Using Algebraic Methods

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Using Algebraic Methods to Solve Linear Systems
3-2
Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 2
2
Warm Up Determine if the given ordered pair is an
element of the solution set of
2x y 5
.
3y x 6
no
2. (1, 1)
1. (3, 1)
yes
Solve each equation for y.
3. x 3y 2x 4y 4
y x 4
4. 6x 5 y 3y 2x 1
y 2x 3
3
Objectives
Solve systems of equations by substitution.
Solve systems of equations by elimination.
4
Vocabulary
substitution elimination
5
The graph shows a system of linear equations. As
you can see, without the use of technology,
determining the solution from the graph is not
easy. You can use the substitution method to find
an exact solution. In substitution, you solve one
equation for one variable and then substitute
this expression into the other equation.
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Example 1A Solving Linear Systems by Substitution
Use substitution to solve the system of equations.
Step 1 Solve one equation for one variable. The
first equation is already solved for y y x 1.
Step 2 Substitute the expression into the other
equation.
x y 7
x (x 1) 7
Substitute (x 1) for y in the other equation.
2x 1 7
Combine like terms.
2x 8
x 4
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Example 1A Continued

• Step 3 Substitute the x-value into one of the
  original equations to solve for y.

y x 1
y (4) 1
Substitute x 4.
y 3
The solution is the ordered pair (4, 3).
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Example 1A Continued
Check A graph or table supports your answer.
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Example 1B Solving Linear Systems by Substitution
Use substitution to solve the system of equations.
Method 2 Isolate x.
Method 1 Isolate y.
2y x 4
2y x 4
First equation.
x 4 2y
Isolate one variable.
3x 4y 7
3x 4y 7
Second equation.
3(4 2y) 4y 7
Substitute the expression into the second
equation.
12 6y 4y 7
3x 2x 8 7
12 10y 7
5x 8 7
Combine like terms.
10y 5
5x 15
x 3
First part of the solution
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Example 1B Continued

• Substitute the value into one of the original
  equations to solve for the other variable.

Method 1
Method 2
Substitute the value to solve for the other
variable.
2y (3) 4

1. x 4

1 x 4
2y 1
x 3
Second part of the solution
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Check It Out! Example 1a
Use substitution to solve the system of equations.
Step 1 Solve one equation for one variable. The
first equation is already solved for y y 2x
1.
Step 2 Substitute the expression into the other
equation.
3x 2y 26
Substitute (2x 1) for y in the other equation.
3x 2(2x1) 26
3x 4x 2 26
Combine like terms.
7x 28
x 4
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Check It Out! Example 1a Continued

• Step 3 Substitute the x-value into one of the
  original equations to solve for y.

y 2x 1
y 2(4) 1
Substitute x 4.
y 7
The solution is the ordered pair (4, 7).
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Check It Out! Example 1a Continued

• Check A graph or table supports your answer.

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Check It Out! Example 1b
Use substitution to solve the system of equations.
Method 1 Isolate y.
Method 2 Isolate x.
2x 2 y
2x 2 y
First equation.
y 2x 2
Isolate one variable.
5x 6y 9
5x 6y 9
Second equation.
Substitute the expression into the second
equation.
5x 6(2x 2) 9
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Check It Out! Example 1b Continued
Method 1 Isolate y.
Method 2 Isolate x.
Combine like terms.
5x 6(2x 2) 9
5x 12x 12 9
10 5y 12y 18
7x 21
10 7y 18
7y 28
x 3
y 4
First part of the solution.
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Example 1b Continued

• Substitute the value into one of the original
  equations to solve for the other variable.

5(3) 6y 9
5x 6(4) 9
Substitute the value to solve for the other
variable.
5x (24) 9
15 6y 9
6y 24
5x 15
x 3
y 4
Second part of the solution
By either method, the solution is (3, 4).
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You can also solve systems of equations with the
elimination method. With elimination, you get rid
of one of the variables by adding or subtracting
equations. You may have to multiply one or both
equations by a number to create variable terms
that can be eliminated.
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Example 2A Solving Linear Systems by Elimination
Use elimination to solve the system of equations.
Step 1 Find the value of one variable.
3x 2y 4
The y-terms have opposite coefficients.
4x 2y 18
7x 14
Add the equations to eliminate y.
First part of the solution
x 2
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Example 2A Continued
Step 2 Substitute the x-value into one of the
original equations to solve for y.
3(2) 2y 4
2y 10
Second part of the solution
y 5
The solution to the system is (2, 5).
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Example 2B Solving Linear Systems by Elimination
Use elimination to solve the system of equations.
Step 1 To eliminate x, multiply both sides of the
first equation by 2 and both sides of the second
equation by 3.
Add the equations.
y 5
First part of the solution
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Example 2B Continued
Step 2 Substitute the y-value into one of the
original equations to solve for x.
Second part of the solution
The solution for the system is (3, 5).
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Example 2B Solving Linear Systems by Elimination
Check Substitute 3 for x and 5 for y in each
equation.
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Check It Out! Example 2a
Use elimination to solve the system of equations.
4x 7y 25
12x 7y 19
Step 1 Find the value of one variable.
The y-terms have opposite coefficients.
8x 6
Add the equations to eliminate y.
First part of the solution
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Check It Out! Example 2a Continued
Step 2 Substitute the x-value into one of the
original equations to solve for y.
3 7y 25
7y 28
Second part of the solution
y 4
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Check It Out! Example 2b
Use elimination to solve the system of equations.
Step 1 To eliminate x, multiply both sides of the
first equation by 8 and both sides of the second
equation by 5.
Add the equations.
49y 196
First part of the solution
y 4
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Check It Out! Example 2b
Step 2 Substitute the y-value into one of the
original equations to solve for x.
Second part of the solution
The solution for the system is (6,4).
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Check It Out! Example 2b
Check Substitute 6 for x and 4 for y in each
equation.
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In Lesson 31, you learned that systems may have
infinitely many or no solutions. When you try to
solve these systems algebraically, the result
will be an identity or a contradiction.
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Example 3 Solving Systems with Infinitely Many
or No Solutions
Classify the system and determine the number of
solutions.
3x y 1
2y 6x 18
Because isolating y is straightforward, use
substitution.
3x y 1
y 1 3x
Solve the first equation for y.
2(1 3x) 6x 18
Substitute (13x) for y in the second equation.
2 6x 6x 18
Distribute.
x
2 18
Simplify.
Because 2 is never equal to 18, the equation is
a contradiction. Therefore, the system is
inconsistent and has no solution.
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Check It Out! Example 3a
Classify the system and determine the number of
solutions.
56x 8y 32
7x y 4
Because isolating y is straightforward, use
substitution.
7x y 4
y 4 7x
Solve the second equation for y.
Substitute (4 7x) for y in the first equation.
56x 8(4 7x) 32
56x 32 56x 32
Distribute.
Simplify.
Because 32 is equal to 32, the equation is an
identity. The system is consistent, dependent and
has infinite number of solutions.
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Check It Out! Example 3b
Classify the system and determine the number of
solutions.
6x 3y 12
2x y 6
Because isolating y is straightforward, use
substitution.
2x y 6
y 6 2x
Solve the second equation.
Substitute (6 2x) for y in the first equation.
6x 3(6 2x) 12
6x 18 6x 12
Distribute.
18 12
x
Simplify.
Because 18 is never equal to 12, the equation
is a contradiction. Therefore, the system is
inconsistent and has no solutions.
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Example 4 Zoology Application
A veterinarian needs 60 pounds of dog food that
is 15 protein. He will combine a beef mix that
is 18 protein with a bacon mix that is 9
protein. How many pounds of each does he need to
make the 15 protein mixture?
Let x present the amount of beef mix in the
mixture.
Let y present the amount of bacon mix in the
mixture.
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Example 4 Continued
Write one equation based on the amount of dog
food
Write another equation based on the amount of
protein
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Example 4 Continued
Solve the system.
First equation
x y 60
y 60 x
Solve the first equation for y.
0.18x 0.09(60 x) 9
Substitute (60 x) for y.
0.18x 5.4 0.09x 9
Distribute.
Simplify.
0.09x 3.6
x 40
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Example 4 Continued
Substitute x into one of the original equations
to solve for y.
Substitute the value of x into one equation.
40 y 60
y 20
Solve for y.
The mixture will contain 40 lb of the beef mix
and 20 lb of the bacon mix.
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Check It Out! Example 4
A coffee blend contains Sumatra beans which cost
5/lb, and Kona beans, which cost 13/lb. If the
blend costs 10/lb, how much of each type of
coffee is in 50 lb of the blend?
Let x represent the amount of the Sumatra beans
in the blend.
Let y represent the amount of the Kona beans in
the blend.
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Check It Out! Example 4 Continued
Write one equation based on the amount of each
bean
Write another equation based on cost of the beans
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Check It Out! Example 4 Continued
Solve the system.
First equation
x y 50
y 50 x
Solve the first equation for y.
5x 13(50 x) 500
Substitute (50 x) for y.
5x 650 13x 500
Distribute.
8x 150
Simplify.
x 18.75
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Check It Out! Example 4 Continued
Substitute x into one of the original equations
to solve for y.
Substitute the value of x into one equation.
18.75 y 50
y 31.25
Solve for y.
The mixture will contain 18.75 lb of the Sumatra
beans and 31.25 lb of the Kona beans.
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Lesson Quiz
Use substitution or elimination to solve each
system of equations.
3x y 1
5x 4y 10
1.
2.
y x 9
3x 4y 2
(2, 7)
(6, 5)
3. The Miller and Benson families went to a
theme park. The Millers bought 6 adult and 15
children tickets for 423. The Bensons bought
5 adult and 9 children tickets for 293. Find
the cost of each ticket.
adult 28 childrens 17