What is A19: Simultaneous equations?

A simultaneous solution is an ordered pair satisfying both equations; graphically it is an intersection point. For two linear equations, make one variable's coefficients equal or opposite, then add or subtract to eliminate it.

How many marks are A19 questions worth?

In verified past-paper appearances, A19 questions on Edexcel GCSE Maths 1MA1 have been worth 1–5 marks.

Which past exam papers has A19 appeared in?

A19 has appeared in 16 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

A19

Solve two simultaneous equations in two variables (linear/linear or linear/quadratic) algebraically; find approximate solutions using a graph

Simultaneous equations

Worked answers, methods and verified real exam appearances for A19 on Edexcel GCSE Maths 1MA1.

Explanation

A simultaneous solution is an ordered pair satisfying both equations; graphically it is an intersection point. For two linear equations, make one variable's coefficients equal or opposite, then add or subtract to eliminate it.
Substitute the first value back to find the second.
At Higher tier, when a line and a quadratic are solved together, substitute the linear expression into the quadratic, solve it, and find the matching second coordinate for every root.
The examiner expects complete ordered pairs and a check in both original equations.
Two $x$ -values alone are incomplete because each belongs to a different point.

Worked example

Solve simultaneously $2x+y=11$ and $x-y=1$ .

1.Add the equations to eliminate $y$ : $3x=12$ .
2.Solve to get $x=4$ .
3.Substitute into $x-y=1$ : $4-y=1$ , so $y=3$ .

Answer: $(x,y)=(4,3)$ .

Common mistakes

Don't add equations whose variable coefficients are not equal or opposite, so no variable is eliminated.
Don't find two possible $x$ -values in a linear-quadratic pair but give no corresponding $y$ -values.

Exam tip

Write each solution as an ordered pair and substitute both coordinates into both equations when the tariff allows a checking mark.

Worked practice

Tier 1 · Easy

Solve simultaneously

y=2x+1

and

x+y=10

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$x=3$ , $y=7$	3	Substitute $y=2x+1$ into $x+y=10$ : $x+2x+1=10$ . Hence $3x=9$ , so $x=3$ and $y=2(3)+1=7$ .

Tier 2 · Standard

On the same axes draw

y=0.6x+1

and

y=5-0.8x

. Use the graph to estimate their point of intersection 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	$(x,y)\approx(2.9,2.7)$	4	The solution is where the two lines cross. Reading the graph gives approximately $x=2.9$ and $y=2.7$ ; algebraically the point is $(20/7,19/7)$ , which confirms those graphical estimates.

Tier 3 · Hard

Higher only: Solve simultaneously

y=x+1

and

x^2+y^2=25

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$(x,y)=(-4,-3)$ or $(3,4)$	5	Substitute $y=x+1$ into the second equation: $x^2+(x+1)^2=25$ . This simplifies to $2x^2+2x-24=0$ , so $x^2+x-12=0=(x+4)(x-3)$ . Thus $x=-4$ or $x=3$ . Using $y=x+1$ gives $(-4,-3)$ and $(3,4)$ .

Tier 1 · Easy

Solve simultaneously

x+y=13

and

x-y=5

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$x=9$ , $y=4$	2	Add the equations to eliminate $y$ : $2x=18$ , so $x=9$ . Substituting into $x+y=13$ gives $y=4$ .

Tier 2 · Standard

Solve simultaneously

2x+3y=22

and

x-y=1

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$x=5$ , $y=4$	3	Multiply $x-y=1$ by $2$ to get $2x-2y=2$ . Subtracting this from $2x+3y=22$ gives $5y=20$ , so $y=4$ . Then $x-4=1$ , giving $x=5$ .

Tier 3 · Hard

Two angles have sizes

p^\circ

and

q^\circ

. Their sizes add to

110^\circ

. Three times

p

40

more than twice

q

. Form two equations and solve them to find

p

and

q

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$p=52$ , $q=58$	4	The equations are $p+q=110$ and $3p-2q=40$ . Multiply the first equation by $2$ to get $2p+2q=220$ . Adding this to the second equation gives $5p=260$ , so $p=52$ . Then $q=110-52=58$ .

Tier 2 · Standard

Solve simultaneously

4x-y=9

and

2x+3y=23

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$x=\dfrac{25}{7}$ , $y=\dfrac{37}{7}$	3	From $4x-y=9$ , $y=4x-9$ . Substitute this into the second equation: $2x+3(4x-9)=23$ . Hence $14x=50$ , so $x=25/7$ . Then $y=4(25/7)-9=37/7$ .

Tier 3 · Hard

Higher only: Solve simultaneously

y=x^2-4

and

y=2x+5

. Give both solutions in exact form.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$(x,y)=(1-\sqrt{10},7-2\sqrt{10})$ or $(1+\sqrt{10},7+2\sqrt{10})$	5	Equating the two expressions for $y$ gives $x^2-4=2x+5$ , so $x^2-2x-9=0$ . The quadratic formula gives $x=1\pm\sqrt{10}$ . Substituting into $y=2x+5$ gives $y=7\pm2\sqrt{10}$ , with matching signs in each ordered pair.

Tier 3 · Hard

Solve simultaneously

3(x-2y)=7

and

2(2x+y)=19

. Give the values of

x

and

y

as fractions in their simplest form.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$x=\dfrac{64}{15}$ , $y=\dfrac{29}{30}$	4	The equations expand to $3x-6y=7$ and $4x+2y=19$ . Multiply the second by $3$ and add the first: $15x=64$ , so $x=64/15$ . Substitution into $4x+2y=19$ gives $256/15+2y=285/15$ , hence $2y=29/15$ and $y=29/30$ .

Q10

Tier 3 · Hard

Higher only: Two numbers

x

and

y

satisfy

x+y=5

and

xy=3

. Solve these equations simultaneously, giving both ordered pairs in exact form.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$(x,y)=\left(\dfrac{5-\sqrt{13}}{2},\dfrac{5+\sqrt{13}}{2}\right)$ or $\left(\dfrac{5+\sqrt{13}}{2},\dfrac{5-\sqrt{13}}{2}\right)$	5	Write $y=5-x$ and substitute into $xy=3$ : $x(5-x)=3$ . Rearranging gives $x^2-5x+3=0$ , so $x=(5\pm\sqrt{13})/2$ . Since $y=5-x$ , the matching $y$ -value has the opposite sign before the surd, giving the two ordered pairs stated.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-11	1F	Q16	5	Non-calculator	Foundation	QP MS
2024-11	2H	Q10	4	Allowed	Higher	QP MS
2022-06	2F	Q28	3	Allowed	Foundation	QP MS
2019-06	1H	Q22	5	Non-calculator	Higher	QP MS
2022-11	1H	Q12	4	Non-calculator	Higher	QP MS
2019-06	3H	Q20	5	Allowed	Higher	QP MS
2019-06	3F	Q30	3	Allowed	Foundation	QP MS
2023-06	1F	Q28	1	Non-calculator	Foundation	QP MS
2022-06	3H	Q22	5	Allowed	Higher	QP MS
2021-11	3H	Q16	4	Allowed	Higher	QP MS
2024-06	1H	Q10	4	Non-calculator	Higher	QP MS
2019-06	1H	Q10	2	Non-calculator	Higher	QP MS
2023-11	2F	Q30	3	Allowed	Foundation	QP MS
2024-11	3H	Q20	4	Allowed	Higher	QP MS
2023-06	2H	Q11	4	Allowed	Higher	QP MS
2021-11	1H	Q20	3	Non-calculator	Higher	QP MS

Other points in A Algebra · sequences and functions

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