What is A21: Forming equations?

Start by defining the unknown in words, with units: let x cm be the width. Write every other quantity as an expression in x from the information given, such as 2x, 2x-6 or x+3.

How many marks are A21 questions worth?

In verified past-paper appearances, A21 questions on Edexcel GCSE Maths 1MA1 have been worth 1–7 marks.

Which past exam papers has A21 appeared in?

A21 has appeared in 51 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

A21

Translate simple situations or procedures into algebraic expressions or formulae; derive an equation (or two simultaneous equations), solve the equation(s) and interpret the solution

Forming equations

Worked answers, methods and verified real exam appearances for A21 on Edexcel GCSE Maths 1MA1.

Definition

Specification A21: translate simple situations or procedures into algebraic expressions or formulae; derive an equation (or two simultaneous equations), solve the equation(s) and interpret the solution.

Explanation

Start by defining the unknown in words, with units: let $x$ cm be the width. Write every other quantity as an expression in $x$ from the information given, such as $2x$ , $2x-6$ or $x+3$ .
The equation comes from the one fact that links them:
a perimeter, an angle sum of $90^\circ$ , $180^\circ$ or $360^\circ$ , a total number of counters, or two expressions for equal sides. Solve it by the usual balancing steps, then answer the question that was asked, which is rarely the value of $x$ itself: substitute back to find the angle, the side length, the area or the ratio.
Schemes give a process mark for the expressions, one for the equation, one for solving and one for using the solution, so each stage must be visible.
A root that gives a negative length or a non-integer count is rejected with a reason. Label the final answer with what it represents.

Worked example

A rectangle has width

x

cm. Its length is

3

cm more than its width. The perimeter of the rectangle is

34

cm. Work out the length of the rectangle.

(3)

(Total for Question 1 is 3 marks)

Working

1.Write the length as an expression in $x$ : $(x+3)$ cm.
2.Form the perimeter equation: $2x+2(x+3)=34$ .
3.Expand and collect: $4x+6=34$ , so $4x=28$ and $x=7$ .
4.Answer the question asked: the length is $x+3=10$ cm.

Answer: $10$ cm

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for forming a correct equation, eg $2x+2(x+3)=34$ or $x+x+3+x+x+3=34$	P1	oe
	(dep) for a process to solve the equation, eg $4x=34-6$ or $x=7$	P1
	for $10$	A1	cao

Common mistakes

Don't form $x+(x+3)=34$ and count only half of the rectangle's perimeter.
Don't stop at $x=7$ without finding and labelling both requested dimensions.
Don't keep a negative algebraic root even though the quantity is a length or age.

Mark-scheme wording to use

“P1 for process to develop 3 algebraic expressions, eg. (R =) n, (S =) 2n, (T =) 2n – 6, oe, at least two must be correct. … P1 for process to sum 3 algebraic expressions and equating to 54, eg. n + “2n” + “2n – 6” = 54 … P1 for start of process to solve the correct linear equation, eg. 5n = 54 + 6 (n = 12) or for 12, 24, 18 P1 for “12” : 2 × “12” – 6 oe eg 12 : 18 oe or 18 : 12 linked to T, R A1 for 1.5 or 3/2 or 1 1/2” — Pearson 1MA1, June 2022 Paper 3H, Q3
“P1 process to start problem solving eg forms an appropriate equation P1 complete process to solve their equation A1 cao” — Pearson 1MA1, Specimen 1 Paper 1H, Q1
“for process to use the perimeter of ABCD, eg 2x + 2× “4.5” = 26” — Pearson 1MA1, June 2019 Paper 1H, Q7

What the examiner wants

An equation before anything else. The November 2024 Paper 2H report says the first mark was for forming an appropriate equation from the diagram, either summing all the sides to $72$ or equating the two equal sides, the second for the process to isolate the terms in $x$ , and that most who reached that point solved correctly.

Exam tip

Define the variable before forming the equation and finish with a sentence interpreting every required quantity.

Worked practice

Tier 1 · Easy

A rectangle has width

x

cm and length

(x+3)

cm. Its perimeter is

34

cm. Form and solve an equation to find both dimensions.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	Width $=7$ cm Length $=10$ cm	3	The perimeter equation is $2x+2(x+3)=34$ . Hence $4x+6=34$ , so $4x=28$ and $x=7$ . The width is $7$ cm and the length is $7+3=10$ cm.

Tier 2 · Standard

A club sells

38

tickets. Adult tickets cost £7 and junior tickets cost £4. The total received is £203. Form two equations and find how many tickets of each type were sold.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$17$ adult tickets and $21$ junior tickets	4	Let $a$ and $j$ be the numbers of adult and junior tickets. Then $a+j=38$ and $7a+4j=203$ . Subtract $4(a+j)=152$ from the money equation to get $3a=51$ , so $a=17$ and $j=21$ .

Tier 3 · Hard

Mira is

4

years older than Theo. In

3

years, the product of their ages will be

192

. Form an equation and find their current ages.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Theo is $9$ years old and Mira is $13$ years old.	5	Let Theo's current age be $x$ , so Mira's is $x+4$ . In three years their ages are $x+3$ and $x+7$ , giving $(x+3)(x+7)=192$ . Thus $x^2+10x-171=0=(x-9)(x+19)$ . The solutions are $x=9$ and $x=-19$ ; reject the negative age. Theo is $9$ and Mira is $13$ .

Tier 1 · Easy

A coach has

r

rows with

s

seats in each row and

6

extra seats at the back. Write down an expression, in terms of

r

and

s

, for the total number of seats on the coach.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$rs+6$	1	The $r$ rows contain $rs$ seats altogether. Including the $6$ extra seats at the back gives $rs+6$ .

Tier 2 · Standard

52

cm ribbon is cut into pieces of lengths

x

cm,

(x+4)

cm and

2x

cm. Form and solve an equation to find the length of each piece.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$12$ cm, $16$ cm and $24$ cm	3	Let the first piece have length $x$ cm. Then the pieces have lengths $x$ , $x+4$ and $2x$ , so $x+(x+4)+2x=52$ . Hence $4x+4=52$ , giving $x=12$ . The three lengths are $12$ cm, $16$ cm and $24$ cm.

Tier 3 · Hard

A right-angled triangle has perpendicular sides of lengths

x

cm and

(x+1)

cm. Its hypotenuse has length

29

cm. Form and solve an equation to find the lengths of the two perpendicular sides.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$20$ cm and $21$ cm	4	Pythagoras' theorem gives $x^2+(x+1)^2=29^2$ . Expanding and simplifying gives $x^2+x-420=0=(x+21)(x-20)$ . Hence $x=-21$ or $x=20$ . A length cannot be negative, so the perpendicular sides are $20$ cm and $21$ cm.

Tier 2 · Standard

A repair company charges a fixed call-out fee of £18 and then £7 for each half-hour of work. A bill is £67. Form and solve an equation to work out the time spent on the repair.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$3.5$ hours	3	Let $h$ be the number of half-hours. The equation is $18+7h=67$ , so $7h=49$ and $h=7$ . Seven half-hours is $3.5$ hours.

Tier 3 · Hard

A rectangular play area has length

(4x-3)

metres and width

(x+2)

metres. Its perimeter is

68

metres. Form and solve an equation to work out the length and width of the play area.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	The length is $25$ m and the width is $9$ m.	4	The perimeter equation is $2(4x-3)+2(x+2)=68$ . Expanding and collecting gives $8x-6+2x+4=68$ , so $10x-2=68$ and $x=7$ . Therefore the length is $4(7)-3=25$ m and the width is $7+2=9$ m. The check $2(25)+2(9)=68$ confirms the result.

Tier 3 · Hard

A two-digit number has tens digit

x

and units digit

(x+3)

. The number is

6

more than four times the sum of its digits. Form and solve an equation to find the number.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$58$	4	The number is $10x+(x+3)=11x+3$ , and the sum of its digits is $2x+3$ . Therefore $11x+3=4(2x+3)+6$ . This gives $11x+3=8x+18$ , so $3x=15$ and $x=5$ . The units digit is $8$ , so the number is $58$ .

Q10

Tier 3 · Hard

A positive integer is multiplied by the integer that is

5

greater than it. The product is

176

. Form and solve an equation to find the two positive integers.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$11$ and $16$	4	Let the smaller positive integer be $n$ . Then $n(n+5)=176$ , so $n^2+5n-176=0$ . Factorising gives $(n+16)(n-11)=0$ , so $n=-16$ or $n=11$ . Reject the negative value because $n$ is positive. The integers are $11$ and $16$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-11	1F	Q16	5	Non-calculator	Foundation	QP MS
2019-06	1F	Q28	4	Non-calculator	Foundation	QP MS
2023-06	1H	Q18	4	Non-calculator	Higher	QP MS
2023-06	2F	Q24	4	Allowed	Foundation	QP MS
2023-11	1H	Q7	5	Non-calculator	Higher	QP MS
2021-11	1H	Q14	3	Non-calculator	Higher	QP MS
2024-06	3H	Q10	3	Allowed	Higher	QP MS
2019-06	1H	Q22	5	Non-calculator	Higher	QP MS
2023-06	1F	Q18	3	Non-calculator	Foundation	QP MS
2023-06	1H	Q10	4	Non-calculator	Higher	QP MS
2022-11	1F	Q16	4	Non-calculator	Foundation	QP MS
2024-11	1F	Q23	4	Non-calculator	Foundation	QP MS
2024-06	3F	Q16	3	Allowed	Foundation	QP MS
2022-06	3F	Q28	3	Allowed	Foundation	QP MS
2023-06	2H	Q5	4	Allowed	Higher	QP MS
2024-11	1H	Q6	4	Non-calculator	Higher	QP MS
2023-11	3F	Q18	5	Allowed	Foundation	QP MS
2021-11	1H	Q11	4	Non-calculator	Higher	QP MS
2022-06	3H	Q3	5	Allowed	Higher	QP MS
2023-11	1F	Q24	5	Non-calculator	Foundation	QP MS
2024-11	3H	Q19	4	Allowed	Higher	QP MS
2024-11	2H	Q7	5	Allowed	Higher	QP MS
2021-11	3F	Q25	3	Allowed	Foundation	QP MS
2019-06	1F	Q13	1	Non-calculator	Foundation	QP MS
2021-11	1F	Q26	4	Non-calculator	Foundation	QP MS
2022-06	2H	Q15	4	Allowed	Higher	QP MS
2019-06	3H	Q5	4	Allowed	Higher	QP MS
2021-11	1F	Q16	4	Non-calculator	Foundation	QP MS
2022-11	2F	Q14	7	Allowed	Foundation	QP MS
2019-11	1H	Q14	4	Non-calculator	Higher	QP MS
2024-11	3F	Q12	2	Allowed	Foundation	QP MS
2023-06	3H	Q24	5	Allowed	Higher	QP MS
2019-11	1F	Q28	3	Non-calculator	Foundation	QP MS
2024-11	2H	Q22	5	Allowed	Higher	QP MS
2024-06	2H	Q7	4	Allowed	Higher	QP MS
2024-06	2F	Q26	4	Allowed	Foundation	QP MS
2023-06	3F	Q11	3	Allowed	Foundation	QP MS
2024-06	1H	Q21	5	Non-calculator	Higher	QP MS
2021-11	3F	Q11	3	Allowed	Foundation	QP MS
2019-06	1H	Q7	4	Non-calculator	Higher	QP MS
2021-11	3H	Q5	3	Allowed	Higher	QP MS
2024-11	2F	Q26	5	Allowed	Foundation	QP MS
2019-11	3F	Q14	2	Allowed	Foundation	QP MS
2024-11	1H	Q13	5	Non-calculator	Higher	QP MS
2019-06	2F	Q7	1	Allowed	Foundation	QP MS
2022-06	3F	Q24	5	Allowed	Foundation	QP MS
2019-06	1H	Q17	4	Non-calculator	Higher	QP MS
2021-11	1H	Q7	4	Non-calculator	Higher	QP MS
2021-11	3H	Q13	5	Allowed	Higher	QP MS
2019-11	2F	Q18	4	Allowed	Foundation	QP MS
2022-06	1H	Q16	4	Non-calculator	Higher	QP MS

Other points in A Algebra · sequences and functions

Want help turning this into marks?

Bring A21 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call