10-2 Solving Equations with Variables on Both Sides

Warm UpWarm Up

Lesson PresentationLesson Presentation

Problem of the DayProblem of the Day

Lesson QuizzesLesson Quizzes

10-2 Solving Equations with Variables on Both Sides

Warm UpSolve.

1. 6n + 8 – 4n = 20

2. –4w + 16 – 4w = –32

3. 25t – 17 – 13t = 67

4. 12 = 2(x + 7) + 4

n = 6

w = 6

t = 7

x = –3

10-2 Solving Equations with Variables on Both Sides

Problem of the Day

You buy 1 cookie on the first day, 2 on the second day, 3 on the third day, and so on for 10 days. Your friend pays $10 for a cookie discount card and then buys 10 cookies at half price. You both pay the same total amount. What is the cost of one cookie? $0.20

10-2 Solving Equations with Variables on Both Sides

Preview of MA.912.A.3.1 Solve linear equations in one variable that include simplifying algebraic equations.

Sunshine State Standards

10-2 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 + 4y

60 = 12y

B. –5b + 72 = –2b

–5b + 72 = –2b

–5b + 5b + 72 = –2b + 5b

72 = 3b

Add 4y to both sides.

Simplify.

60 – 4y = 8y

Add 5b to both sides.

Simplify.

10-2 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 + 2y

40 = 8y

B. –8b + 24 = –5b

–8b + 24 = –5b

–8b + 8b + 24 = –5b + 8b

24 = 3b

Add 2y to both sides.

Simplify.

40 – 2y = 6y

Add 8b to both sides.

Simplify.

Check It Out: Example 1

10-2 Solving Equations with Variables on Both Sides

Solve.

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

7c = 2c + 55

7c = 2c + 55

7c – 2c = 2c – 2c + 55

5c = 55

5c = 555 5

c = 11

Subtract 2c from both sides.Simplify.

Divide both sides by 5.

10-2 Solving Equations with Variables on Both Sides

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

Solve.

49 – 3m = 4m + 14

49 – 3m = 4m + 14

49 – 3m + 3m = 4m + 3m + 14

49 = 7m + 14

49 – 14 = 7m + 14 – 14

35 = 7m

35 = 7m7 7

5 = m

Add 3m to both sides.

Simplify.Subtract 14 fromboth sides.

Divide both sides by 7.

10-2 Solving Equations with Variables on Both Sides

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

25

x = 15

x – 12

25

x = 15

x – 12

25

x 15

– x = 1 5

x – 121 5

x–

15

x –12=

15

x (5)(–12)=(5)

x = –60

Subtract 15

x from bothsides.

Simplify.

Multiply both sides by 5.

Solve.

10-2 Solving Equations with Variables on Both Sides

Solve.8f = 3f + 65

8f = 3f + 65

8f – 3f = 3f – 3f + 65

5f = 65

5f = 655 5

f = 13

Subtract 3f from both sides.Simplify.

Divide both sides by 5.

Check It Out: Example 2A

10-2 Solving Equations with Variables on Both Sides

Solve.

54 – 3q = 6q + 9

54 – 3q = 6q + 9

54 – 3q + 3q = 6q + 3q + 9

54 = 9q + 9

54 – 9 = 9q + 9 – 9

45 = 9q

45 = 9q9 9

5 = q

Add 3q to both sides.

Simplify.Subtract 9 from both sides.

Divide both sides by 9.

Check It Out: Example 2B

10-2 Solving Equations with Variables on Both Sides

23

w = 13

w – 9

23

w = 13

w – 9

23 w 1

3

– w = 1 3

w – 91 3

w–

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 2C

Solve.

10-2 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

10-2 Solving Equations with Variables on Both Sides

Additional Example 3 Continued

18.25d = 136.5 + 8.5d

18.25d – 8.5d = 136.5 + 8.5d – 8.5d

9.75d = 136.5

9.75d = 136.5

9.75 9.75d = 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.

10-2 Solving Equations with Variables on Both Sides

Check 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?

10-2 Solving Equations with Variables on Both Sides

Check It Out: Example 3 Continued

Let m represent the number of minutes.

75 = 40 + 0.10m75 – 40 = 40 – 40 + 0.10m

350 = m

Subtract 40from both sides.Simplify.

A person who subscribes to the first plan would have to use 350 minutes to pay as much as a person who subscribes to the unlimited plan.

Divide both sides by 0.10.

35 = 0.10m

35 0.10m 0.10 0.10

=

10-2 Solving Equations with Variables on Both Sides

Standard Lesson Quiz

Lesson Quizzes

Lesson Quiz for Student Response Systems

10-2 Solving Equations with Variables on Both Sides

Lesson 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. –

4k = –12

3n = 81

y = –11

x = 1634 x = 18 x – 14

10-2 Solving Equations with Variables on Both Sides

Lesson 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

10-2 Solving Equations with Variables on Both Sides

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

19p = 10p + 54

A. 9p = –54

B. 29p = –54

C. 29p = 54

D. 9p = 54

Lesson Quiz for Student Response Systems

10-2 Solving Equations with Variables on Both Sides

2. Solve.

73 + 4m = –7m – 26

A. m = –9

B. m = –7

C. m = 9

D. m = 7

Lesson Quiz for Student Response Systems

10-2 Solving Equations with Variables on Both Sides

3. Matthew orders T-shirts from an online store for $25 each and an additional $20 for shipping. Mike buys the same kind of T-shirts at a local shop for $27 each. If Matthew and Mike spent the same amount, how many T-shirts did each of them buy?

A. 5 T-shirts

B. 10 T-shirts

C. 15 T-shirts

D. 20 T-shirts

Lesson Quiz for Student Response Systems