Solving simultaneous equations 93
Problem 6. Solve
x
8
+
5
2
= y
(1)
13 −
y
3
= 3x
(2)
Whenever fractions are involved in simultaneous equations it is often easier to firstly remove them. Thus,
multiplying equation (1) by 8 gives
8
x
8
+ 8
5
2
= 8y
i.e.
x + 20 = 8y
(3)
Multiplying equation (2) by 3 gives
39 − y = 9x
(4)
Rearranging equations (3) and (4) gives
x − 8y = −20
(5)
9x + y = 39
(6)
Multiplying equation (6) by 8 gives
72x + 8y = 312
(7)
Adding equations (5) and (7) gives
73x + 0 = 292
x =
292
73
= 4
Substituting x = 4 into equation (5) gives
4 − 8y = −20
4 + 20 = 8y
24 = 8y
y =
24
8
= 3
Checking, substituting x = 4 and y = 3 in the original
equations, gives
(1): LHS =
4
8
+
5
2
=
1
2
+ 2
1
2
= 3 = y = RHS
(2): LHS = 13 −
3
3
= 13 − 1 = 12
RHS = 3x = 3(4) = 12
Hence, the solution is x = 4, y = 3.
Problem 7. Solve
2.5x + 0.75 − 3y = 0
1.6x = 1.08 − 1.2y
It is often easier to remove decimal fractions. Thus,
multiplying equations (1) and (2) by 100 gives
250x + 75 − 300y = 0
( 1 )
160x = 108 − 120y
(2)
Rearranging gives
250x − 300y = −75
(3)
160x + 120y = 108
(4)
Multiplying equation (3) by 2 gives
500x − 600y = −150
(5)
Multiplying equation (4) by 5 gives
800x + 600y = 540
(6)
Adding equations (5) and (6) gives
1300x + 0 = 390
x =
390
1300
=
39
130
=
3
10
= 0.3
Substituting x = 0.3 into equation (1) gives
250(0.3) + 75 − 300y = 0
75 + 75 = 300y
150 = 300y
y =
150
300
= 0.5
Checking, by substituting x = 0.3 and y = 0.5 in equation (2), gives
LHS = 160(0.3) = 48
RHS = 108 − 120(0.5) = 108 − 60 = 48
Hence, the solution is x = 0.3, y = 0.5