Question 2: How do you solve a system of three equations in two variables?
To solve a system of three equations in two variables, we extend the Substitution and
Elimination Methods introduced earlier. A third equation in two variables simply adds a
third line to the system. A solution to the system is an ordered pair that solves all of the
equations in the system.
Figure 3 – Two systems of three equations in two variables. In Figure 4a, the solution to the system is
(4,1). In Figure 4b, there is no solution since all three lines do not intersect at a single point.
Graphically, this is a point of intersection where all of the lines intersect. In Figure 3a,
three lines are graphed corresponding to the three lines in a linear system. All of the
lines intersect at the point  4,1 which means it satisfies each of the equations in the
system. If you consider only two of the lines in the system and use the Substitution
Method or the Elimination Method, the resulting solution will also satisfy the third line in
the system.
The system in Figure 3b has several points, such as  2, 3 , where two of the lines in the
system intersect. If you use the Substitution Method or the Elimination Method on these
two lines, you would find a solution of x  2 and y  3 . However, if this ordered pair is
substituted in the other equation in the system, it will not be satisfied. This means that
there is no ordered pair that satisfies all three equations simultaneously so the system is
inconsistent.
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Example 3
Find the Solution to the System
Solve the system
x  2y  2
2 x  8 y  16
x y 5
Solution Although we could use the Substitution Method to solve this
system, the Elimination Method will be used in this example since it
generalizes to larger systems much more easily. The leading coefficient
of the first equation is already a 1, so we need to eliminate x from the
other two equations.
 2 x  4 y  4
2 x  8 y  16
12 y  12
 x  2 y  2
x y 5
3y  3
Second equation
-2 times the first equation
New second equation
-1 times the first equation
Third equation
New third equation
Replace the second and third equations in the original system of
equations with these new equations to give
x  2y  2
12 y  12
3y  3
Multiply the second equation by
1
12
to give the equivalent system of
equations,
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x  2y  2
y 1
3y  3
To complete the Elimination Method, we need to eliminate y from the
first and third equations:
2y  2
2 times the second equation
x  2y  2
4
x
3 y  3
3y  3
First equation
New first equation
-3 times the second equation
Third equation
00
New third equation
This helps us to write the equivalent system of equations,
x4
y 1
00
The last equation is an identity which means it is always true. The first
two equations indicate that the solution to two of the equations is
 x, y    4,1 . This and the identity indicate that this ordered pair also
solves the third equation. We can check this by substituting  4,1 into
each equation:
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?
First equation: 4  2 1  2
2  2 TRUE
?
Second equation: 2  4   8 1  16
16  16 TRUE
?
Third equation: 4  1  5
5  5 TRUE
Since  x, y    4,1 satisfies each equation in the system, it is the
solution to the system of equations. Furthermore, since the first two
equation specify a unique solution and not a relationship between the
variables, this is the only solution to the original system of equations.
Example 4
Find the Solution to the System
Solve the system
10 x  y  44
 12 x  y  2
x  2y  8
Solution Start by solving any two of the equations for a solution. The
Elimination Method provides a systematic approach for solving this
system. To make the leading coefficient on the first equation a 1,
interchange the first and third equation to yield
x  2y  8
 12 x  y  2
10 x  y  44
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Now eliminate x from the second and third equations:
1
2
x y 4
 12 x  y  2
2y  6
10 x  20 y  80
10 x 
y  44
 21y  36
1
2
times the first equation
Second equation
New second equation
-10 times the first equation
Third equation
New third equation
Replace the second and third equations with these new equations to
give an equivalent system of equations,
x  2y  8
2y  6
21y  36
Multiply the second equation by
1
2
to change the leading coefficient of
the second equation to a 1. This leaves us with the equivalent system of
equations,
x  2y  8
y 3
21y  36
Finally, eliminate x in the first and third equations:
 2 y  6
x  2y  8
x
2
-2 times the second equation
First equation
New first equation
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21 y  63
21 times the second equation
21 y  36
New third equation
0  27
Third equation
Use these new equations to write the equivalent system of equations,
x2
y3
0  27
This indicates that  x, y    2,3 satisfies two of the equations, but not
the third. The third equation is a contradiction. So the system is
inconsistent and has no solutions. If we substitute  x, y    2,3 into the
original system of equations, we can discover which equations are
satisfied and which are not:
?
First equation: 10  2   3  44
17  44 FALSE
?
Second equation:  12  2   3  2
2  2 TRUE
?
Third equation: 2  2  3  8
8  8 TRUE
The strategies in Example 3 and 0 can be applied to a system of any number of
equations in two variables. If you use the Substitution Method, find the solution to two of
the equations and then check it in each of the other equations. A solution to the system
will satisfy all of the equations in the system. If the solution from any two equations does
not work in all of the other equations in the system, the system does not have any
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solutions. If you use the Elimination Method, follow the strategy and look for the
transformations to yield new equations that are contradictions (the original system has
no solutions) or identities (the original system has many solutions). Remember,
contradictions are equations like 0  27 that are never true and identities are equations
like 5  5 that are always true.
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