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Chapter 1
Section 8-7
Systems of Equations and Applications
8-7-1
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Systems of Equations and
Applications
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Linear Systems in Two Variables
Elimination Method
Substitution Method
Applications of Linear Systems
8-7-2
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Linear System in Two Variables
When multiple equations are considered
together the set of equations is called a system
of equations. For example
y = 3x + 4
y = –2x + 2
The point where the graphs intersect is a
solution of each of the individual equations.
It is also the solution of the system of
equations.
8-7-3
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Graphs a Linear System (Three
Possibilities)
1. The two graphs intersect in a single point. The
system is consistent and the equations are
independent.
y
One
solution
x
8-7-4
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Graphs a Linear System (Three
Possibilities)
2. The graphs are parallel lines. In this case, the
system is inconsistent and the equations are
independent. There is no solution common to both
equations.
y
No
solution
x
8-7-5
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Graphs a Linear System (Three
Possibilities)
3. The graphs are the same line. The system is
consistent and the equations are dependent, because
any solution of one equation is also a solution of the
other. The solution set is an infinite set of ordered
y
pairs.
Infinite
number of
x solutions
8-7-6
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Elimination Method
We can use algebraic methods to solve
systems. One method is called the
elimination method. The elimination method
involves combining the two equations of the
system so that one variable is eliminated. We
use the fact that
If a = b and c = d then a + c = b + d.
8-7-7
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Solving Linear Systems by
Elimination
Step 1
Write both equations in standard form
Ax + By = C.
Step 2
Make the coefficients of one pair of
variable terms opposites. Multiply one
or both equations by appropriate
numbers so that the sum of the
coefficients of either x or y is zero.
Step 3
Add the new equations to eliminate a
variable. The sum should be an
equation with just one variable.
© 2008 Pearson Addison-Wesley. All rights reserved
8-7-8
Solving Linear Systems by
Elimination
Step 4
Solve the equation from Step 3.
Step 5
Find the other value. Substitute the
result of Step 4 into either of the given
equations and solve for the other
variable.
Step 6
Find the solution set. Check the
solution in both of the given equations.
Then write the solution set.
8-7-9
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Elimination
Solve the system.
3 x  2 y  4 (1)
2 x  y  5 (2)
Solution
Step 1 Both equations are in standard form.
Step 2 Multiply equation (2) by 2 to get opposite
coefficients on y.
3x  2 y  4
4 x  2 y  10
8-7-10
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Elimination
Solution (continued)
Step 3 Add the equations to eliminate y.
7 x  14
Step 4 Solve to find x = 2.
Step 5 Substitute in one of the original equations
to find y. 2(2)  y  5
y 1
Step 6 The solution (2, 1) checks in both original
equations.
© 2008 Pearson Addison-Wesley. All rights reserved
8-7-11
Substitution Method
Linear systems can also be solved by the
substitution method. This method is useful
for solving linear systems in which one
variable has coefficient 1 or –1. It is also the
best choice for solving many nonlinear
systems in advanced algebra courses.
8-7-12
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Solving Linear Systems by
Substitution
Step 1
Solve for one variable in terms of the
other. Solve one of the equations for
either variable.
Step 2
Substitute for that variable in the other
equation. The result should be an
equation with just one variable.
Step 3
Solve the equation from Step 2.
8-7-13
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Solving Linear Systems by
Substitution
Step 4
Find the other value. Substitute the
result of Step 3 into the equation from
Step 1 to find the value of the other
variable.
Step 5
Find the solution set. Check the
solution in both of the given equations.
Then write the solution set.
8-7-14
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Substitution
Solve the system.
3 x  2 y  4 (1)
2 x  y  5 (2)
Solution
Step 1 Solve equation (2) for y.
y = –2x + 5
Step 2 Substitute –2x + 5 for y in equation (1).
3x  2(2 x  5)  4
8-7-15
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Example: Substitution
Solution (continued)
Step 3 Solve the equation
3x  4x 10  4
7 x  14
x2
Step 4 Substitute to find y.
y  2(2)  5  1
Step 5 The solution (2, 1) checks in both original
equations.
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8-7-16
Special Cases – Solving a System
Solving the system
3x  2 y  4
6 x  4 y  7
leads to a false statement such as 0 = 15. This
indicates that the two equations have no solutions in
common. The system is inconsistent with the empty
set as the solution set.
8-7-17
© 2008 Pearson Addison-Wesley. All rights reserved
Special Cases – Solving a System
Solving the system 3x  2 y  4
6 x  4 y  8
leads to a true statement such as 0 = 0. This
indicates that the two equations are dependent.
Choose one equation, solve for x and the solution set
can be written as
 2 y  4  
, y .


 3
8-7-18
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Applications of Linear Systems
Many problems involve more than one
unknown quantity. Although some problems
with two unknowns can be solved using just
one variable, many times it is easier to use
two variables.
8-7-19
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Solving an Applied Problem by
Writing a System of Equations
Step 1
Read the problem carefully until you
understand what is given and what is to
be found.
Step 2
Assign variables to represent the
unknown values, using diagrams or
tables as needed. Write down what each
variable represents.
Step 3
Write a system of equations that relates
the unknowns.
8-7-20
© 2008 Pearson Addison-Wesley. All rights reserved
Solving an Applied Problem by
Writing a System of Equations
Step 4
Solve the system of equations.
Step 5
State the answer to the problem. Does it
seem reasonable?
Step 6
Check the answer in the words of the
original problem.
8-7-21
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Example: Solving a Perimeter
Problem
A particular field has a perimeter of 320 yards. Its
length measures 40 yards more than its width.
What are the dimensions of the field?
Solution
Step 1
We need to find the dimensions of the field.
Step 2
Assign variables. Let L = length and
W = width.
8-7-22
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Example: Solving a Perimeter
Problem
Solution (continued)
Step 3
Perimeter: 2L + 2W = 320
and L = W + 40.
Step 4
Solve using substitution to get
W = 60 and L = 100.
Step 5
The length is 100 yards and the width is
60 yards.
Step 6
Check: 2(100) + 2(60) = 320 and
100 – 40 = 60.
8-7-23
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