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Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson Quiz

Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

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Page 1: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution6-2 Solving Systems by Substitution

Holt Algebra 1

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Page 2: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Warm UpSolve each equation for x.

1. y = x + 3

2. y = 3x – 4

Simplify each expression.

x = y – 3

2x – 103. 2(x – 5)

4. 12 – 3(x + 1) 9 – 3x

Page 3: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Warm Up ContinuedEvaluate each expression for the given value of x.

5. x + 8 for x = 6

6. 3(x – 7) for x =10

12

9

Page 4: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Solve linear equations in two variables by substitution.

Objective

Page 5: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

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.

Page 6: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Solving Systems of Equations by Substitution

Step 2

Step 3

Step 4

Step 5

Step 1Solve for one variable in at least one equation, if necessary.

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.

Page 7: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Solve the system by substitution.

Example 1A: Solving a System of Linear Equations by Substitution

y = 3x

y = x – 2

Step 1 y = 3xy = x – 2

Both equations are solved for y.

Step 2 y = x – 23x = x – 2

Substitute 3x for y in the second equation.

Solve for x. Subtract x from both sides and then divide by 2.

Step 3 –x –x2x = –22x = –22 2x = –1

Page 8: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Solve the system by substitution.

Example 1A Continued

Step 4 y = 3x Write one of the original equations.

Substitute –1 for x. y = 3(–1)y = –3

Step 5 (–1, –3)

Check Substitute (–1, –3) into both equations in the system.

Write the solution as an ordered pair.

y = 3x–3 3(–1)

–3 –3

y = x – 2–3 –1 – 2

–3 –3

Page 9: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Solve the system by substitution.

Example 1B: Solving a System of Linear Equations by Substitution

y = x + 1

4x + y = 6

Step 1 y = x + 1 The first equation is solved for y.

Step 2 4x + y = 64x + (x + 1) = 6

Substitute x + 1 for y in the second equation.

Subtract 1 from both sides. Step 3 –1 –1

5x = 5 5 5

x = 1

5x = 5

5x + 1 = 6 Simplify. Solve for x.

Divide both sides by 5.

Page 10: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Solve the system by substitution.

Example1B Continued

Step 4 y = x + 1 Write one of the original equations.

Substitute 1 for x. y = 1 + 1y = 2

Step 5 (1, 2)

Check Substitute (1, 2) into both equations in the system.

Write the solution as an ordered pair.

y = x + 1

2 1 + 12 2

4x + y = 6

4(1) + 2 66 6

Page 11: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Solve the system by substitution.

Example 1C: Solving a System of Linear Equations 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.

Step 2 x – y = 5(–2y – 1) – y = 5

Substitute –2y – 1 for x in the second equation.

–3y – 1 = 5 Simplify.

−2y −2yx = –2y – 1

Page 12: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Example 1C Continued

Step 3 –3y – 1 = 5Add 1 to both sides. +1 +1

–3y = 6

–3y = 6–3 –3

y = –2

Solve for y.

Divide both sides by –3.

Step 4 x – y = 5 x – (–2) = 5

x + 2 = 5 –2 –2

x = 3

Step 5 (3, –2)

Write one of the original equations.

Substitute –2 for y.

Subtract 2 from both sides.Write the solution as an

ordered pair.

Page 13: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Check It Out! Example 1a

Solve the system by substitution.

y = x + 3

y = 2x + 5

Both equations are solved for y.Step 1 y = x + 3 y = 2x + 5

Substitute 2x + 5 for y in the first equation.

Solve for x. Subtract x and 5 from both sides. –x – 5 –x – 5

x = –2

Step 3 2x + 5 = x + 3

Step 22x + 5 = x + 3

y = x + 3

Page 14: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Check It Out! Example 1a Continued

Solve the system by substitution.

Step 4 y = x + 3 Write one of the original equations.

Substitute –2 for x. y = –2 + 3

y = 1

Step 5 (–2, 1) Write the solution as an ordered pair.

Page 15: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

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.

(2y – 4) + 8y = 16

x + 8y = 16 Step 2

Step 3 10y – 4 = 16Add 4 to both sides. +4 +4

10y = 20

y = 2

10y 2010 10= Divide both sides by 10.

Page 16: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Check It Out! Example 1b Continued

Solve the system by substitution.

Step 4 x + 8y = 16 Write one of the original equations.

Substitute 2 for y. x + 8(2) = 16

x + 16 = 16

x = 0 – 16 –16

Simplify.Subtract 16 from both

sides.

Step 5 (0, 2) Write the solution as an ordered pair.

Page 17: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

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.

Substitute –y – 7 for x in the first equation.

Distribute 2.

2(–y – 7) + y = –4

x = –y – 7 Step 2

Step 1 x + y = –7– y – y

x = –y – 7

2(–y – 7) + y = –4

–2y – 14 + y = –4

Page 18: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Combine like terms. Step 3

+14 +14

–y = 10

Check It Out! Example 1c Continued

Solve the system by substitution.

–2y – 14 + y = –4

Add 14 to each side.

–y – 14 = –4

y = –10

Step 4 x + y = –7 Write one of the original equations.

Substitute –10 for y. x + (–10) = –7

x – 10 = – 7

Page 19: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Check It Out! Example 1c Continued

Solve the system by substitution.

x – 10 = –7 Step 5+10 +10

x = 3

Add 10 to both sides.

Step 6 (3, –10) Write the solution as an ordered pair.

Page 20: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

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.

Page 21: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

When you solve one equation for a variable, you must substitute the value or expression into the other original equation, not the one that had just been solved.

Caution

Page 22: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Example 2: Using the Distributive Property

y + 6x = 11

3x + 2y = –5 Solve by substitution.

Solve the first equation for y by subtracting 6x from each side.

Step 1 y + 6x = 11– 6x – 6x

y = –6x + 11

Substitute –6x + 11 for y in the second equation.

Distribute 2 to the expression in parenthesis.

3x + 2(–6x + 11) = –5

3x + 2y = –5 Step 2

3x + 2(–6x + 11) = –5

Page 23: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Step 3

Example 2 Continued

y + 6x = 11

3x + 2y = –5 Solve by substitution.

3x + 2(–6x) + 2(11) = –5

–9x + 22 = –5

Simplify. Solve for x.

Subtract 22 from both sides.–9x = –27

– 22 –22

Divide both sides by –9.

–9x = –27–9 –9

x = 3

3x – 12x + 22 = –5

Page 24: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Step 4 y + 6x = 11

Substitute 3 for x.y + 6(3) = 11

Subtract 18 from each side.y + 18 = 11

–18 –18

y = –7

Step 5 (3, –7) Write the solution as an ordered pair.

Simplify.

Example 2 Continued

y + 6x = 11

3x + 2y = –5 Solve by substitution.

Write one of the original equations.

Page 25: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Check It Out! Example 2

–2x + y = 8

3x + 2y = 9Solve by substitution.

Solve the first equation for y by adding 2x to each side.

Step 1 –2x + y = 8+ 2x +2x

y = 2x + 8

Substitute 2x + 8 for y in the second equation. 3x + 2(2x + 8) = 9

3x + 2y = 9 Step 2

Distribute 2 to the expression in parenthesis.

3x + 2(2x + 8) = 9

Page 26: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Step 3 3x + 2(2x) + 2(8) = 9

7x + 16 = 9

Simplify. Solve for x.

Subtract 16 from both sides.7x = –7

–16 –16

Divide both sides by 7.

7x = –77 7x = –1

Check It Out! Example 2 Continued

–2x + y = 8

3x + 2y = 9Solve by substitution.

3x + 4x + 16 = 9

Page 27: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Step 4 –2x + y = 8

Substitute –1 for x.–2(–1) + y = 8

y + 2 = 8

–2 –2

y = 6

Step 5 (–1, 6) Write the solution as an ordered pair.

Check It Out! Example 2 Continued

–2x + y = 8

3x + 2y = 9Solve by substitution.

Subtract 2 from each side.

Simplify.

Write one of the original equations.

Page 28: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

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.

Page 29: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Example 2 Continued

Total paid is

sign-up fee plus

paymentamount

for eachmonth.

Option 1 t = $50 + $20 m

Option 2 t = $30 + $25 m

Step 1 t = 50 + 20mt = 30 + 25m

Both equations are solved for t.

Step 2 50 + 20m =30 + 25m Substitute 50 + 20m for t in the second equation.

Page 30: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Step 3 50 + 20m = 30 + 25m Solve for m. Subtract 20m from both sides.–20m – 20m

50 = 30 + 5m Subtract 30 from both sides. –30 –30

20 = 5m Divide both sides by 5.

Write one of the original equations.

Step 4 t = 30 + 25m

t = 30 + 25(4)

t = 30 + 100

t = 130

Substitute 4 for m.Simplify.

Example 2 Continued

5 5

m = 4

20 = 5m

Page 31: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Step 5 (4, 130)Write the solution as an

ordered pair.

In 4 months, the total cost for each option would be the same $130.

Jenna should choose the first plan because it costs $290 for the year and the second plan costs $330.

Example 2 Continued

Option 1: t = 50 + 20(12) = 290

Option 2: t = 30 + 25(12) = 330

If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.

Page 32: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

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.

Page 33: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Total paid is fee plus

paymentamount

for eachmonth.

Option 1 t = $60 + $80 m

Option 2 t = $160 + $70 m

Check It Out! Example 3 Continued

Step 1 t = 60 + 80mt = 160 + 70m

Both equations are solved for t.

Step 2 60 + 80m = 160 + 70m Substitute 60 + 80m for t in the second equation.

Page 34: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Step 3 60 + 80m = 160 + 70m Solve for m. Subtract 70m from both sides. –70m –70m

60 + 10m = 160 Subtract 60 from both sides.

Divide both sides by 10.

–60 –6010m = 10010 10

m = 10

Write one of the original equations.

Step 4 t = 160 + 70m

t = 160 + 70(10)

t = 160 + 700

t = 860

Substitute 10 for m.Simplify.

Check It Out! Example 3 Continued

Page 35: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

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.

The first option is cheaper for the first six months.

Check It Out! Example 3 Continued

Option 1: t = 60 + 80(6) = 540

Option 2: t = 160 + 270(6) = 580

b. If you plan to move in 6 months, which is the cheaper option? Explain.

Page 36: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

Lesson Quiz: Part I

Solve each system by substitution.

1.

2.

3.

(1, 2)

(–2, –4)y = 2x

x = 6y – 11

3x – 2y = –1

–3x + y = –1

x – y = 4

Page 37: Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 1

6-2 Solving Systems by Substitution

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.