solving systems by substitutionsolving systems by …...6-2 solving systems by substitution solve...

Holt Algebra 1

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

Holt Algebra 1

Warm Up

Lesson Presentation

Lesson Quiz

Holt Algebra 1

6-2 Solving Systems by Substitution

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

Holt Algebra 1


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

12

9

Holt Algebra 1


Solve linear equations in two variables by substitution.

Objective

Holt Algebra 1


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.

Holt Algebra 1


Solving Systems of Equations by Substitution

Step 2

Step 3

Step 4

Step 5

Step 1 Solve 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.

Holt Algebra 1


Solve the system by substitution.

Example 1A: Solving a System of Linear Equations by

Substitution

y = 3x

y = x – 2

Step 1 y = 3x

y = x – 2

Both equations are solved for y.

Step 2 y = x – 2

3x = 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 –x

2x = –2 2x = –2

2 2 x = –1

Holt Algebra 1



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

Holt Algebra 1



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 = 6

4x + (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.

Holt Algebra 1



Example1B Continued

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

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

y = 2

Step 5 (1, 2)

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


y = x + 1

2 1 + 1

2 2

4x + y = 6

4(1) + 2 6

6 6

Holt Algebra 1



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 −2y

x = –2y – 1

Holt Algebra 1


Example 1C Continued

Step 3 –3y – 1 = 5 Add 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.


Holt Algebra 1


Check It Out! Example 1a


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 2

2x + 5 = x + 3

y = x + 3

Holt Algebra 1


Check It Out! Example 1a Continued


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.

Holt Algebra 1


Check It Out! Example 1b


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 = 16

Add 4 to both sides. +4 +4 10y = 20

y = 2

10y 20

10 10 = Divide both sides by 10.

Holt Algebra 1


Check It Out! Example 1b Continued


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.

Holt Algebra 1


Check It Out! Example 1c


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

Holt Algebra 1


Combine like terms. Step 3

+14 +14

–y = 10

Check It Out! Example 1c Continued


–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

Holt Algebra 1


Check It Out! Example 1c Continued


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.

Holt Algebra 1


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.

Holt Algebra 1


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

Holt Algebra 1


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

Holt Algebra 1


Step 3

Example 2 Continued

y + 6x = 11


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

Holt Algebra 1


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


Write one of the original

equations.

Holt Algebra 1


Check It Out! Example 2

–2x + y = 8

3x + 2y = 9 Solve 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

Holt Algebra 1


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 = –7

7 7

x = –1

Check It Out! Example 2 Continued

–2x + y = 8


3x + 4x + 16 = 9

Holt Algebra 1


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.


–2x + y = 8


Subtract 2 from each side.

Simplify.


equations.

Holt Algebra 1


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.

Holt Algebra 1


Example 2 Continued

Total paid is

sign-up fee plus

payment amount

for each month.

Option 1 t = $50 + $20 m

Option 2 t = $30 + $25 m

Step 1 t = 50 + 20m

t = 30 + 25m

Both equations are solved

for t.

Step 2 50 + 20m = 30 + 25m Substitute 50 + 20m for t in

the second equation.

Holt Algebra 1


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.


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

Holt Algebra 1


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.

Holt Algebra 1


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.

Holt Algebra 1


Total paid is fee plus

payment amount

for each month.

Option 1 t = $60 + $80 m

Option 2 t = $160 + $70 m


Step 1 t = 60 + 80m

t = 160 + 70m

Both equations are solved

for t.

Step 2 60 + 80m = 160 + 70m Substitute 60 + 80m for t in

the second equation.

Holt Algebra 1


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 –60

10m = 100

10 10 m = 10


equations.

Step 4 t = 160 + 70m

t = 160 + 70(10)

t = 160 + 700

t = 860

Substitute 10 for m.

Simplify.


Holt Algebra 1


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.


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.

Holt Algebra 1


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

Holt Algebra 1


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.

solving systems by substitutionsolving systems by …...6-2 solving systems by substitution solve...

Documents