Algebra 3
3c + 2c × 4c + c ÷ 5c − 8c
= 3c + 2c × 4c +
c
5c
− 8c
= 3c + 8c
2
+
1
5
− 8c
= 8c
2
− 5c +
1
5
or c(8c − 5) +
1
5
Problem 10. Simplify
(2a − 3) ÷ 4a + 5 × 6 −3a.
(2a − 3) ÷ 4a + 5 × 6 − 3a
=
2a − 3
4a
+ 5 × 6 − 3a
=
2a − 3
4a
+ 30 − 3a
=
2a
4a
−
3
4a
+ 30 − 3a
=
1
2
−
3
4a
+ 30 − 3a = 30
1
2
−
3
4a
− 3a
Now try the following exercise
Exercise 2 Further problems on brackets,
factorization and precedence
1. Simplify 2( p + 3q − r) − 4(r − q + 2 p) + p.
[−5 p + 10q − 6r]
2. Expand and simplify (x + y)(x − 2y).
[x 2 − x y − 2y 2 ]
3. Remove the brackets and simplify:
24 p − [2{3(5 p − q) − 2( p + 2q)} + 3q].
[11q − 2 p]
4. Factorize 21a 2 b 2 − 28ab.
[7ab(3ab − 4)]
5. Factorize 2x y 2 + 6x 2 y + 8x 3 y.
[2x y(y + 3x + 4x 2 )]
6. Simplify 2y + 4 ÷ 6y + 3 × 4 − 5y.
2
3y
− 3y + 12
7. Simplify 3 ÷ y + 2 ÷ y − 1.
5
y
− 1
8. Simplify a 2 − 3ab × 2a ÷ 6b + ab.
[ab]
1.3 Revision of equations
(a) Simple equations
Problem 11. Solve 4 − 3x = 2x − 11.
Since 4 − 3x = 2x − 11 then 4 + 11 = 2x + 3x
i.e. 15 = 5x from which, x =
15
5
= 3
Problem 12. Solve
4(2a − 3) − 2(a − 4) = 3(a − 3) − 1.
Removing the brackets gives:
8a − 12 − 2a + 8 = 3a − 9 − 1
Rearranging gives:
8a − 2a − 3a = −9 − 1 + 12 − 8
i.e.
3a = −6
and
a =
−6
3
= −2
Problem 13. Solve
3
x − 2
=
4
3x + 4
.
By ‘cross-multiplying’:
3(3x + 4) = 4(x − 2)
Removing brackets gives:
9x + 12 = 4x − 8
Rearranging gives:
9x − 4x = −8 − 12
i.e.
5x = −20
and
x =
−20
5
= −4
Problem 14. Solve
√
t + 3
√
t
= 2.