Download Example 1 Determine if a System of Equations is Inconsistent

Question 1: Does every system have a unique solution?
For systems of two equations in two variables, it is possible to draw two lines that do not
intersect. A system like this is called an inconsistent system.
Figure 2 - The graphs of parallel lines do not cross.
For an inconsistent system, there are no ordered pairs that simultaneously solve both
systems since the graphs never cross. On a graph this may be difficult to diagnose. Two
lines that are almost parallel might be indistinguishable from two lines that are parallel.
Luckily when we attempt to find a solution with the substitution method or the
elimination method, parallel lines are easy to distinguish from lines that are almost
parallel.
Example 1
Determine if a System of Equations is Inconsistent
Decide if the system of equations
x
3 y  27
0.25 x  0.75 y  6.75
is inconsistent by attempting to solve the system.
Solution This is the same system we solved in Section 2.1 Example 4.
In that example, we graphed each equation and showed that lines were
parallel. In this example, we’ll solve the equations by one of two
possible strategies, substitution or elimination. We’ll look at each of
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these strategies to ensure that you understand how you could solve
either way.
Substitution Method
It is easy to solve for x in the first equation by adding 3y to both sides of
the equation. This yields
x  3 y  27
Now replace the x in the second equation with 3 y  27 :
0.25  3 y  27   0.75 y  6.75
We can simplify the left hand side of the equation to give us
0.75 y  6.75  0.75 y  6.75
6.75  6.75
Remove the parentheses
Combine like terms
Notice that the variable has dropped out of the equation. The resulting
equation makes no sense since -6.75 cannot equal 6.75. This
contradiction tells us that this system has no solutions.
Elimination Method
The leading coefficient in the first equation is a 1, we begin by
eliminating x in the second equation. Either variable can be chosen, but
in this system it is easier to eliminate x. Multiply the first equation by
0.25 and add it to the second equation:
0.25 x  0.75 y  6.75
0.25 x  0.75 y  6.75
0  13.5
Second equation
0.25 times the first equation
New second equation
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Notice that y is eliminated at the same time x is eliminated. The sum of
these equations results in a contradiction so the system has no
solutions.
In each method, solving the system results in a contradiction.
Contradictions such as these indicate that there are no solutions.
So far we have seen that a system may have a unique solution (one ordered pair solves
the system) or no solutions at all. Other systems of equations have nonunique solutions.
For these systems, there are many ordered pairs that satisfy each equation in the
system. We may still use the Substitution Method or the Elimination Method to solve
systems of equations with nonunique solutions.
Example 2
Does the System of Equations have a Unique Solution?
Decide if the system of equations
y  2 x  10
x  12 y  5
has a unique solution by attempting to solve the system.
Solution As in the previous example, we can solve this system
algebraically using the Substitution Method or the Elimination Method.
In this example we’ll solve the system with both methods. In general,
you only need to use one of these strategies to solve the system.
Substitution Method
This system is ideal for the Substitution Method since the first equation
is already solved for one of the variables. Substitute 2 x  10 in place of
y in the second equation to yield
x  12  2 x  10   5
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Now simplify the left hand side to solve for x:
x  12  2 x   12 10   5
Remove the parentheses
x x5  5
Simplify each term
55
Combine like terms
All the variables have dropped out and the resulting equation is always
true. This signals that there is not a unique solution to the system. Many
ordered pairs will satisfy both equations in the system of equations.
Elimination Method
To use the Elimination Method, we need to have all of the terms with
variables on one side of the equals sign and the constant terms on the
other side. The first equation does not have this format so add 2x to
both sides of the equation to give
2 x  y  10
Replacing the first equation in the system with this equivalent equation
leads to
2 x  y  10
x  12 y  5
We could multiply the first equation by
1
2
to make the leading coefficient
a 1. But the second equation has a leading coefficient of 1 so we’ll
interchange the equations to give an equivalent system,
x  12 y  5
2 x  y  10
To eliminate x from the second equation, multiply the first equation by 2 and add it to the second equation:
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2 x  y  10
2 x  y  10
0 0
-2 times the first equation
Second equation
New second equation
This new equation helps us to rewrite the system as the equivalent
system
x  12 y  5
00
As in the Substitution Method, when all of the variables drop out and
leave us with a true statement, there is not a unique solution. Since
there is not an equation that specifies y to be a unique number, it can
be chosen to be any number. Then the corresponding x value is found
using the other equation in the system, x  12 y  5 . The easiest way to
do this is to solve for x first to give x   12 y  5 .
We could also use the other equation in the original system,
y  2 x  10 , to find solutions to this system. In this case, we could
specify x to be any number. Then the corresponding value of y is found
from the equation y  2 x  10 .
In either case, the ordered pairs we find lie on a line. Any ordered pair
on that line is a solution to the system. To locate points on this line,
simply substitute a value for x in the equation y  2 x  10 or substitute
a value for y in the equation x   12 y  5 . We can pick any reasonable
value for x and use the equation y  2 x  10 to find the corresponding
value for y:
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If we pick x
equal to
then the value for y is
and the corresponding
solution to the system is
0
y  2  0   10  10
(0, 10)
10
y  2 10   10  10
(10, -10)
12
y  2 12   10  14
(12, -14)
We can also pick a value for y and use the equation x   12 y  5 to find
the corresponding value for x:
If we pick y
equal to
then the value for x is
and the corresponding
solution to the system is
10
x   12 10   5  0
(0, 10)
-10
x   12  10   5  10
(10, -10)
-14
x   12  14   5  12
(12, -14)
In either case, we find the exact same ordered pairs are solutions to the
system. To write this solution formally, we say that the solution is all
ordered pairs  x, y  where y  2 x  10 and x is any real number. You
can think of each ordered pair corresponding to an arbitrary x value and
a y value that is calculated from the equation of the line. Or you can
think of each ordered pair corresponding to an arbitrary y value and an x
value that is calculated from the equation of the line.
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