Solving Systems by Substitution

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6-2
Solving Systems by Substitution
Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 1
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Warm Up Solve each equation for x. 1. y x
3 2. y 3x 4 Simplify each expression.
x y 3
2x 10
3. 2(x 5)
4. 12 3(x 1)
9 3x
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Warm Up Continued Evaluate each expression for
the given value of x. 5. x 8 for x 6 6.
3(x 7) for x 10
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Objective
Solve linear equations in two variables by
substitution.
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Sometimes it is difficult to identify the exact
solution to a system by graphing. In this case,
you can use a method called substitution.
The goal when using substitution is to reduce the
system to one equation that has only one
variable. Then you can solve this equation by the
methods taught in Chapter 2.
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Solve for one variable in at least one equation,
if necessary.
Step 1
Substitute the resulting expression into the
other equation.
Solve that equation to get the value of the first
variable.
Substitute that value into one of the original
equations and solve.
Write the values from steps 3 and 4 as an ordered
pair, (x, y), and check.
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Example 1A Solving a System of Linear Equations
by Substitution
Solve the system by substitution.
y 3x
y x 2
Step 1 y 3x
Both equations are solved for y.
y x 2
Substitute 3x for y in the second equation.
Solve for x. Subtract x from both sides and then
divide by 2.

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Example 1A Continued
Solve the system by substitution.
Write one of the original equations.
Substitute 1 for x.
Write the solution as an ordered pair.
Check Substitute (1, 3) into both equations
in the system.
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Example 1B Solving a System of Linear Equations
by Substitution
Solve the system by substitution.
y x 1
4x y 6
The first equation is solved for y.
Step 1 y x 1
Substitute x 1 for y in the second equation.
5x 1 6
Simplify. Solve for x.
Subtract 1 from both sides.
Divide both sides by 5.
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Example1B Continued
Solve the system by substitution.
Write one of the original equations.
Substitute 1 for x.
Write the solution as an ordered pair.
Check Substitute (1, 2) into both equations in
the system.
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Example 1C Solving a System of Linear Equations
by Substitution
Solve the system by substitution.
x 2y 1
x y 5
Step 1 x 2y 1
Solve the first equation for x by subtracting 2y
from both sides.
Substitute 2y 1 for x in the second equation.
3y 1 5
Simplify.
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Example 1C Continued
Step 3
3y 1 5
Solve for y.
Add 1 to both sides.
Divide both sides by 3.
Write one of the original equations.
Substitute 2 for y.
Subtract 2 from both sides.
Write the solution as an ordered pair.
Step 5 (3, 2)
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Check It Out! Example 1a
Solve the system by substitution.
y x 3
y 2x 5
Both equations are solved for y.
Substitute 2x 5 for y in the first equation.
Solve for x. Subtract x and 5 from both sides.
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Check It Out! Example 1a Continued
Solve the system by substitution.
Write one of the original equations.
Substitute 2 for x.
Write the solution as an ordered pair.
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Check It Out! Example 1b
Solve the system by substitution.
x 2y 4
x 8y 16
The first equation is solved for x.
Step 1 x 2y 4
Substitute 2y 4 for x in the second equation.
Simplify. Then solve for y.
Add 4 to both sides.
Divide both sides by 10.
y 2
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Check It Out! Example 1b Continued
Solve the system by substitution.
Write one of the original equations.
x 8(2) 16
Substitute 2 for y.
x 16 16
Simplify.
Subtract 16 from both sides.
Write the solution as an ordered pair.
Step 5 (0, 2)
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Check It Out! Example 1c
Solve the system by substitution.
2x y 4
x y 7
Solve the second equation for x by subtracting y
from each side.
Step 1 x y 7
Substitute y 7 for x in the first equation.
2(y 7) y 4
Distribute 2.
2y 14 y 4
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Check It Out! Example 1c Continued
Solve the system by substitution.
Combine like terms.
Step 3
2y 14 y 4
y 14 4
Add 14 to each side.
y 10
Write one of the original equations.
x (10) 7
Substitute 10 for y.
x 10 7
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Check It Out! Example 1c Continued
Solve the system by substitution.
x 10 7
Step 5
Add 10 to both sides.
x 3
Step 6
(3, 10)
Write the solution as an ordered pair.
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Sometimes you substitute an expression for a
variable that has a coefficient. When solving for
the second variable in this situation, you can
use the Distributive Property.
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Example 2 Using the Distributive Property
y 6x 11
Solve by substitution.
3x 2y 5
Solve the first equation for y by subtracting 6x
from each side.
Substitute 6x 11 for y in the second equation.
Distribute 2 to the expression in parenthesis.
3x 2(6x 11) 5
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Example 2 Continued
y 6x 11
Solve by substitution.
3x 2y 5
Step 3
Simplify. Solve for x.
3x 2(6x) 2(11) 5
3x 12x 22 5
9x 22 5
Subtract 22 from both sides.
Divide both sides by 9.
x 3
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Example 2 Continued
y 6x 11
Solve by substitution.
3x 2y 5
Write one of the original equations.
Substitute 3 for x.
y 6(3) 11
Simplify.
y 18 11
Subtract 18 from each side.
Step 5
(3, 7)
Write the solution as an ordered pair.
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Check It Out! Example 2
2x y 8
Solve by substitution.
3x 2y 9
Step 1 2x y 8
Solve the first equation for y by adding 2x to
each side.
Substitute 2x 8 for y in the second equation.
Distribute 2 to the expression in parenthesis.
3x 2(2x 8) 9
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Check It Out! Example 2 Continued
2x y 8
Solve by substitution.
3x 2y 9
Step 3
3x 2(2x) 2(8) 9
Simplify. Solve for x.
3x 4x 16 9
7x 16 9
Subtract 16 from both sides.
Divide both sides by 7.
x 1
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Check It Out! Example 2 Continued
2x y 8
Solve by substitution.
3x 2y 9
Write one of the original equations.
Substitute 1 for x.
2(1) y 8
y 2 8
Simplify.
Subtract 2 from each side.
Step 5
(1, 6)
Write the solution as an ordered pair.
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Example 2 Consumer Economics Application
Jenna is deciding between two cell-phone plans.
The first plan has a 50 sign-up fee and costs
20 per month. The second plan has a 30 sign-up
fee and costs 25 per month. After how many
months will the total costs be the same? What
will the costs be? If Jenna has to sign a
one-year contract, which plan will be cheaper?
Explain.
Write an equation for each option. Let t
represent the total amount paid and m represent
the number of months.
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Example 2 Continued
Total paid
sign-up fee
for each month.
payment amount
is
plus
Both equations are solved for t.
Substitute 50 20m for t in the second equation.
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Example 2 Continued
Solve for m. Subtract 20m from both sides.
50 30 5m
Subtract 30 from both sides.
20 5m
Divide both sides by 5.
Write one of the original equations.
t 30 25(4)
Substitute 4 for m.
Simplify.
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Example 2 Continued
Write the solution as an ordered pair.
Step 5
(4, 130)
In 4 months, the total cost for each option would
be the same 130.
If Jenna has to sign a one-year contract, which
plan will be cheaper? Explain.
Option 1 t 50 20(12) 290
Option 2 t 30 25(12) 330
Jenna should choose the first plan because it
costs 290 for the year and the second plan costs
330.
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Check It Out! Example 3
One cable television provider has a 60 setup fee
and 80 per month, and the second has a 160
equipment fee and 70 per month.
a. In how many months will the cost be the same?
What will that cost be.
Write an equation for each option. Let t
represent the total amount paid and m represent
the number of months.
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Check It Out! Example 3 Continued
Total paid
for each month.
payment amount
is
fee
plus
Both equations are solved for t.
Substitute 60 80m for t in the second equation.
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Check It Out! Example 3 Continued
Step 3
60 80m
160 70m
Solve for m. Subtract 70m from both sides.
Subtract 60 from both sides.
10m 100
Divide both sides by 10.
Write one of the original equations.
t 160 70(10)
Substitute 10 for m.
Simplify.
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Check It Out! Example 3 Continued
Step 5
(10, 860)
Write the solution as an ordered pair.
In 10 months, the total cost for each option
would be the same, 860.
b. If you plan to move in 6 months, which is the
cheaper option? Explain.
Option 1 t 60 80(6) 540
Option 2 t 160 270(6) 580
The first option is cheaper for the first six
months.
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Lesson Quiz Part I
Solve each system by substitution. 1. 2.
3.
y 2x
(2, 4)
x 6y 11
(1, 2)
3x 2y 1
3x y 1
x y 4
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Lesson Quiz Part II
4. Plumber A charges 60 an hour. Plumber B
charges 40 to visit your home plus 55 for each
hour. For how many hours will the total cost for
each plumber be the same? How much will that cost
be? If a customer thinks they will need a plumber
for 5 hours, which plumber should the customer
hire? Explain.
8 hours 480 plumber A plumber A is cheaper
for less than 8 hours.