2 Higher Engineering Mathematics
Problem 5. Simplify
(x 2 √ y)(
√
x
3
y 2 )
(x 5 y 3 )
1
2
(x 2 √ y)(
√
x
3
y 2 )
(x 5 y 3 )
1
2
=
x 2 y
1
2 x
1
2 y
2
3
x
5
2 y
3
2
= x
2+ 1
2 − 5
2 y
1
2 + 2
3 − 3
2
= x
0 y
− 1
3
= y
−
1
3
or
1
y
1
3
or
1
3
√
y
Now try the following exercise
Exercise 1 Revision of basic operations
and laws of indices
1. Evaluate 2ab + 3bc − abc when a = 2,
b = −2 and c = 4.
[−16]
2. Find the value of 5 pq 2 r 3 when p =
2
5 ,
q = −2 and r = −1.
[−8]
3. From 4x − 3y + 2z subtract x + 2y − 3z.
[3x − 5y + 5z]
4. Multiply 2a − 5b + c by 3a + b.
[6a 2 − 13ab + 3ac − 5b 2 + bc]
5. Simplify (x
2 y
3 z)(x
3 yz
2
) and evaluate when
x =
1
2 , y = 2 and z = 3.
[x 5 y 4 z 3 , 13
1
2 ]
6. Evaluate (a
3
2 bc −3 )(a
1
2 b
− 1
2 c) when a = 3,
b = 4 and c = 2.
[±4
1
2 ]
7. Simplify
a 2 b + a 3 b
a 2 b 2
1 + a
b
8. Simplify
(a 3 b
1
2 c
− 1
2 )(ab)
1
3
(
√
a 3
√
b c)
a
11
6 b
1
3 c
−
3
2
or
6
√
a
11 3
√
b
√
c 3
(b) Brackets, factorization and precedence
Problem 6. Simplify a 2 − (2a − ab) − a(3b + a).
a
2
− (2a − ab) − a(3b + a)
= a
2
− 2a + ab − 3ab − a
2
= −2a − 2ab or −2a(1 + b)
Problem 7. Remove the brackets and simplify the
expression:
2a − [3{2(4a − b) − 5(a + 2b)} + 4a].
Removing the innermost brackets gives:
2a − [3{8a − 2b − 5a − 10b} + 4a]
Collecting together similar terms gives:
2a − [3{3a − 12b} + 4a]
Removing the ‘curly’ brackets gives:
2a − [9a − 36b + 4a]
Collecting together similar terms gives:
2a − [13a − 36b]
Removing the square brackets gives:
2a − 13a + 36b = −11a + 36b or
36b − 11a
Problem 8. Factorize (a) x y − 3xz
(b) 4a 2 + 16ab 3 (c) 3a 2 b − 6ab 2 + 15ab.
(a) x y − 3xz = x( y − 3z)
(b) 4a 2 + 16ab 3 = 4a(a + 4b
3 )
(c) 3a 2 b − 6ab 2 + 15ab = 3ab(a − 2b + 5)
Problem 9. Simplify 3c + 2c × 4c + c ÷ 5c − 8c.
The order of precedence is division, multiplication, addition and subtraction (sometimes remembered
by BODMAS). Hence
Problem 5. Simplify
(x 2 √ y)(
√
x
3
y 2 )
(x 5 y 3 )
1
2
(x 2 √ y)(
√
x
3
y 2 )
(x 5 y 3 )
1
2
=
x 2 y
1
2 x
1
2 y
2
3
x
5
2 y
3
2
= x
2+ 1
2 − 5
2 y
1
2 + 2
3 − 3
2
= x
0 y
− 1
3
= y
−
1
3
or
1
y
1
3
or
1
3
√
y
Now try the following exercise
Exercise 1 Revision of basic operations
and laws of indices
1. Evaluate 2ab + 3bc − abc when a = 2,
b = −2 and c = 4.
[−16]
2. Find the value of 5 pq 2 r 3 when p =
2
5 ,
q = −2 and r = −1.
[−8]
3. From 4x − 3y + 2z subtract x + 2y − 3z.
[3x − 5y + 5z]
4. Multiply 2a − 5b + c by 3a + b.
[6a 2 − 13ab + 3ac − 5b 2 + bc]
5. Simplify (x
2 y
3 z)(x
3 yz
2
) and evaluate when
x =
1
2 , y = 2 and z = 3.
[x 5 y 4 z 3 , 13
1
2 ]
6. Evaluate (a
3
2 bc −3 )(a
1
2 b
− 1
2 c) when a = 3,
b = 4 and c = 2.
[±4
1
2 ]
7. Simplify
a 2 b + a 3 b
a 2 b 2
1 + a
b
8. Simplify
(a 3 b
1
2 c
− 1
2 )(ab)
1
3
(
√
a 3
√
b c)
a
11
6 b
1
3 c
−
3
2
or
6
√
a
11 3
√
b
√
c 3
(b) Brackets, factorization and precedence
Problem 6. Simplify a 2 − (2a − ab) − a(3b + a).
a
2
− (2a − ab) − a(3b + a)
= a
2
− 2a + ab − 3ab − a
2
= −2a − 2ab or −2a(1 + b)
Problem 7. Remove the brackets and simplify the
expression:
2a − [3{2(4a − b) − 5(a + 2b)} + 4a].
Removing the innermost brackets gives:
2a − [3{8a − 2b − 5a − 10b} + 4a]
Collecting together similar terms gives:
2a − [3{3a − 12b} + 4a]
Removing the ‘curly’ brackets gives:
2a − [9a − 36b + 4a]
Collecting together similar terms gives:
2a − [13a − 36b]
Removing the square brackets gives:
2a − 13a + 36b = −11a + 36b or
36b − 11a
Problem 8. Factorize (a) x y − 3xz
(b) 4a 2 + 16ab 3 (c) 3a 2 b − 6ab 2 + 15ab.
(a) x y − 3xz = x( y − 3z)
(b) 4a 2 + 16ab 3 = 4a(a + 4b
3 )
(c) 3a 2 b − 6ab 2 + 15ab = 3ab(a − 2b + 5)
Problem 9. Simplify 3c + 2c × 4c + c ÷ 5c − 8c.
The order of precedence is division, multiplication, addition and subtraction (sometimes remembered
by BODMAS). Hence