Chapter 9
Basic algebra
9.1
Introduction
We are already familiar with evaluating formulae using
a calculator from Chapter 4.
For example, if the length of a football pitch is L and its
width is b, then the formula for the area A is given by
A= L b
This is

Chapter 6
Ratio and proportion
6.1
Introduction
Ratio is a way of comparing amounts of something; it
shows how much bigger one thing is than the other.
Some practical examples include mixing paint, sand
and cement, or screen wash. Gears, map scales, food

44 Basic Engineering Mathematics
7. Charless law states that volume is directly
proportional to thermodynamic temperature
for a given mass of gas at constant pressure.
A gas occupies a volume of 4.8 litres at 330 K.
Determine (a) the temperature when the

Ratio and proportion
Here are some worked examples to help us understand
more about direct proportion.
Problem 9. 3 energy saving light bulbs cost
7.80. Determine the cost of 7 such light bulbs
(i) 3 light bulbs cost 7.80
7.80
= 2.60
3
Hence, 7 light bulb

58 Basic Engineering Mathematics
Hence, 42 105 = 4.2 106 in engineering
notation
(b) Enter 3100 1012 into the calculator. Press =
Now press ENG and the answer is 3.1 109.
= 4.2 M in prefix form.
From the table of prexes on page 55, 109
corresponds to nano

Revision Test 2 : Decimals, calculators and percentages
This assignment covers the material contained in Chapters 35. The marks available are shown in brackets at the
end of each question.
1.
Convert 0.048 to a proper fraction.
(2)
2.
Convert 6.4375 to a

66 Basic Engineering Mathematics
Using law (3) of indices gives
p1/2 q 2r 2/3
and evaluate
p1/4 q 1/2r 1/6
when p = 16, q = 9 and r = 4, taking positive roots
only
Problem 25.
d 2 e2 f 1/2
d 2e2 f 1/2
=
(d 3/2 e f 5/2 )2
d 3 e2 f 5
Simplify
Using law (2)

46 Basic Engineering Mathematics
(working at the same rate) would take 6 hours. Half the
men, twice the time. (Note that 4 3 = 2 6.)
Here are some worked examples on inverse proportion.
Problem 17. It is estimated that a team of four
designers would take

Revision Test 3 : Ratio, proportion, powers, roots, indices and units
This assignment covers the material contained in Chapters 68. The marks available are shown in brackets at the
end of each question.
1. In a box of 1500 nails, 125 are defective. Expres

76 Basic Engineering Mathematics
38y = 114
i.e.
114
38y
=
38
38
y = 3
Dividing both sides by 38 gives
Cancelling gives
Taking the square root of both sides gives
Problem 12.
Solve the equation
x =2
Squaring both sides gives
2
x = (2)2
x=4
which is the so

Ratio and proportion
Thus, 128 :48 is equivalent to 64 :24 which is equivalent
to 8 :3 and 8 : 3 is the simplest form.
Problem 3. A wooden pole is 2.08 m long. Divide
it in the ratio of 7 to 19
(i) Since the ratio is 7 :19, the total number of parts
is 7

42 Basic Engineering Mathematics
Thus, 45 p as a ratio of 7.65 is 1 : 17
45 : 765, 9 :153, 3 :51 and 1 : 17 are equivalent ratios
and 1 :17 is the simplest ratio.
2. In a laboratory, acid and water are mixed in the
ratio 2 :5. How much acid is needed to m

64 Basic Engineering Mathematics
(i) (iv) (vii)
Alternatively, the answer may be expressed as
2
2
x xy + y
3
x + y x + 0 + 0 + y3
x3 + x2 y
x 2 y
4b 3
4a 3 6a 2 b + 5b3
= 2a 2 2ab b2 +
2a b
2a b
+ y3
Now try the following Practice Exercise
x 2 y x y 2
Pra

56 Basic Engineering Mathematics
23. What does the prefix p mean?
24. What is the symbol and meaning of the prefix
mega?
8.4
3
= 0.375, and expressing it in standard form
8
gives
0.375 = 3.75 101
.
2
(b) 19 = 19. 6 = 1.97 10 in standard form, cor3
rect to

Transposing formulae
Taking the cube root of both sides gives
3V
3
3
3
r =
2
3V
i.e.
r =3
2
(b) When V = 32cm3,
3V
3 32
=3
= 2.48 cm.
radius r = 3
2
2
87
14. Transpose Z = R 2 + (2 f L)2 for L and
evaluate L when Z = 27.82, R = 11.76 and
f = 50.
12.4 Mo

Solving simple equations
7. An alloy contains 60% by weight of copper,
the remainder being zinc. How much copper
must be mixed with 50 kg of this alloy to give
an alloy containing 75% copper?
9. Applying the principle of moments to a beam
results in the f

Solving simple equations
Now try the following Practice Exercise
Practice Exercise 44 Practical problems
involving simple equations (answers on
page 344)
1.
2.
3.
4.
A formula used for calculating resistance of a
L
cable is R =
. Given R = 1.25, L = 2500

Chapter 7
Powers, roots and laws of
indices
7.1
Introduction
The manipulation of powers and roots is a crucial underlying skill needed in algebra. In this chapter, powers and
roots of numbers are explained, together with the laws
of indices.
Many worked e

Further algebra
Problem 7. Remove the brackets from the
expression and simplify 2[x 2 3x(y + x) + 4x y]
2[x 2 3x(y + x) + 4x y] = 2[x 2 3x y 3x 2 + 4x y]
(Whenever more than one type of brackets is involved,
always start with the inner brackets)
13. (8x +

Chapter 8
Units, prefixes and
engineering notation
8.1
Introduction
Quantity
Unit
Symbol
Length
metre
m (1 m = 100 cm
= 1000 mm)
Mass
kilogramkg (1 kg = 1000 g)
80 kV = 80 103 V, which means 80 000 volts
Time
second s
25 mA = 25 103 A,
which means 0.025 a

Basic algebra
(2)
am
= amn
an
(3) (am )n = amn
n
m
(4) a n =
(5) an =
am
1
an
(6) a0 = 1
c5
= c52 = c3
c2
3
For example, d 2 = d 23 = d 6
For example,
4
For example, x 3 =
3
For example, 32 =
x4
1
1
=
32
9
a 3 b 2 c4
Problem 20. Simplify
and evaluate whe

52 Basic Engineering Mathematics
Dividing each term by the HCF (i.e. 22 ) gives
26 34
28 34
26
=
=
22 33 23
33 2
34 (33 2)
3 2
4
3
3
5
Problem 23. Simplify
giving
3
2
5
the answer with positive indices
Raising a fraction to a power means that both the n

Chapter 10
Further algebra
10.1
3(2x 3y) (3x y) = 3 2x 3 3y 3x y
Introduction
In this chapter, the use of brackets and factorization
with algebra is explained, together with further practice
with the laws of precedence. Understanding of these
topics is of

38 Basic Engineering Mathematics
100 + 6.25
3600
100
106.25
=
3600
100
= 1.0625 3600
Value after 1 year =
annum. Calculate the value of the investment
after 2 years.
8.
An electrical contractor earning 36 000 per
annum receives a pay rise of 2.5%. He
pa

80 Basic Engineering Mathematics
8.40x = 319.20 226.80 = 92.40
92.40
x=
= 11
8.40
Thus, 11 hours overtime would have to be worked to
earn 319.20 per week. Hence, the total number of
hours worked is 36 + 11, i.e. 47 hours.
Problem 27. A formula relating in

PRELIMINARY CALCULUS
Since
1
1
dt
x
x
1 + t2
= sec2
=
1 + tan2
=
,
dx
2
2
2
2
2
the required relationship is
dx =
Evaluate the integral
I=
2
dt.
1 + t2
(2.34)
2
dx.
1 + 3 cos x
Rewriting cos x in terms of t and using (2.34) yields
2
2
dt
2
1+t
1 + 3 (1

2.2 INTEGRATION
Evaluate the integral
I=
We can write the integral in the form
I=
1
dx.
x2 + 4x + 7
1
dx.
(x + 2)2 + 3
Substituting y = x + 2, we nd dy = dx and hence
1
I=
dy,
y2 + 3
Hence, by comparison with the table of standard integrals (see subsectio

2.1 DIFFERENTIATION
The pattern emerging is clear and strongly suggests that the results generalise to
f (n) =
n
r=0
n!
n
u(r) v (nr) =
Cr u(r) v (nr) ,
r!(n r)!
n
(2.14)
r=0
where the fraction n!/[r!(n r)!] is identied with the binomial coecient n Cr
(se

PRELIMINARY CALCULUS
C
c
r + r
O
Q
r
p + p
P
p
Figure 2.13 The coordinate system described in exercise 2.20.
2.20
2.21
A two-dimensional coordinate system useful for orbit problems is the tangentialpolar coordinate system (gure 2.13). In this system a cur

2.2 INTEGRATION
y
C
d
( + d)
dA
()
O
B
x
Figure 2.9 Finding the area of a sector OBC dened by the curve () and
the radii OB, OC, at angles to the x-axis 1 , 2 respectively.
dA = 12 2 d, as illustrated in gure 2.9, and hence the total area between two
ang

3.6 APPLICATIONS TO DIFFERENTIATION AND INTEGRATION
3.6 Applications to dierentiation and integration
We can use the exponential form of a complex number together with de Moivres
theorem (see section 3.4) to simplify the dierentiation of trigonometric fun

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS
Express sin 3 and cos 3 in terms of powers of cos and sin .
Using de Moivres theorem,
cos 3 + i sin 3 = (cos + i sin )3
= (cos3 3 cos sin2 ) + i(3 sin cos2 sin3 ).
(3.28)
We can equate the real and imaginary coecie