Download Solve linear systems by substitution.

Chapter 4
Section 2
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4.2
1
2
3
Solving Systems of Linear
Equations by Substitution
Solve linear systems by substitution.
Solve special systems by substitution.
Solve linear systems with fractions.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Objective 1
Solve linear systems by
substitution.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4.2 - 3
Solve linear systems by substitution.
Graphing to solve a system of equations has a serious
drawback. It is difficult to find an accurate solution, such as
5
1
,



6  , from a graph. One algebraic method for solving a system
3
of equations is the substitution method.
This method is particularly useful for solving systems in which
one equation is already solved, or can be solved quickly, for one
of the variables.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4.2 - 4
Solve linear systems by substitution. (cont’d)
To solve a system by substitution, follow these steps.
Step 1: Solve one equation for either variable. If one of the
variables has coefficient 1 or −1, choose it, since it
usually makes the substitution method easier.
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.
Step 4: Substitute the result from Step 3 into the equation
from Step 1 to find the value of the other variable.
Step 5: Check the solution in both of the original equations.
Then write the solution set.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4.2 - 5
EXAMPLE 1
Using the Substitution Method
Solve the system by the substitution method.
2 x  7 y  12
x  2 y
Solution: 2  2 y   7 y  12
4 y  7 y  12
3 y 1 2

3
3
y  4
x  2 y
x  2  4
x 8
8, 4
The solution set found by the substitution method will be the same as
the solution found by graphing. The solution set is the same; only the
method is different.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4.2 - 6
EXAMPLE 2
Using the Substitution Method
Solve the system by the substitution method.
2 x  7 y  12
x  3 2y
Solution:
2  3  2 y   7 y  12
6  4 y  7 y  12
6  3 y  6  12  6
3 y 1 8

3
3
x  3  2  6
x  3 12
x  15
15, 6
y  6
Be careful when you write the ordered-pair solution of a system. Even though
we found y first, the x-coordinate is always written first in the ordered pair.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4.2 - 7
EXAMPLE 3
Using the Substitution Method
Use substitution to solve the system.
x  1  4 y
2 x  5 y  11
Solution:
x  1  1  4 y  1
x  4 y  1
x  4  1 1
x  4 1
x 3
2  4 y 1  5 y  11
8 y  2  5 y  2  11  2
13 y 13

13 13
y  1
3, 1
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Slide 4.2 - 8
Objective 2
Solve special systems by
substitution.
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Slide 4.2 - 9
EXAMPLE 4
Solving an Inconsistent System
by Substitution
Use substitution to solve the system.
y  8x  4
16 x  2 y  8
Solution: 16 x  2 8x  4  8
16x 16x  8  8
8  8
Since the statement is false, the solution set is Ø.
It is a common error to give “false” as the solution of an inconsistent
system. The correct response is Ø.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4.2 - 10
EXAMPLE 5
Solving a System with
Dependent Equations by
Substitution
Solve the system by the substitution method.
x  3 y  7
4 x  12 y  28
Solution:
x  3 y  3 y  7  3 y
x  7  3 y
4  7  3 y   12 y  28
28  12 y  12 y  28
28  28
Since the statement is true every solution of one equations is also a
solution to the other, so the system has an infinite number of
solutions and the solution set is {(x,y)|x + 3y = −7}.
It is a common error to give “true” as the solution of a system of dependent
equations. Remember to give the solution set in set-builder notation using
the equation in the system that is in standard form with integer coefficients
that have no common factor (except 1).
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Slide 4.2 - 11
Objective 3
Solve linear systems with
fractions.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4.2 - 12
EXAMPLE 6
Using the Substitution Method
with Fractions as Coefficients
Solve the system by the substitution method.
Solution:
1
1
1
x
y 
2
3
3
1
x  2 y  2
2
1

2
x

2
y
 
  2  2 
2

1 
1
1
 6  x  y     6
3 
3
2
3 x  2 y  2
3  4  4 y   2 y  2
12  12 y  2 y  12  2  12
10 y
10

10
10
y  1
 0, 1
x  4 y  4 y  4  4 y
x  4  4 y
x  4  4  1
x  4  4
x0
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Slide 4.2 - 13