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Linear Systems of Equations in two variables
Example 1
:
Question
: Solve the system of equations and interpret the solution graphically:
{
=
y
+
x
2
=
+
x
3
y
18
.
Solution
:
The problem is to find two numbers
x
and
y
such that both of the given equations hold simultaneously. We consider three different methods
of obtaining the solution.
Method I
:
Subsitution method.
{
=
y
+
x
2

( )
1
=
+
x
3
y
18

( )
2
Substituting
=
y
+
x
2
from equation (1) in equation (2) gives:
=
+
x
3 (
)
+
x
2
18
<=>
=
+
+
x
3
x
6
18
<=>
=
4
x
12
<=>
=
x
3
When
=
x
3,
equation (1) gives
=
y
5.
The two values
=
x
3
and
=
y
5
can be checked in equation (2).
The
solution
is given by
=
x
3
and
=
y
5.
Method II
:
First elimination method.
Arrange the equations with the unknowns
x
and
y
on the left side with the
x
term before the
y
term.
{
=
y
+
x
2

( )
1
=
+
x
3
y
18

( )
2
<=>
{
=
 +
x
y
2

( )
3
=
+
x
3
y
18

( )
4
The variable
x
can be "
eliminated
" by adding the equations (3) and (4) together (adding the two lefthand sides together and the two
righthand sides together) to give
=
4
y
20.
This gives
=
y
5.
Then
x
can be obtained by substituting this value in equation (1) to give
=
5
+
x
2, so that
=
x
3.
Alternatively, multiplying both sides of equation (3) by 3 will mean that the
y
terms are the same in the two equations.
<=>
{
=

+
3
x
3
y
6

( )
5
=
+
x
3
y
18

( )
6
Subtracting equation (5) from equation (6) ( subtracting the lefthand side of equation (5) from the lefthand side of (6) and the righthand
side of equation (5) from the righthand
side of (6) )
eliminates
y
to give:
=
4
x
12.
Hence
=
x
3.
The
solution
is given by
=
x
3
and
=
y
5.
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View Full Document Method III
:
Second elimination method.
{
=
y
+
x
2

( )
1
=
+
x
3
y
18

( )
2
Arrange each of the equations in the standard form
=
y
+
m x
b
for the equation of a straight line with slope
m
and
y
intercept
b
.
The first equation already has this form, and the second equation is equivalent to
=
3
y
 +
x
18
which gives
=
y

1
3
+
x
6.
=
y
+
x
2

( )
7
=
y

+
x
3
6

( )
8
The variable
y
can be eliminated between equations (7) and (8) simply by equating the two right hand sides, to give
=
+
x
2

1
3
+
x
6
 (9)
This equation is equivalent to
+
x
1
3
=
x
4
that is,
4
3
=
x
4
Hence
=
x
3
1
so that
=
x
3.
Substituting this value for
x
in equation (7) gives
=
y
5.
The
solution
is given by:
=
x
3
and
=
y
5.
Graphical illustration of the solution
.
Consider the system in the form used for the third solution method in which both equations are arranged in the standard form
=
y
+
m x
b
for the equation of a straight line.
=
y
+
x
2

( )
7
=
y

+
x
3
6

( )
8
Each pair of real numbers (
)
,
x y
that satisfy the equation
=
y
+
x
2 corresponds a point on the straight line graph of this first equation, and
similarly, the each pair of real numbers (
)
,
x y
that satisfy the equation
=
y

1
3
+
x
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