Chapter 12 lesson 2 and 3

12-2 Solving Multi-Step Equations

Course 2

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation

Warm UpSolve.

1. –8p – 8 = 56

2. 13d – 5 = 60

3. 9x + 24 = 60

4.

p = –8

d = 5

x = 4

Course 2


k7 + 4 = 11

5. 19 + z4 = 24

k = 49

z = 20

Problem of the Day

Without a calculator, multiply 2.637455 by 6, add 12, divide the result by 3, subtract 4, and then multiply by 0.5. What number will you end with? (Hint: If you start with x, what do you end with?)

2.637455 (the number you started with)

Course 2


Learn to solve multi-step equations.

Course 2


Solve 12 – 7b + 10b = 18.

Additional Example 1: Combining Like Terms to Solve Equations

12 – 7b + 10b = 18

12 + 3b = 18

– 12 – 12

3b = 6

3b = 63 3

b = 2

Combine like terms.

Subtract 12 from both sides.

Divide both sides by 3.

Course 2


Solve 14 – 8b + 12b = 62.

14 – 8b + 12b = 62

14 + 4b = 62

– 14 – 14

4b = 48

4b = 484 4

b = 12

Combine like terms.

Subtract 14 from both sides.


Check It Out: Example 1

Course 2


You may need to use the Distributive Property to solve an equation that has parentheses. Multiply each term inside the parentheses by the factor that is outside the parentheses. Then combine like terms.

Course 2


Solve 5(y – 2) + 6 = 21

Additional Example 2: Using the Distributive Property to Solve Equations

5(y – 2) + 6 = 21

5(y) – 5(2) + 6 = 21

+ 4 + 4

5y = 25

5 5

y = 5

Distribute 5 on the left side.

Simplify and combine like terms.

Add 4 to both sides.

Course 2


5y – 4 = 21


Solve 3(x – 3) + 4 = 28


3(x – 3) + 4 = 28

3(x) – 3(3) + 4 = 28

+ 5 + 5

3x = 33

3 3

x = 11

Distribute 3 on the left side.

Simplify and combine like terms.


Course 2


3x – 5 = 28


Troy owns three times as many trading cards as Hillary. Subtracting 9 from the number of trading cards Troy owns and then dividing by 6 gives the number of cards owns. If Sean owns 24 trading cards, how many trading cards does Hillary own?

Additional Example 3: Problem Solving Application

Course 2


Additional Example 3 Continued

11 Understand the Problem

Rewrite the question as a statement.

• Find the number of trading cards that Hillary owns.

List the important information:

• Troy owns 3 times as many trading cards as Hillary has.

• Subtracting 9 from the number of trading cards that Troy has and then dividing by 6 gives the number cards Sean owns.

• Sean owns 24 trading cards.

Course 2


22 Make a Plan

Let c represent the number of trading cards Hillary owns. Then 3c represents the number Troy has, and3c – 9

6 represents the number Sean owns, which

equals 24.

3c – 9 6Solve the equation = 24 for c to find the

number of cards Hillary owns.


Course 2


Solve33

3c – 9 6

= 24

3c – 9 6 = (6)24(6)

3c – 9 = 144

3c – 9 + 9 = 144 + 9

3c = 153

3c = 1533 3c = 51

Hillary owns 51 cards.

Multiply both sides by 6.




Course 2


Look Back44

If Hillary owns 51 cards, then Troy owns 153 cards. When you subtract 9 from 153, you get 144. And 144 divided by 6 is 24, which is the number of cards that Sean owns. So the answer is correct.


Course 2



Insert Lesson Title Here

John is twice as old as Helen. Subtracting 4 from John’s age and then dividing by 2 gives William’s age. If William is 24, how old is Helen?

Course 2


Check It Out: Example 3 Continued


11 Understand the Problem

Rewrite the question as a statement.

• Find Helen’s age.

List the important information:

• John is 2 times as old as Helen.

• Subtracting 4 from John’s age and then dividing by 2 gives William’s age.

• William is 24 years old.

Course 2




22 Make a Plan

Let h stand for Helen’s age. Then 2h represents

John’s age, and 2h – 42

which equals 24.

represents William’s age,

2h – 42

Solve the equation = 24 for h to find Helen’s age.

Course 2




Solve332h – 4 2

2h – 4 2

= 24

= (2)24(2)

2h – 4 = 48

2h – 4 + 4 = 48 + 4

2h = 52

2h = 5222

h = 26

Helen is 26 years old.




Course 2




Look Back44

If Helen is 26 years old, then John is 52 years old. When you subtract 4 from 52 you get 48. And 48 divided by 2 is 24, which is the age of William. So the answer is correct.

Course 2


Lesson Quiz

Solve.

1. c + 21 + 5c = 63

2. –x – 11 + 17x = 53

3. 59 = w – 16 = 4w

4. 4(k – 3) + 1 = 33

x = 4

c = 7


15 = w

k = 11

5. Kelly swam 4 times as many laps as Kathy. Adding 5 to the number of laps Kelly swam gives you the number of laps Julie swam. If Julie swam 9 laps, how many laps did Kathy swim? 1 lap

Course 2


12-3 Solving Equations with Variables on Both Sides

Course 2

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation

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

Course 2


Learn to solve equations that have variables on both sides.

Course 2


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.

Course 2



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.


Course 2


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.


Course 2


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.


Course 2


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.


Course 2


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.


Course 2


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.


Check It Out: Example 2A

Course 2


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.


Check It Out: Example 2B

Course 2


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.


Check It Out: Example 2C

Solve.

Course 2


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

Course 2



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.

Course 2



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?


Course 2




Let m represent the number of minutes.

75 = 40 + 0.10m75 – 40 = 40 – 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

Course 2


35 0.10m 0.10 0.10

=

Lesson Quiz: Part I


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

Course 2


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


Course 2


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Chapter 12 lesson 2 and 3

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