94 Basic Engineering Mathematics
Now try the following Practice Exercise
Practice Exercise 50 Solving simultaneous
equations (answers on page 345)
Solve the following simultaneous equations and
verify the results.
1. 7 p + 11 + 2q = 0
2.
x
2
+
y
3
= 4
−1 = 3q − 5 p
x
6
−
y
9
= 0
3.
a
2
− 7 = −2b
4.
3
2
s − 2t = 8
12 = 5a +
2
3
b
s
4
+ 3y = −2
5.
x
5
+
2y
3
=
49
15
6. v − 1 =
u
12
3x
7
−
y
2
+
5
7
= 0
u +
v
4
−
25
2
= 0
7. 1.5x − 2.2y = −18
8. 3b − 2.5a = 0.45
2.4x + 0.6y = 33
1.6a + 0.8b = 0.8
13.4 Solving more difficult
simultaneous equations
Here are some further worked problems on solving more
difficult simultaneous equations.
Problem 8. Solve
2
x
+
3
y
= 7
( 1 )
1
x
−
4
y
= −2
( 2 )
In this type of equation the solution is easier if a
substitution is initially made. Let
1
x
= a and
1
y
= b
Thus equation (1) becomes 2a + 3b = 7
( 3 )
and equation (2) becomes
a − 4b = −2
(4)
Multiplying equation (4) by 2 gives
2a − 8b = −4
( 5 )
Subtracting equation (5) from equation (3) gives
0 + 11b = 11
i.e.
b = 1
Substituting b = 1 in equation (3) gives
2a + 3 = 7
2a = 7 − 3 = 4
i.e.
a = 2
Checking, substituting a = 2 and b = 1 in equation (4),
gives
LHS = 2 − 4(1) = 2 − 4 = −2 = RHS
Hence, a = 2 and b = 1.
However, since
1
x
= a,
x =
1
a
=
1
2
or 0.5
and since
1
y
= b,
y =
1
b
=
1
1
= 1
Hence, the solution is x = 0.5, y = 1.
Problem 9. Solve
1
2a
+
3
5b
= 4
( 1 )
4
a
+
1
2b
= 10.5
( 2 )
Let
1
a
= x and
1
b
= y
then
x
2
+
3
5
y = 4
( 3 )
4x +
1
2
y = 10.5
( 4 )
To remove fractions, equation (3) is multiplied by 10,
giving
10
x
2
+ 10
3
5
y
= 10(4)
i.e.
5x + 6y = 40
(5)
Multiplying equation (4) by 2 gives
8x + y = 21
(6)
Multiplying equation (6) by 6 gives
48x + 6y = 126
(7)