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Chapter 8
Systems of Linear
Equations and
Problem Solving
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8.2
Solving by Substitution or
Elimination
• The Substitution Method
• The Elimination Method
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The Substitution Method
Algebraic (nongraphical) methods for solving
systems are often superior to graphing,
especially when fractions are involved. One
algebraic method, the substitution method,
relies on having a variable isolated.
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Example
Solve the system x  y  2,
(1)
y  x  1.
(2)
The equations are
numbered for
reference.
Solution
Equation (2) says that y and x – 1 name the same
number. Thus we can substitute x – 1 for y in
equation (1):
x+y=2
Equation (1)
x + (x – 1) = 2
Substituting x – 1 for y
We solve the last equation:
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Solution (continued)
x + (x – 1) = 2
2x – 1 = 2
Now return to the
original pair of
equations and
substitute 3/2 for x
in either equation
so that we can
solve for y
2x = 3
3
x= .
2
y  x 1
y = 3/2 – 1
Equation (2)
Substituting 3/2 for x
1
y= .
2
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Solution (continued)
3 1

We obtain the ordered pair  ,  .
2 2
We can substitute the ordered pair into the
original pair of equations to check that it is
the solution.
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Example
Solve the system 3x  y  5,
(1)
2 x  3 y  7. (2)
Solution
First, select an equation to solve for one
variable. To isolate y, subtract 3x from both
sides of equation (1):
(1)
3x  y  5
y  5  3x.
(3)
Next, proceed as in the last example, by
substituting:
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Solution (continued)
2x – 3y = 7
2x – 3(5 – 3x) = 7
Equation (2)
Substituting 5 – 3x for y
2x – 15 + 9x = 7
11x = 22
x = 2.
We can substitute 2 for x in either equation (1), (2), or
(3). It is easiest to use (3) because it has already been
solved for y:
y  5  3x
y = 5 – 3(2)
y = –1.
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Solution (continued)
We obtain the ordered pair (2, –1).
We can substitute the ordered pair into the
original pair of equations to check that it is
the solution.
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Example
Solve the system x  7 y  5, (1)
x  7 y  2. (2)
Solution
We can substitute 7y + 2 for x in equation (1)
and solve:
7y + 2 = 7y + 5 Substituting 7y + 2 for x
Subtracting 7y from both sides
2 = 5.
When the y terms drop out, the result is a
contradiction. We state that the system has no
solution.
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The Elimination Method
The elimination method for solving systems of
equations makes use of the addition principle:
If a = b, then a + c = b + c.
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Example
Solve the system
x  y  5,
 x  y  9.
(1)
(2)
Solution
Note that according to equation (2), –x + y and
9 are the same number. Thus we can work
vertically and add –x + y to the left side of
equation (1) and 9 to the right side:
x + y = 5 (1)
–x + y = 9 (2)
0x + 2y = 14.
Adding
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Solution (continued)
This eliminates the variable x, and leaves an
equation with just one variable, y for which we
solve: 2y = 14
y=7
Next, we substitute 7 for y in equation (1)
and solve for x:
x+7=5
Substituting. We also could
have used equation (2).
x = –2
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Solution (continued)
We obtain the ordered pair (–2, 7).
We can substitute the ordered pair into the
original pair of equations to check that it is
the solution.
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Example
Solve the system
x  2 y  6,
1
1
x  y  1.
4
3
(1)
(2)
Solution
To clear the fractions in equation (2), we
multiply both sides of equation (2) by 12 to get
equation (3):
1 
1
12  x  y   12(1)
3 
4
(3)
3x  4 y  12.
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Solution (continued)
(1)
x

2
y


6,
Now we solve the system
3x  4 y  12. (3)
Notice that if we add equations (1) and (3), we will
not eliminate any variables. If the –2y in equation
(1) were changed to –4y, we would. To accomplish
this change, we multiply both sides of equation (1)
by 2:
2x – 4y = –12 Multiply eqn. (1) by 2
3x + 4y = 12
(3)
Adding
5x + 0y = 0
Solving for x
x = 0.
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Solution (continued)
Then
3(0) + 4y = 12
Substituting 0 for x in
equation (3)
4y = 12
y=3
We obtain the ordered pair (0, 3).
We can plug the ordered pair into the original
pair of equations to check that it is the
solution.
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Example
Solve the system 2 x  3 y  2,
Solution
(1)
4 x  6 y  4. (2)
4x – 6y = 4
– 4x + 6y = –4
Multiply equation (1) by 2
Add
0=0
Note that what remains is an identity. Any pair
that is a solution of equation (1) is also a solution
of equation (2). The equations are dependent
and the solution set is infinite:
( x, y) | 2 x  3 y  2, or equivalently ( x, y) | 4 x  6 y  4.
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Rules for Special Cases
When solving a system of two linear equations
in two variables:
1. If an identity is obtained, such as 0 = 0, then
the system has an infinite number of
solutions. The equations are dependent
and, since a solution exists, the system is
consistent.
2. If a contradiction is obtained, such as 0 = 7,
then the system has no solution. The
system is inconsistent.
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