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Chapter 9

Basic algebra
9
...

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 an algebraic equation
...

The total resistance, RT , of resistors R1 , R2 and R3
connected in series is given by
RT = R1 + R2 + R3
This is an algebraic equation
...
3 k, R2 = 2
...
5 k, then
RT = 6
...
4 + 8
...
2 k
The temperature in Fahrenheit, F, is given by
9
F = C + 32
5
where C is the temperature in Celsius
...

9
If C = 100◦C, then F = × 100 + 32
5
= 180 + 32 = 212◦F
...


9
...

If, say, a, b, c and d represent any four numbers then in
algebra:
DOI: 10
...
00009-0

(a) a + a + a + a = 4a
...

(b) 5b means 5 × b
...

(c) 2a + 3b + a − 2b = 2a + a + 3b − 2b = 3a + b
Only similar terms can be combined in algebra
...

In addition, with terms separated by + and − signs,
the order in which they are written does not matter
...
(Note that the first term, i
...

2a, means +2a
...
(Note
that − × − = +)
(e) (a)(c)(d) means a × c × d
Brackets are often used instead of multiplication
signs
...

(f ) ab = ba
If a = 2 and b = 3 then 2 × 3 is exactly the same
as 3 × 2, i
...
6
...
For
32 = 3 × 3 = 9
...

Here are some worked examples to help get a feel for
basic operations in this introduction to algebra
...
2
...


Find the sum of 4x, 3x, −2x and −x

4x + 3x + −2x + −x = 4x + 3x − 2x − x
(Note that + ×− = −)
= 4x
Problem 2
...


Simplify 3 + x + 5x − 2 − 4x
...


Add x − 2y + 3 to 3x + 4y − 1
...


Subtract a − 2b from 4a + 3b
...


From a + b − 2c take 3a + 2b − 4c
...
From x 2 + x y − y 2 take x y − 2x 2
...
2
...


Simplify bc × abc

5x+3y + z + −3x + −4y + 6z
= 5x + 3y + z − 3x − 4y + 6z

bc × abc = a × b × b × c × c

= 5x − 3x + 3y − 4y + z + 6z

= a × b 2 × c2

= 2x − y + 7z

= ab2 c 2

Note that the order can be changed when terms are separated by + and − signs
...


Problem 6
...


Simplify 4x 2 − x

− 2y + 5x + 3y
Problem 7
...


ab × b2c × a = a × a × b × b × b × c
= a 2 × b3 × c

Simplify 3x y − 7x + 4x y + 2x

3x y − 7x + 4x y + 2x = 3x y + 4x y + 2x − 7x
= 7xy − 5x

Simplify ab × b2c × a

= a2 b3 c
Problem 8
...


Find the sum of 4a, −2a, 3a and −8a
...


Find the sum of 2a, 5b, −3c, −a, −3b and
7c
...


Simplify 2x − 3x 2 − 7y + x + 4y − 2y 2
...


Simplify 5ab − 4a + ab + a
...


Simplify 2x − 3y + 5z − x − 2y + 3z + 5x
...


5 pq 2r 3 = 5 × p × q 2 × r 3
 2  3
1
2
= 5×2×
× 2
5
2

Basic algebra
 2  3
2
5
= 5×2×
×
5
2

1 5
since 2 =
2 2

5 2 2 2 5 5 5
× × × × × ×
1 1 5 5 2 2 2
1 1 1 1 1 5 5
= × × × × × ×
1 1 1 1 1 1 1
= 5×5

Problem 13
...
Multiply 2a + 3b by a + b
Each term in the first expression is multiplied by a, then
each term in the first expression is multiplied by b and
the two results are added
...

2a + 3b
a+b
Multiplying by a gives
Multiplying by b gives

2a 2 + 3ab
2ab + 3b2

Adding gives

2a 2 + 5ab + 3b2

Thus, (2a + 3b)(a + b) = 2a2 + 5ab + 3b2
Problem 11
...
The usual layout is shown below with the
dividend and divisor both arranged in descending
powers of the symbols
...


...
e
...

(iii) The divisor is then multiplied by 2x, i
...

2x(x − 1) = 2x 2 − 2x, which is placed under the
dividend as shown
...

(iv) The process is then repeated, i
...
the first term
of the divisor, x, is divided into 3x, giving +3,
which is placed above the dividend as shown
...
The remainder, on subtraction, is zero,
which completes the process
...
Simplify 2x ÷ 8x y
2x
8x y
2x
2×x
=
8x y 8 × x × y
1×1
=
4×1× y
1
=
4y

Problem 14
...


2x ÷ 8x y means

(A check can be made on this answer by
multiplying (2x + 3) by (x − 1), which equals
2x 2 + x − 3
...
Simplify

x 3 + y3
x+y

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

Practice Exercise 36 Basic operations in
algebra (answers on page 343)

x y2 + y3
x y2 + y3

...


1
...


2
...


(i)

x into x 3 goes x 2
...


3
...


(ii)

x 2 (x + y) = x 3 + x 2 y

4
...


5
...

(iv)

x into −x y goes −x y
...

2

(v) −x y(x + y) = −x 2 y − x y 2

6
...

(vii)

x into x y 2 goes y 2
...


(viii)

y 2 (x + y) = x y 2 + y 3

(ix) Subtract
...

Thus,
x+y
The zeros shown in the dividend are not normally shown,
but are included to clarify the subtraction process and
to keep similar terms in their respective columns
...


Divide 4a 3 − 6a 2 b + 5b3
2a 2 −

by 2a − b

8
...


9
...


25x 2 y z 3
5x yz

10
...

11
...

12
...

13
...

14
...


2ab − b2


2a − b 4a 3 − 6a 2 b
4a 3 − 2a2 b

+ 5b3

−4a 2 b

+ 5b3

15
...

16
...

17
...


−4a 2 b + 2ab2
−2ab2 + 5b3
−2ab2 + b3
4b3
4a 3 − 6a 2 b + 5b3
= 2a 2 − 2ab − b2 , remain2a − b
der 4b3
...
3

Laws of indices

The laws of indices with numbers were covered in
Chapter 7; the laws of indices in algebraic terms are
as follows:
(1) am ×an = am + n

For example, a 3 × a 4 = a 3+4 = a 7

Basic algebra
(2)

am
= am−n
an

(3) (am )n = amn
√
n

m

(4) a n =
(5) a−n =

am

1
an

(6) a0 = 1

c5
= c5−2 = c3
c2
 3
For example, d 2 = d 2×3 = d 6

For example,

4

For example, x 3 =

√
3

For example, 3−2 =

x4

1
1
=
32
9

a 3 b 2 c4
Problem 20
...

Problem 17
...
Simplify ( p3 )2 (q 2 )4

a 2 × a 1 × b 3 × b 2 × c1 × c5
Using law (3) of indices gives

Using law (1) of indices gives
a 2+1 × b3+2 × c1+5 = a 3 × b5 × c6
a 2 b3c × ab2 c 5 = a 3 b5 c 6

i
...


1

3

1

1

Problem 18
...
Simplify

Problem 22
...
Using law (3) of indices
gives
(mn 2 )3
m 1×3 n 2×3
m3n6
=
=
(m 1/2 n 1/4)4
m (1/2)×4n (1/4)×4 m 2 n 1
m3 n6
= m 3−2 n 6−1 = mn5
m2 n1
Problem 23
...
Simplify

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
...

1
...


a × a2 × a5

3
...


b4 × b7

5
...


c5 × c3 ÷ c4
(x 2 )(x)
x6
 2 −3
y
 −7 −2
c
 4
1
b3
1
 3
s3

9
...

13
...


17
...

12
...

16
...


x 3 y2 z
x 5 y z3

19
...

2
a 5 bc3

and
20
...


Simplify

when

Here are some further worked examples on the laws of
indices

x 2 y3 + x y2
xy
a +b
can be split
c

Algebraic expressions of the form
a b
into +
...

Problem 27
...

Dividing each term by x y gives

x2 y
evaluate

1

= (2)(33 )(2) = 108

Practice Exercise 37 Laws of indices
(answers on page 343)

8
...


When p = 16, q = 9 and r = 4,

x y2 − x y

Problem 28
...
Dividing
each term by a 1/2b gives
a 2b
a3/2
a 1/2b
=
=
2
1/2
3
2
1/2
3
ab − a b
a b
ab
a1/2 b−b2
−
a 1/2b
a 1/2b
a2b

√ √
√ √
3
Problem 29
...
Hence,
7

7

a2 b6 c

11
2

=


 
6
a7 b7 c11

1
When a = , b = 64 and c = 1,
4

 7 √
√

 

1
6
6
7
7
11
a b c =
647
111
4
 7
1
(2)7 (1) = 1
=
2
√ 
(x 2 y 1/2 )( x 3 y 2 )
Problem 30
...

Now try the following Practice Exercise
Practice Exercise 38 Laws of indices
(answers on page 343)



1
...

In problems 2 to 5, simplify the given expressions
...


a 2b + a 3 b
a 2b2

3
...

5
...


(abc)2
(a 2 b−1c−3 )3
p3 q 2
pq 2 − p2 q
√  √
√  
3
( x y 3 z 2 )( x y 3 z 3 )

7
...


8
...

The n’th term is: a + (n − 1)d
n
Sum of n terms, Sn = [2a + (n − 1)d]
2

Geometric progression:
If a = first term and r = common ratio, then the geometric progression is: a, ar, ar2 ,
...

4
...

10
...

16
...


Exercise 5 (page 11)

19 kg
2
...
479 mm
−66
5
...
−225
−2136
8
...
£10 7701
−4
11
...
5914
189 g
14
...
$15 333
89
...

3
...

5
...

9
...

9
...

17
...

(a) 8613 kg (b) 584 kg
(a) 351 mm (b) 924 mm
(a) 10 304 (b) −4433
6
...

(a) 8067 (b) 3347
10
...

4
...

8
...


(a) 48 m (b) 89 m
(a) 1648 (b) 1060
18 kg

1
...
14
7
...
88
8
...
1016/B978-1-85617-697-2
...


3 4 1 3 5
, , , ,
7 9 2 5 8
9
10
...
1
15
16
18
...


6
...

21
...
2
3
...
11
8
...

13
...

18
...

11
...

19
...

8
...

16
...


4
...
2

5
...
5

5
12
1
9
...

23
4
...
15

3
28
8
10
...


15
...
400 litres
22
...

15

1
...
59
6
...
−1

2
5

Exercise 6 (page 13)

11
...
7

(a) £1827 (b) £4158

Exercise 3 (page 6)
1
...

5
...

9
...
2

11
...
4
20
17
12
...
−

3
...
2

1
6

3
4
1
9
...
4

13
20
1
10
...


Answers to practice exercises
Exercise 14 (page 25)

Chapter 3

1
...
571
5
...
96
8
...
0871

Exercise 8 (page 17)
1
...
23
8
...


13
20
21
6
...
(a) 1
50
5
...
0
...


4
...
6

7
16

(e) 16

17
80

1
...

7
...

13
...
182
2
...
122
3
...
82
0
...
0
...
2
...
273
8
...
256
9
...
30366
6
1
...
3
...
37
...
2 × 10
14
...
767 ×10
15
...
32 ×106

12
...
6875 13
...
21875 14
...
1875

Exercise 16 (page 27)
1
...
4667

Exercise 9 (page 18)
1
...
18
5
...
297

2
...
785
3
...
38
6
...
000528

2
...
3
6
...
3

4
...
27

3
...
54
7
...
52 mm

4
...
83

13
14

3
...
458

6
...
7083

7
...
2
...
3

p(b + 2c)
4d(d − 3 f 5)
2q(q + 4n)
bc(a + b2 )
3x y(x y 3 − 5y + 6)
7ab(3ab − 4)


2x y x − 2y 2 + 4x 2 y 3
3x
17
...
(a + b)(y + 1)
22
...

3
...

7
...

11
...

15
...
0 19
...
( p + q)(x + y)
23
...

4
...

8
...

12
...

16
...
1

2
...
1

7
...
2
16
...
−4

3
...

2
13
...
−3

12
...
6

9
...
−2

2
...
4 + 3a
6
...
10y 2 − 3y +
9
...


1
− x − x2
5

1
7

1
4

8
...
5

2
...
−4

6
...
−4

8
...
−10

12
...
9

17
...
±12

22
...
−15t
12
...
2
...
2

5
...
2

10
...
3

14
...
−6

18
...
4

20
...
±3

24
...

4
...

6
...
8 m/s2
3
...
472
(a) 1
...
30 m/s2

Exercise 45 (page 80)
1
...
45◦ C
7
...
0
...
50
8
...
30
6
...
3
...
d = c − e − a − b

1
− 4x
3

10
...
2x + 8x 2
3
...

− 4x
2

5
...
R =
I
c
7
...
v =

y
7
v −u
4
...
y = (t − x)
3
y−c
8
...
x =

Answers to practice exercises
I
PR
E
11
...
C = (F − 32)
9
9
...
L =

XL
2π f

12
...
x = a(y − 3)
14
...
64 mm

1
2π CX C

*

1
√

Z2 −

14
...
1 × 10−6

aμ
ρCZ 4 n

2

Chapter 13
Exercise 47 (page 87)

Exercise 49 (page 92)

S −a
a
1
...
x =

yd
d
(y + λ) or d +
λ
λ

3
...
D =

AB 2
5E y

5
...
R2 =

R R1
R1 − R

E −e
E − e − Ir
or R =
−r
I
I

y

ay
8
...
x = 
2
4ac
(y 2 − b2 )

7
...
R =
πθ
√
Z 2 − R2
, 0
...
L =
2π f
10
...
u =




xy
1
...
r =
(1 − x − y)
c
5
...
b =
2( p2 + q 2 )
9
...
L =

8S 2
3d

Q
, 55
mc
+ d, 2
...
L =
μ−m


x−y
6
...
R = 4

uf
, 30
u− f


2dgh
, 0
...
v =
0
...
v =

x = 4, y = 2
x = 2, y = 1
...
5, n = 0
...

4
...

8
...

12
...

16
...
a = N 2 y − x

Exercise 48 (page 89)

1
...

5
...

9
...

13
...


1
...

5
...


p = −1, q = −2
a = 2, b = 3
x = 3, y = 4
x = 10, y = 15

2
...

6
...


x = 4, y = 6
s = 4, t = −1
u = 12, v = 2
a = 0
...
40

Exercise 51 (page 96)
1
1
1
...
p = , q =
4
5
5
...
x = 5, y = 1
4

1
1
2
...
x = 10, y = 5
1
6
...
1

Exercise 52 (page 99)
1
...

5
...

8
...
2, b = 4
u = 12, a = 4, v = 26
m = −0
...
00426, R0 = 22
...
I1 = 6
...
62
4
...
a = 12, b = 0
...
F1 = 1
...
5

Exercise 53 (page 100)
1
...
x = 5, y = −1, z = −2

2
...
x = 4, y = 0, z = 3

346 Basic Engineering Mathematics
5
...

9
...

11
...
x = 1, y = 6, z = 7
x = 5, y = 4, z = 2 8
...
5, y = 2
...
5
i1 = −5, i2 = −4, i3 = 2
F1 = 2, F2 = −3 F3 = 4

Exercise 57 (page 109)
1
...

6
...

10
...
191 s 2
...
345 A or 0
...
619 m or 19
...
066 m
1
...
165 m
12 ohms, 28 ohms

3
...

7
...


7
...
0133
86
...
4 or −4
4
...
5 or 1
...

10
...

16
...
−2 or −

2
3

2
...
0 or −
3
8
...
−3 or −7
14
...
−3
20
...
5

1
...


7
...

9
...
(a) 8 x + 8 x3 +
x +c
5

7x 2
+c
2
3
(b) t 8 + c
8
5 4
x +c
(b)
24
3
(b) 2t − t 4 + c
4
(b)

(b) 4θ + 2θ 2 +

5
(a) θ 2 − 2θ + θ 3 + c
2
3
2
3
(b) x 4 − x 3 + x 2 − 2x + c
4
3
2
4
1
(a) − + c
(b) − 3 + c
3x
4x
4√ 5
1√
4 9
(a)
x +c
(b)
x +c
5
9
15 √
10
5
(b)
x +c
(a) √ + c
7
t

7
(b) − cos 3θ + c
3
1
(b) 18 sin x + c
3
−2
(b)
+c
15e5x

13
...
(a) 4x + c

3
sin 2x + c
2
1
11
...
(a) e + c
8
10
...
(a) 1
...
5

2
...
5

3
...
333

4
...
75 (b) 0
...
(a) 10
...
1667

6
...
(a) 1 (b) 4
...
(a) 0
...
638

9
...
09 (b) 2
...
(a) 0
...
099

Exercise 141 (page 334)
1
...
37
...
1
...
proof
6
...
5
9
...
67

3
...
1
10
...
29

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