76 Basic Engineering Mathematics
i.e.
38y = −114
Dividing both sides by 38 gives
38y
38
=
−114
38
Cancelling gives
y = −3
which is the solution of the equation
2y
5
+
3
4
+ 5 =
1
20
−
3y
2
Problem 12. Solve the equation
√
x = 2
Whenever square root signs are involved in an equation,
both sides of the equation must be squared.
Squaring both sides gives
√
x
2 = (2)
2
i.e.
x = 4
which is the solution of the equation
√
x = 2.
Problem 13. Solve the equation 2
√
d = 8
Whenever square roots are involved in an equation, the
square root term needs to be isolated on its own before
squaring both sides.
Cross-multiplying gives
√
d =
8
2
Cancelling gives
√
d = 4
Squaring both sides gives
√
d
2 = (4)
2
i.e.
d = 16
which is the solution of the equation 2
√
d = 8.
Problem 14. Solve the equation
√
b + 3
√
b
= 2
Cross-multiplying gives
√
b + 3 = 2
√
b
Rearranging gives
3 = 2
√
b −
√
b
i.e.
3 =
√
b
Squaring both sides gives
9 = b
which is the solution of the equation
√
b + 3
√
b
= 2.
Problem 15. Solve the equation x 2 = 25
Whenever a square term is involved, the square root of
both sides of the equation must be taken.
Taking the square root of both sides gives
x 2 =
√
25
i.e.
x = ±5
which is the solution of the equation x
2
= 25.
Problem 16. Solve the equation
15
4t 2 =
2
3
We need to rearrange the equation to get the t
2 term on
its own.
Cross-multiplying gives
15(3) = 2(4t
2
)
i.e.
45 = 8t
2
Dividing both sides by 8 gives 45
8
=
8t 2
8
By cancelling
5.625 = t
2
or
t
2
= 5.625
Taking the square root of both sides gives
t 2 =
√
5.625
i.e.
t = ±2.372
correct to 4 significant figures, which is the solution of
the equation
15
4t 2 =
2
3
Now try the following Practice Exercise
Practice Exercise 43 Solving simple
equations (answers on page 344)
Solve the following equations.
1.
1
5
d + 3 = 4
2. 2 +
3
4
y = 1 +
2
3
y +
5
6
3.
1
4
(2x − 1) + 3 =
1
2
4.
1
5
(2 f − 3) +
1
6
( f − 4) +
2
15
= 0
5.
1
3
(3m − 6) −
1
4
(5m + 4) +
1
5
(2m − 9) = −3
6.
x
3
−
x
5
= 2
7. 1 −
y
3
= 3 +
y
3
−
y
6