Head to savemyexams.co.uk for more awesome resources
Simultaneous Linear Equations
Difficulty: Easy
Model Answers 1
Level
IGCSE
Subject
Maths (0580/0980)
Exam Board
CIE
Topic
Algebra & Graphs
Sub-Topic
Simultaneous Linear Equations
Paper
Paper 2
Difficulty
Easy
Booklet
Model Answers 1
Time allowed:
32 minutes
Score:
/25
Percentage:
/100
Grade Boundaries:
CIE IGCSE Maths (0580)
A*
>88%
A
76%
B
63%
C
51%
D
40%
E
30%
CIE IGCSE Maths (0980)
9
>94%
1
8
85%
7
77%
6
67%
5
57%
4
47%
Model answers are copyright. © 2019 Save My Exams Ltd. All rights reserved.
3
35%
Question 1
Head to savemyexams.co.uk for more awesome resources
Solve the simultaneous equations.
You must show all your working.
Rearrange Ⓑ to give:
And substitute into Ⓐ:
Collect like terms, subtract 18:
(and change sign):
Substitute back into Ⓑ:
To get:
And finally CHECK in Ⓐ:
2x + 3y = 13
x + 2y = 9
Ⓐ
Ⓑ
𝑥𝑥 = 9 − 2𝑦𝑦
2 9 − 2𝑦𝑦 + 3𝑦𝑦 = 13
18 − 4𝑦𝑦 + 3𝑦𝑦 = 13
+𝑦𝑦 = +5
𝑥𝑥 + 2 × 5 = 9
𝑥𝑥 = −1
2 × −1 + 3 × 5 = 13 👍👍
𝒙𝒙 = −𝟏𝟏, 𝒚𝒚 = 𝟓𝟓
2
Model answers are copyright. © 2019 Save My Exams Ltd. All rights reserved.
[3]
Question 2
Head to savemyexams.co.uk for more awesome resources
Solve the simultaneous equations.
You must show all your working.
1
x2
8y = 1
x + 2y = 6
1
2
[3]
We have two equations and two unknowns.
1
𝑥𝑥 − 8𝑦𝑦 = 1
2
𝑥𝑥 + 2𝑦𝑦 = 6
1
2
Multiply the first equation by 2 and subtract it from the second equation:
𝑥𝑥 + 2𝑦𝑦 − 2 ×
Sum the x and y terms
1
1
𝑥𝑥 − 8𝑦𝑦 = 6 − 2 × (1)
2
2
1
𝑥𝑥 + 2𝑦𝑦 − 𝑥𝑥 + 16𝑦𝑦 = 6 − 2
2
1
Convert the mixed number: 6 2 =
1
18𝑦𝑦 = 6 − 2
2
6×2+1
2
=
13
2
18𝑦𝑦 =
13
−2
2
18𝑦𝑦 =
9
2
Divide both sides by 18.
We get one of the two unknowns.
3
𝑦𝑦 =
9
36
𝒚𝒚 =
𝟏𝟏
𝟒𝟒
Model answers are copyright. © 2019 Save My Exams Ltd. All rights reserved.
Head to savemyexams.co.uk for more awesome resources
Plug the value of y into the first equation.
1
1
𝑥𝑥 − 8 × = 1
2
4
Add 2 to both sides of the equation.
1
𝑥𝑥 − 2 = 1
2
1
𝑥𝑥 = 3
2
Multiply both sides by 2 to get the second unknown.
𝒙𝒙 = 𝟔𝟔
4
Model answers are copyright. © 2019 Save My Exams Ltd. All rights reserved.
Question 3
3
Head to savemyexams.co.uk for more awesome resources
Solve the simultaneous equations.
2x – y = 7
3x + y = 3
2𝑥𝑥 − 𝑦𝑦 = 7 (1)
3𝑥𝑥 + 𝑦𝑦 = 3 (2)
Add (1) and (2) together
Divide through by 5
Sub this into (1)
5𝑥𝑥 = 10
𝒙𝒙 = 𝟐𝟐
4 − 𝑦𝑦 = 7
→ 𝒚𝒚 = −𝟑𝟑
5
Model answers are copyright. © 2019 Save My Exams Ltd. All rights reserved.
[2]
Question 4
Head to savemyexams.co.uk for more awesome resources
15 Find the co-ordinates of the point of intersection of the two lines.
2x – 7y = 2
4x + 5y = 42
2𝑥𝑥 − 7𝑦𝑦 = 2 … (1)
4𝑥𝑥 + 5𝑦𝑦 = 42 … (2)
To find point of intersection, solve the simultaneous equations:
Equation 1:
𝑥𝑥 =
Sub (3) into (2):
4
2 + 7𝑦𝑦
… 3
2
2 + 7𝑦𝑦
+ 5𝑦𝑦 = 42
2
4 + 14𝑦𝑦 + 5𝑦𝑦 = 42
4 + 19𝑦𝑦 = 42
19𝑦𝑦 = 38
𝑦𝑦 = 2
Sub y = 2 into (3):
𝑥𝑥 =
2 + 14
2
𝑥𝑥 = 8
Therefore, the coordinates are:
(𝟖𝟖, 𝟐𝟐)
6
Model answers are copyright. © 2019 Save My Exams Ltd. All rights reserved.
[3]
Question 5
Head to savemyexams.co.uk for more awesome resources
Solve the simultaneous equations.
3x + 5y = 24
x + 7y = 56
We have
3𝑥𝑥 + 5𝑦𝑦 = 24 (1)
𝑥𝑥 + 7𝑦𝑦 = 56 (2)
Multiply (2) by 3 then subtract (1) from the result
2 × 3 − 1 ∶ 3𝑥𝑥 + 21𝑦𝑦 − 3𝑥𝑥 − 5𝑦𝑦 = 168 − 24
Divide through by 16
Sub this into (2)
→ 16𝑦𝑦 = 144
→ 𝒚𝒚 = 𝟗𝟗
𝑥𝑥 + 7 9 = 56
→ 𝑥𝑥 + 63 = 56
Subtract 63 from both sides
𝒙𝒙 = −𝟕𝟕
7
Model answers are copyright. © 2019 Save My Exams Ltd. All rights reserved.
[3]
Question 6
3
Head to savemyexams.co.uk for more awesome resources
Solve the simultaneous equations.
x + 5y = 22
x + 3y = 12
x + 5y = 22
x + 3y = 12
To solve the 2 simultaneous equations, we subtract the
second one from the first one.
x + 5y – x – 3y = 22 – 12
2y = 10
y=5
x + 3y = 12
We substitute y = 5 in this equation to work out x.
x + 3 x 5 = 12
x + 15 = 12
x = -3
8
Model answers are copyright. © 2019 Save My Exams Ltd. All rights reserved.
[2]
Question 7
Head to savemyexams.co.uk for more awesome resources
10 Solve the simultaneous equations.
3x + y = 30
2x – 3y = 53
Multiply the first equation by 3 to make the coefficient ofy equal.
9𝑥𝑥 + 3𝑦𝑦 = 90
2𝑥𝑥 − 3𝑦𝑦 = 53Add the two equations to eliminate y and get an equation
which only depends only onx.
9𝑥𝑥 + 3𝑦𝑦 + 2𝑥𝑥 − 3𝑦𝑦 = 90 + 53
11𝑥𝑥 = 143
𝒙𝒙 = 𝟏𝟏𝟏𝟏
Substitute the value of x into one of the first equation and solvefor y.
9 × 13 + 3𝑦𝑦 = 90
3𝑦𝑦 = 90 − 117
𝑦𝑦 = −
27
3
𝒚𝒚 = −𝟗𝟗
9
Model answers are copyright. © 2019 Save My Exams Ltd. All rights reserved.
[3]
Question 8
Head to savemyexams.co.uk for more awesome resources
12 Solve the simultaneous equations.
x – 5y = 0
15x + 10y = 17
Write down the two equations.
𝑥𝑥 − 5𝑦𝑦 = 0
and
15𝑥𝑥 + 10𝑦𝑦 = 17
Multiply the first equation by 2 to have to same factor in front of y
(although with different signs).
2𝑥𝑥 − 10𝑦𝑦 = 0
and
15𝑥𝑥 + 10𝑦𝑦 = 17
Now we can eliminate y by adding the two equations together.
17𝑥𝑥 = 17
With only one unknown, we solve for x.
𝒙𝒙 = 𝟏𝟏
Substitute the value of x into the first (original) equation.
1 − 5𝑦𝑦 = 0
Again, we have only one unknown so we solve for y.
1 = 5𝑦𝑦
𝟏𝟏
𝒚𝒚 = 𝟓𝟓
10
Model answers are copyright. © 2019 Save My Exams Ltd. All rights reserved.
[3]
Question 9
8
Head to savemyexams.co.uk for more awesome resources
Solve the simultaneous equations.
x + 2y = 3
2x – 3y = 13
Write down the two equations.
𝑥𝑥 + 2𝑦𝑦 = 3
and
2𝑥𝑥 − 3𝑦𝑦 = 13
Multiply the first equation by 2 to have to same factor in front of x.
2𝑥𝑥 + 4𝑦𝑦 = 6
and
2𝑥𝑥 − 3𝑦𝑦 = 13
Now we can eliminate x by subtracting the two equations together.
2𝑥𝑥 + 4𝑦𝑦 − 2𝑥𝑥 − 3𝑦𝑦 = 6 − 13
7𝑦𝑦 = −7
With only one unknown, we solve for y.
𝒚𝒚 = −𝟏𝟏
Substitute the value of y into the first (original) equation.
𝑥𝑥 + 2 × (−1) = 3
Again, we have only one unknown so we solve for x.
𝑥𝑥 − 2 = 3
𝒙𝒙 = 𝟓𝟓
11
Model answers are copyright. © 2019 Save My Exams Ltd. All rights reserved.
[3]