1.
y
1
=
−
2
x
+ 1
y
2 =
−
6
x
+ 3
2.
y
1
= 3
x
+ 6
y
2
= 3
x
+ 2
3.
y
1
=
x
−
3
y
2
=
−
2
x
+ 12
may have one solution, no solution, or an infinite number of solutions.
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Case I:
One solution
. The two graphs intersect at one point, which is the solution. We say that the equations are
independent.
It is a
consistent system
of equations. There is a point (an ordered pair)
consistent
with both equations.
Case II:
No solution
. The two graphs are parallel and so do not intersect. We say that the system of equations is
inconsistent
because there
is no point consistent with both equations.
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Case III:
An infinite number of solutions
. The graphs of each equation yield the same line. Every ordered pair on this line is a solution to
both of the equations. We say that the equations are
dependent.
EXAMPLE 7
If possible, solve the system.
2
x
+ 8
y
= 16
(1)
4
x
+ 16
y
=
−
8
(2)
Solution
To eliminate the variable y, we’ll multiply equation
(1)
by
−
2.
−
2(2
x
) + (
−
2)(8
y
) = (
−
2)(16)
−
4
x
−
16
y
=
−
32 (3)
We now have the following equivalent system.
−
4
x
−
16
y
=
−
32 (3)
4
x
+ 16
y
=
−
8 (2)
When we add equations
(3)
and
(2)
, we get
0 =
−
40,
which, of course, is false. Thus, we conclude that this system of equations is inconsistent, and
there is no solution.
Therefore, equations
(1)
and
(2)
do not intersect, as we can see on the graph to the right.
If we had used the substitution method to solve this system, we still would have obtained a false statement. When you try to solve an
inconsistent system of linear equations by any method, you will always obtain a mathematical equation that is not true.
Student Practice 7
If possible, solve the system.
4
x
−
2
y
= 6
−
6
x
+ 3
y
= 9
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281282
EXAMPLE 8
If possible, solve the system.
0.5
x
−
0.2
y
= 1.3
(1)
−
1.0
x
+ 0.4
y
=
−
2.6
(2)
Solution
Although we could work directly with the decimals, it is easier to multiply each equation by the appropriate power of 10 (10, 100, and so
on) so that the coefficients of the new system are integers. Therefore, we will multiply equations
(1)
and
(2)
by 10 to obtain the following
equivalent system.

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