marks for isolating the terms.

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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