Higher Engineering Mathematics, Sixth Edition

Algebra 3

3 c+ 2 c× 4 c+c÷ 5 c− 8 c

= 3 c+ 2 c× 4 c+

(c

5 c

)

− 8 c

= 3 c+ 8 c^2 +

1

5

− 8 c

= 8 c^2 − 5 c+

1

5

or c(8c−5)+

1

5

Problem 10. Simplify

( 2 a− 3 )÷ 4 a+ 5 × 6 − 3 a.

( 2 a− 3 )÷ 4 a+ 5 × 6 − 3 a

=

2 a− 3

4 a

+ 5 × 6 − 3 a

=

2 a− 3

4 a

+ 30 − 3 a

=

2 a

4 a

−

3

4 a

+ 30 − 3 a

=

1

2

−

3

4 a

+ 30 − 3 a= 30

1

2

−

3

4 a

− 3 a

Now try the following exercise

Exercise 2 Further problems on brackets,

factorization and precedence

1. Simplify2(p+ 3 q−r)− 4 (r−q+ 2 p)+p.
   [− 5 p+ 10 q− 6 r]


2. Expand and simplify(x+y)(x− 2 y).
   [x^2 −xy− 2 y^2 ]


3. Remove the brackets and simplify:

24 p−[2{ 3 ( 5 p−q)− 2 (p+ 2 q)}+ 3 q].

[11q− 2 p]

4. 
   

   Factorize 21a^2 b^2 − 28 ab.[7ab( 3 ab− 4 )]

   
   


5. 
   

   Factorize 2xy^2 + 6 x^2 y+ 8 x^3 y.
   [2xy(y+ 3 x+ 4 x^2 )]

   
   


6. 
   

   Simplify 2y+ 4 ÷ 6 y+ 3 × (^4) [− 5 y.
   2
   3 y
   − 3 y+ 12
   ]

   
   


7. 
   

   Simplify 3÷y+ 2 ÷y−1.

   
   

[

5

y

− 1

]

8. Simplifya^2 − 3 ab× 2 a÷ 6 b+ab.[ab]

1.3 Revision of equations

(a) Simple equations

Problem 11. Solve 4− 3 x= 2 x− 11.

Since 4− 3 x= 2 x−11 then 4+ 11 = 2 x+ 3 x

i.e. 15= 5 xfrom which,x=

15

5

= 3

Problem 12. Solve

4 ( 2 a− 3 )− 2 (a− 4 )= 3 (a− 3 )− 1.

Removing the brackets gives:

8 a− 12 − 2 a+ 8 = 3 a− 9 − 1

Rearranging gives:

8 a− 2 a− 3 a=− 9 − 1 + 12 − 8

i.e. 3 a=− 6

and a=

− 6

3

=− 2

Problem 13. Solve

3

x− 2

=

4

3 x+ 4

.

By ‘cross-multiplying’: 3( 3 x+ 4 )= 4 (x− 2 )

Removing brackets gives: 9x+ 12 = 4 x− 8

Rearranging gives: 9 x− 4 x=− 8 − 12

i.e. 5 x=− 20

and x=

− 20

5

=− 4

Problem 14. Solve

(√

t+ 3

√

t

)

=2.