Holt CA Course 1

11-4 Solving Equations with Variables on Both Sides

Warm UpWarm UpCalifornia StandardsCalifornia StandardsLesson PresentationLesson Presentation

PreviewPreview

Holt CA Course 1

11-4 Solving Equations with Variables on Both Sides

Warm UpSolve.

1. 6n + 8 – 4n = 202. –4w + 16 – 4w = –323. 25t – 17 – 13t = 674. 12 = 2(x + 7) + 4

n = 6w = 6t = 7x = –3

Holt CA Course 1

11-4 Solving Equations with Variables on Both Sides

Preview of Algebra 1 5.0 Students solve multistep problems, including word problems, involving linear equations and linear inequalities in one variable and provide justification for each step.

California Standards

Holt CA Course 1

11-4 Solving Equations with Variables on Both Sides

Group the terms with variables on one side of the equal sign, and simplify.

Additional Example 1: Using Inverse Operations to Group Terms with Variables

A. 60 – 4y = 8y

60 – 4y + 4y = 8y + 4y60 = 12y

B. –5b + 72 = –2b–5b + 72 = –2b

–5b + 5b + 72 = –2b + 5b72 = 3b

Add 4y to both sides.Simplify.

60 – 4y = 8y

Add 5b to both sides.Simplify.

Holt CA Course 1

11-4 Solving Equations with Variables on Both Sides

Group the terms with variables on one side of the equal sign, and simplify.

A. 40 – 2y = 6y

40 – 2y + 2y = 6y + 2y40 = 8y

B. –8b + 24 = –5b–8b + 24 = –5b

–8b + 8b + 24 = –5b + 8b24 = 3b

Add 2y to both sides.Simplify.

40 – 2y = 6y

Add 8b to both sides.Simplify.

Check It Out! Example 1

Holt CA Course 1

11-4 Solving Equations with Variables on Both Sides

Solve.

Additional Example 2A: Solving Equations with Variables on Both Sides

7c = 2c + 557c = 2c + 55

5c = 555c = 555 5c = 11

Subtract 2c from both sides.Simplify.Divide both sides by 5.

Check 7c = 2c + 557(11) = 2(11) + 55?

77 = 77

– 2c – 2c

Substitute 11 for c.

11 is the solution.

Holt CA Course 1

11-4 Solving Equations with Variables on Both SidesAdditional Example 2B: Solving Equations with Variables

on Both SidesSolve.

49 – 3m = 4m + 1449 – 3m = 4m + 14

49 = 7m + 1449 = 7m + 14

35 = 7m35 = 7m7 75 = m

Add 3m to both sides.Simplify.Subtract 14 fromboth sides.

Divide both sides by 7.

+ 3m + 3m

– 14 – 14

Holt CA Course 1

11-4 Solving Equations with Variables on Both SidesAdditional Example 2C: Solving Equations with Variables

on Both Sides 2

5 x = 15 x – 12

25 x = 1

5 x – 12

15 x –12=15 x (5)(–12)=(5)

x = –60

Subtract 15x from both

sides.

Simplify.

Multiply both sides by 5.

15

– x 15

– x

Holt CA Course 1

11-4 Solving Equations with Variables on Both Sides

Solve.8f = 3f + 658f = 3f + 65

5f = 655f = 655 5f = 13

Subtract 3f from both sides.

Simplify.

Divide both sides by 5.

Check It Out! Example 2A

–3f –3f

Holt CA Course 1

11-4 Solving Equations with Variables on Both Sides

Solve.54 – 3q = 6q + 954 – 3q = 6q + 9

54 = 9q + 954 = 9q + 9

45 = 9q45 = 9q9 95 = q

Add 3q to both sides.

Simplify.

Subtract 9 from both sides.

Divide both sides by 9.

Check It Out! Example 2B

+ 3q + 3q

– 9 – 9

Holt CA Course 1

11-4 Solving Equations with Variables on Both Sides

23 w = 1

3 w – 9

23w = 1

3 w – 9

13 w –9=13 w (3)(–9)=(3)

w = –27

Subtract 13w from both

sides.

Simplify.

Multiply both sides by 3.

Check It Out! Example 2CSolve.

13

– w – w13

Holt CA Course 1

11-4 Solving Equations with Variables on Both Sides

Christine can buy a new snowboard for $136.50. She will still need to rent boots for $8.50 a day. She can rent a snowboard and boots for $18.25 a day. How many days would Christine need to rent both the snowboard and the boots to pay as much as she would if she buys the snowboard and rents only the boots for the season?

Additional Example 3: Consumer Math Application

Holt CA Course 1

11-4 Solving Equations with Variables on Both SidesAdditional Example 3 Continued

18.25d = 136.5 + 8.5d

9.75d = 136.59.75d = 136.59.75 9.75

d = 14

Let d represent the number of days.

Subtract 8.5dfrom both sides.Simplify.Divide both sides by 9.75.

Christine would need to rent both the snowboard and the boots for 14 days to pay as much as she would have if she had bought the snowboard and rented only the boots.

– 8.5d – 8.5d

Holt CA Course 1

11-4 Solving Equations with Variables on Both SidesCheck It Out! Example 3

A local telephone company charges $40 per month for services plus a fee of $0.10 a minute for long distance calls. Another company charges $75.00 a month for unlimited service. How many minutes does it take for a person who subscribes to the first plan to pay as much as a person who subscribes to the unlimited plan?

Holt CA Course 1

11-4 Solving Equations with Variables on Both SidesCheck It Out! Example 3 Continued

Let m represent the number of minutes.75 = 40 + 0.10m

350 = m

Subtract 40from both sides.

Simplify.

If you are going to use more than 350 minutes, it will be cheaper to subscribe to the unlimited plan.

Divide both sides by 0.10.

35 = 0.10m 35 0.10m 0.10 0.10=

–40 –40

Holt CA Course 1

11-4 Solving Equations with Variables on Both SidesLesson Quiz: Part I

Group the terms with variables on one side of the equal sign, and simplify.

1. 14n = 11n + 81

2. –14k + 12 = –18k

Solve.

3. 58 + 3y = –4y – 19

4. –6x = x – 112

4k = –123n = 81

y = –11

x = 16

Holt CA Course 1

11-4 Solving Equations with Variables on Both SidesLesson Quiz Part II

5. Mary can purchase ice skates for $57 and then pay a $6 entry fee at the ice skating rink. She can also rent skates there for $3 and pay the entry fee. How many times must Mary skate to pay the same amount whether she purchases or rents the skates?19 times