is to rearrange the equation that is easier to rearrange. In the present case
by looking at the two equations it seems that equation (1) might be easier to
rearrange.
3
1
6
2
6
2
3
1
2
4
6
4
6
2
(3)
2
2
4
2
2
4
4
x
x
x
y
y
x
y
y
x

+
=
⇒
=

⇒
=
=

⇒
=

Now let us put (3) into (2): what we have to do here is to replace
3
1
2
2
y by
x

:
3
1
9
3
3
9
4
3
3
4
3
3
4
3
4
3
2
2
2
2
2
2
2
(
4
)
3
2
3
9
8
3
6
9
2
2
2
2
5
15
15
5
15
3
(4)
2
2
5
x
y
x
x
x
x
x
x
y
x
x
x
x
x
x
x


=
⇒ 


=
⇒ 

+
=
⇒ 
+
=
+
× 
+
×
+

+
+
⇒
=
⇒
=

⇒
=
⇒ 
=
⇒
=
= 

14243
Now putting (4) in (3) we get:
3
1
3
3
6
( 3)
3
2
2
2
2
2
y
y
y
=

× 
⇒
=
+
⇒
=
=
Therefore, the solution to our system of equations is
3
3
x
and y
= 
=
.
Note
: Now if you hate fractions you can write
3
9
1.5
4.5
2
2
x as
x and
as
. So you
will have:
7.5
4
1.5
3
4.5
2.5
7.5
3
2.5
x
x
x
x

+
=
+
⇒ 
=
⇒
=
= 

.
According to (3),
3
1
1.5
0.5
2
2
y
x
y
x
=

⇒
=

.
3
we knowthat x
= 
1.5
0.5( 3)
1.5
1.5
3
therefore y
=


=
+
=
.
Subscribe to view the full document.
30
2.3.2 The elimination method
In this method one has to subtract (or add) the equations from (or to) one
another. Again let us try to solve our first example using the elimination
method.
4
0
3
7
x
y
x
y

=
+
=
As before we have to label these equations:
4
0
(1)
3
7
(2)
x
y
x
y

=
+
=
Now we need to match up the numbers in front of the
'
'
x s or y s
so that
when we subtract (or add) one equation from (or to) the other, one of the
variable disappear. Again we need to choose the easiest option. Looking at
the system we can see that by adding (1) to (2) the
'
y s
disappear (or they
are eliminated)
4
0
(1)
3
7
(2)
7
7
0
7
7
7
1
7
x
y
x
y
x
x
x

=
+
+
=
+
=
⇒
=
⇒
=
=
We need now to replace
x
by its value in any of the equations (it does not
matter which one; but always choose the easiest one to calculate). Looking at
the system the easiest one to calculate is the first equation, (1).
4
0
we found that
1
4
1
0
4
0
4
x
y
and
x
y
y
y

=
=
⇒
×

=
⇒

=
⇒
=
As you can see we arrived to the same results as using the substitution
method.
We can also try our second example using the elimination approach.
2
4
6
4
3
3
x
y
x
y
+
=


=
Again we have to label these equations: