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
- Simplify2(p+ 3 q−r)− 4 (r−q+ 2 p)+p.
[− 5 p+ 10 q− 6 r] - Expand and simplify(x+y)(x− 2 y).
[x^2 −xy− 2 y^2 ] - Remove the brackets and simplify:
24 p−[2{ 3 ( 5 p−q)− 2 (p+ 2 q)}+ 3 q].
[11q− 2 p]
Factorize 21a^2 b^2 − 28 ab.[7ab( 3 ab− 4 )]
Factorize 2xy^2 + 6 x^2 y+ 8 x^3 y.
[2xy(y+ 3 x+ 4 x^2 )]
Simplify 2y+ 4 ÷ 6 y+ 3 × (^4) [− 5 y.
2
3 y
− 3 y+ 12
]
Simplify 3÷y+ 2 ÷y−1.
[
5
y
− 1
]
- 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.