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Chapter 1

Algebra
1
...
4
to 1
...
However, before this, some essential algebra
revision on basic laws and equations is included
...
elsevier
...
2

Multiply by x → 3x 2 + 2x y
Multiply by −y →

3x 2 − xy − 2y 2

Adding gives:
Alternatively,

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

Revision of basic laws

= 3x 2 − xy − 2y 2

(a) Basic operations and laws of indices
The laws of indices are:
(i) a m × a n = a m+n
(iii)

(a m )n

(v)

a −n

=

a m×n

1
= n
a

am
(ii)
= a m−n
an
√
m
(iv) a n = n a m
(vi)

a0

Problem 3
...


When a = 3, b =
Problem 1
...
Multiply 3x + 2y by x − y
...
Simplify

x 2 y3 + x y2
xy

x 2 y3 + x y2
x 2 y3 x y2
=
+
xy
xy
xy
= x 2−1 y 3−1 + x 1−1 y 2−1
= xy 2 + y or y(xy + 1)

2 Higher Engineering Mathematics

Problem 5
...


2

x2 y 2 x 2 y 3

=

x y

a 2 − (2a − ab) − a(3b + a)
= a 2 − 2a + ab − 3ab − a 2
= −2a − 2ab or −2a(1 + b)

= x 2+ 2 − 2 y 2 + 3 − 2
1

5

1

2

3

= x 0 y− 3
1

Problem 7
...


1
y

1
3

1
or √
3 y

Remove the brackets and simplify the

2a − [3{2(4a − b) − 5(a + 2b)} + 4a]
...
Evaluate 2ab + 3bc − abc when a = 2,
b = −2 and c = 4
...
Find the value of 5 pq 2r 3 when p = 25 ,
q = −2 and r = −1
...
From 4x − 3y + 2z subtract x + 2y − 3z
...
Multiply 2a − 5b + c by 3a + b
...
Simplify (x y z)(x yz ) and evaluate when
[x 5 y 4 z 3 , 13 12 ]
x = 12 , y = 2 and z = 3
...
Simplify

− 21

c) when a = 3,
[±4 12 ]


a2b + a3b
a 2 b2
1
2

− 12

(a 3 b c )(ab)
√ √
( a 3 b c)


11

36b − 11a

2

6
...

7
...
Factorize (a) x y − 3x z
(b) 4a 2 + 16ab3 (c) 3a 2 b − 6ab 2 + 15ab
...


Simplify 3c + 2c × 4c + c ÷ 5c − 8c
...
Hence

Algebra
3c + 2c × 4c + c ÷ 5c − 8c
c 
= 3c + 2c × 4c +
− 8c
5c
1
= 3c + 8c2 + − 8c
5
1
1
= 8c2 − 5c +
or c(8c − 5) +
5
5


7
...

8
...


1
...
Simplify
(2a − 3) ÷ 4a + 5 × 6 −3a
...
Solve 4 − 3x = 2x − 11
...
e
...
Solve

3
2a
−
+ 30 − 3a
=
4a 4a

4(2a − 3) − 2(a − 4) = 3(a − 3) − 1

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