Solving simple equations 77
8.
2
a
=
3
8
9.
1
3n
+
1
4n
=
7
24
10.
x + 3
4
=
x − 3
5
+ 2
11.
3t
20
=
6 − t
12
+
2t
15
−
3
2
12.
y
5
+
7
20
=
5 − y
4
13.
v − 2
2v − 3
=
1
3
14.
2
a − 3
=
3
2a + 1
15.
x
4
−
x + 6
5
=
x + 3
2
16. 3
√
t = 9
17. 2
√ y = 5
18. 4 =
3
a
+ 3
19.
3
√
x
1 −
√
x
= −6
20. 10 = 5
x
2
− 1
21. 16 =
t 2
9
22.
y + 2
y − 2
=
1
2
23.
6
a
=
2a
3
24.
11
2
= 5 +
8
x 2
11.3 Practical problems involving
simple equations
There are many practical situations in engineering in
which solving equations is needed. Here are some
worked examples to demonstrate typical practical
situations
Problem 17. Applying the principle of moments
to a beam results in the equation
F × 3 = (7.5 − F ) × 2
where F is the force in newtons. Determine the
value of F
Removing brackets gives
3F = 15 − 2F
Rearranging gives
3F + 2F = 15
i.e.
5F = 15
Dividing both sides by 5 gives
5F
5
=
15
5
from which, force, F = 3N.
Problem 18. A copper wire has a length L of
1.5 km, a resistance R of 5 and a resistivity of
17.2 × 10 −6 mm. Find the cross-sectional area, a,
of the wire, given that R =
ρ L
a
Since R =
ρ L
a
then
5 =
(17.2 × 10 −6 mm)(1500 × 10 3 mm)
a
.
From the units given, a is measured in mm 2 .
Thus, 5a = 17.2 × 10
−6
× 1500 × 10
3
and
a =
17.2 × 10 −6 × 1500 × 10 3
5
=
17.2 × 1500 × 10 3
10 6 × 5
=
17.2 × 15
10 × 5
= 5.16
Hence, the cross-sectional area of the wire is 5.16 mm
2 .
Problem 19. PV = m RT is the characteristic gas
equation. Find the value of gas constant R when
pressure P = 3 × 10 6 Pa, volume V = 0.90 m 3 ,
mass m = 2.81 kg and temperature T = 231 K
Dividing both sides of PV = m RT by mT gives
PV
mT
=
m RT
mT
Cancelling gives
PV
mT
= R
Substituting values gives
R =
3 × 10 6 (0.90)
(2.81)(231)