Simultaneous equations Revision
1
Solve the following pair of simultaneous
equations using the substitution method.
x  4  2y
5 x  2 y  32
x  4  2y
5 x  2 y  32
1
2
Substitute 1 into  2 
5  4  2 y   2 y  32
20  10 y  2 y  32
12 y  12
y  1
Substitute y  1 into (1)
x  4  2  1
x 42
x 6
Solution: (6, 1)
2
Solve the following pair of simultaneous
equations using the substitution method
y  3x  7
2 x  5 y  48
y  3x  7
2 x  5 y  48
Substitute 1 into  2 
2 x  5  3 x  7   48
2 x  15 x  35  48
13x  13
x  1
Substitute y  1 into 1
y  3  1  7
y  10
Solution: (1, 10)
Page 1
1
2
5
Solve the following pair of simultaneous
equations using the substitution method.
y  2 x  22
3 x  4 y  53
y  2 x  22
3x  4 y  53
1
2
Substitute (1) into  2 
3x  4  2 x  22   53
3x  8 x  88  53
5 x  35
x7
Substitute x  7 into 1
y  2  7  22
y  14  22
y  8
Solution: (7, 8)
6
Solve the following pair of simultaneous
equations using the substitution method.
1
x   1  3 y 
4
6x  2 y  5
1
 1  3 y 
4
6x  2 y  5
x
Substitute (1) into
6
1
2
2
1
 1  3 y   2 y  5
4
6  1  3 y   8 y  20
6  18 y  8 y  20
26 y  26
y  1
Substitute y  1 into 1
1
 1  3  1
4
1
x   1  3
4
1
x  2
4
1
x
2
1

Solution:  ,  1 
2

x
Page 2
7
Solve the following pair of simultaneous
equations using the elimination method.
4 x  y  12
2 x  y  12
4 x  y  12
2 x  y  12
1   2  :
1
2
6 x  24
x4
Substitute x  4 into  2 
2  4  y  12
8  y  12
y4
Solution: (4, 4)
8
Solve the following pair of simultaneous
equations using the elimination method.
4 x  y  5
 4 x  2 y  14
4 x  y  5
4 x  2 y  14
1   2  :
1
2
3y  9
y3
Substitute y  3 into 1
4 x  3  5
4 x  8
x  2
Solution: (2, 3)
9
Solve the following pair of simultaneous
equations using the elimination method.
5 x  2 y  24
3 x  y  1
(2)  2:
(1)  (3):
5 x  2 y  24
(1)
3x  y  1
( 2)
(3)
6 x  2 y  2
11x  22
x2
Substitute x  2 into  2 
3  2  y  1
6  y  1
y  7
Solution: (2, –7)
Page 3
10
Solve the following pair of simultaneous
equations using the elimination method.
8x  9 y  0
14 x  12 y  3
8x  9 y  0
14 x  12 y  3
(1)  4: 32 x  36 y  0
(2)  3: 42 x  36 y  9
(4)  (3):
1
2
 3
4
10 x  9
9
x
10
9
Substitute x 
into 1
10
9
8  9y  0
10
72
 9y  0
10
36
9y  
5
4
y
5
4
9
Solution:  ,  
5
 10
Page 4
3
Solve the following pair of simultaneous
equations using a graphical method:
y  1  4x
y  2x  7
Use Analysis and G-solve to find the
Intersection
The two lines intersect at (1, 5).
Page 5
4
Solve the following pair of simultaneous
equations using the substitution method:
x  1 y
2 x  3 y  17
1
2
x  1 y
2 x  3 y  17
Substitute (1) into  2 
2 1  y   3 y  17
2  2 y  3 y  17
5 y  15
y  3
Substitute y  3 into 1
x  1   3
x4
Solution: (4, 3)
5
Solve the following pair of simultaneous
equations using the elimination method:
x  4y  6
3 x  6 y  15
(1)  3:
(2)  (3):
x  4y  6
(1)
3x  6 y  15
( 2)
3 x  12 y  18
(3)
6 y  3
y 
1
2
1
into 1
2
1
x 4  6
2
x26
Substitute y  
x4
1

Solution:  4,  
2

Page 6
6
Solve the following pair of simultaneous
equations using the elimination method:
4 x  9 y  7
3 x  6 y  4
4 x  9 y  7 (1)
3 x  6 y  4 ( 2 )
(1)  3 : 12 x  27 y  21 (3)
(2)  4 : 12 x  24 y  16 ( 4)
(3)  (4) :
3 y  5
5 2
y   1 
3 3
5
substitute y   into 1
3
5
4 x  9    7
3
4 x  15  7
4x  8
x 2
2

Solution :  2,  1 
3

7
Find two numbers whose sum is 23 and whose
difference is 27.
Let the two numbers be x and y.
x  y  23 1
x  y  27
(1)  (2) :
2
2 x  50
x  25
Substitute x  25 into 1
25  y  23
y  2
The two numbers are 25 and 2.
8
A rectangular swimming pool has a perimeter Let x be the width of the pool.
of 50 metres. The length is 9 metres more than The length is x + 9.
the width. What are the dimensions of the pool? Perimeter = 50
50  2 x  2 x  9 
50  2 x  2 x  18
32  4 x
x8
Width is 8 metres.
Length is 17 metres.
Page 7
9
A moneybox contains only 10c and 20c coins. Let x be the number of 10c coins.
If there are 54 coins altogether, totalling $8.60, Let y be the number of 20c coins.
how many of each type of coin are there?
x  y  54 1
10 x  20 y  860
(2)  10: x  2 y  86
(3)  (1):
2
 3
y  32
Substitute y  32 into 1
x  32  54
x  22
Moneybox contains 22 × 10c coins and
32 × 20c coins.
10
Nicholas buys 4 pears and 3 apples for $2.75.
Nadia buys 7 pears and 5 apples for $4.70.
How much does each type of fruit cost?
Let p represent the cost of a pear and a
represent the cost of an apple (in cents).
4 p  3a  275 1
2
20 p  15a  1375  3
21 p  15a  1410  4 
7 p  5a  470
(1)  5:
(2)  3:
(4)  (3):p  35
Substitute p  35 into 1
4  35  3a  275
140  3a  275
3a  135
a  45
A pear costs 35 cents and an apple 45 cents.
Page 8
11
Three adults and 5 children pay $62.30 to enter Let a represent an adult’s entry fee and
the zoo, whereas 4 adults and 3 children pay
c represent a child’s entry fee (in cents).
$60.70. What was the entry fee for each adult
3a  5c  6230
1
and each child?
4a  3c  6070
2
(1)  3: 9a  15c  18 690
(2)  5: 20a  15c  30350
(4)  (3) :
 3
4
11a  11 660
a  1060
Substitute a  1060 into 1
3  1060  5c  6230
3180  5c  6230
5c  3050
c  610
Entrance fees are adults $10.60 and children
$6.10.
Page 9