after expansion. Now let us try to factorise
2
6
4
x
xy
+
.
{
}
{
}
{
}
{
}
(
29
(
29
2
2
4
6
6
4
6
4
6
4
3
2
2
2
2
3
2
2
3
2
×
×
+
=
×
+
×
=
×
×
×
+
×
×
×
=
×
×
+
×
=
+
E5555F
E5555F
&&&
&&&
E555555555F
E555555555F
xy
x
x x
x y
x
xy
x
x
x
y
x
x
x
y
x
x
y
x
x
y
.
You don’t have to go through these steps to get your answer.
This is just
to show you how we arrived to our solution. This question could have been
answered in one go, as bellow.
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16
(
29
2
6
4
2
3
2
x
xy
x
x
y
+
=
+
.
Now let us try to factorise (a)
2
2
x y
xy

and (b)
3
2
7
x
x
+
.
(a)
{
}
{
}
(
29
(
29
2
2
2
2


=
×
×
+
× 
×
=

=

&&&
&&&
&&&
E5555555F
E55555555F
x y
xy
x y
xy
x
x
y
x
y
y
x
y
x
y
xy x
y
.
Or in one go:
(
29
2
2
x y
xy
xy x
y

=

.
(b)
{
}
{
}
(
29
(
29
3
2
3
2
2
7
7
7
7
7
+
=
×
×
+
×
×
=
+
=
+
&&&
&&&
&&&
E5555555F
E5555555F
x
x
x
x
x
x
x
x
x
x
x
x
x
x
Or in one go
(
29
3
2
2
7
7
x
x
x
x
+
=
+
Extra Exercises for Practice
Factorise
2
2
3
2
2
2
2
2
( )
2
( )
( )
1
(
)
2
18
(
) 2
9

+


+


a
a
ab
b
b
x
x
c z
d
p
p
e
q
q
Exercise 1.3.1
Factorise
(
) 5
25
a
x

;
(
) 27
18
b
x
+
;
( )
8
24
c
x

+
;
2
(
)
4
d
x
x
+
;
2
2
( )
e x y
xy
+
;
2
(
)
4
d
x
x
+
;
2
2
( )
e x y
xy
+
;
2
(
) 4
4
f
a x
a

;
2
(
) 24
8
g
a x
a
+
17
2. Solving Equations
We will try to solve three types of equations: linear, quadratic, and
simultaneous. For this section we might need to use some of the tools we
have learned in the previous sections.
2.1 Solving Linear Equations
A linear equation is one that can be written in the form
0
ax
b
+
=
where a and
b are numbers and the unknown quantity is x. For example
4
1
0
x
+
=
and
1
3
0
2
x

=
are both linear equations. Linear equations may also appear in the
following forms (which appeared to be different from
0
ax
b
+
=
but are
equivalent):
2
1
3
x

=
;
3
4
1
x
x

+
=
+
; and
(
29
3
2
6
0
x

+
=
. These equations can be
rearranged to get the form
0
ax
b
+
=
, as shown bellow.
2
1
3
x

=
2
1
3
3
3
x


=

2
4
0
x

=
Now let us move to the second expression:
3
4
1
x
x

+
=
+
3
4
1
x
x

+
=
+
Subtract 3 from both sides so that the right hand side becomes 0
Simplify this expression
This expression is of the form
0
ax
b
+
=
Simplify this expression by bearing in mind that when you move
on the other side of the “=” you change sign change
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18
4
1
3
5
2
x
x
x


=

⇒ 
= 
5
2
2
2
x

+
=  +
5
2
0
x

+
=
Finally,
(
29
3
2
6
0
x

+
=
can also be expressed in the form
0
ax
b
+
=
.
(
29
3
2
6
0
x

+
=
6
18
0
x

+
=
Now let us try to solve some linear equations: (a)
10
0
x
+
=
; (b)
5
100
0

+
=
x
;
(c)
4
6
3
x

= 
; (d)
(
29
3
2
5
x
+
=
; and (e)
5
2
2
3
x
x
+
= 

.
(a)
10
0
10
x
x
+
=
⇒
= 
(as you can see if you replace x by –10 you get 0)
(b)
100
5
100
0
5
100
20
5
x
x
x


+
=
⇒ 
= 
⇒
=
=

(c)
3
4
6
3
4
3
6
4
3
4
x
x
x
x

=  ⇒
= 
+
⇒
=
⇒
=
(d)
(
29
1
3
2
5
3
6
5
3
5
6
3
1
3
x
x
x
x
x

+
=
⇒
+
=
⇒
=

⇒
=  ⇒
=
(e)
5
5
2
2
3
5
2
3
2
7
5
7
x
x
x
x
x
x

+
= 

⇒
+
=  
⇒
=  ⇒
=
.
Add +2 to both sides so that the right hand side is 0
Simplify
This expression is of the form
0
ax
b
+
=
Expand the brackets
This expression is of the form
0
ax
b
+
=