1. Warm Up
Problem of the Day
Lesson Presentation
Lesson Quizzes

2. Warm Up
Solve.
1. 3x = 102
= 15
3. z – 100 = 21
w = 98.6
x = 34 y 15
y = 225
z = 121
w = 19.5

3. Problem of the Day
Ana has twice as much money as Ben, and Ben has three times as much as Clio. Together they have $160. How much does each person have?
Ana, $96; Ben, $48; Clio, $16

4. Learn to solve multi-step equations.

5. To solve a multi-step equation, you may have to simplify the equation first by combining like terms or by using the Distributive Property.

6. Additional Example 1A: Solving Equations That Contain Like Terms
Solve.
8x x – 2 = 37
11x + 4 = 37 Combine like terms.
– 4 – 4 Subtract 4 from both sides.
11x = 33 33 11
11x =
Divide both sides by 11.
x = 3

7. Additional Example 1A Continued
Check
8x x – 2 = 37
8(3) (3) – 2 = 37 ?
Substitute 3 for x.
– 2 = 37 ?
37 = 37 ? 

8. Additional Example 1B: Solving Equations That Contain Like Terms
Solve.
4(x – 6) + 7 = 11
4(x – 6) + 7 = 11 Distributive Property
4(x) – 4(6) + 7 = 11
Simplify by multiplying:
4(x) = 4x and 4(6) = 24.
4x – = 11
4x – 17 = 11
Simplify by adding: – = 17.
Add 17 to both sides.
4x = 28
Divide both sides by 4. 4
x = 7

9. Check It Out: Example 1
Solve.
9x x – 2 = 42
13x + 3 = 42 Combine like terms.
– 3 – 3 Subtract 3 from both sides.
13x = 39 39 13
13x =
Divide both sides by 13.
x = 3

10. Check It Out: Example 1 Continued
9x x – 2 = 42
9(3) (3) – 2 = 42 ?
Substitute 3 for x.
– 2 = 42 ?
42 = 42 ? 

11. If an equation contains fractions, it may help to multiply both sides of the equation by the least common denominator (LCD) of the fractions. This step results in an equation without fractions, which may be easier to solve.

12. The least common denominator (LCD) is the smallest number that each of the denominators will divide into.
Remember!

13. Additional Example 2: Solving Equations That Contain Fractions
Solve.
+ – = x 2 7x 9 17 2 3
The LCD is 18.
( ) ( ) x 2 3 7x 9 17
– = 18
Multiply both sides by 18.
18( ) + 18( ) – 18( ) = 18( ) 7x 9 x 2 17 3
Distributive Property.
14x + 9x – 34 = 12
23x – 34 = 12 Combine like terms.

14. Additional Example 2 Continued
23x – 34 = Combine like terms.
Add 34 to both sides.
23x = 46 =
23x 23 46
Divide both sides by 23.
x = 2

15. Additional Example 2 Continued
Check x 2 7x 9 17 2 3
+ – = 2 3
Substitute 2 for x.
7(2) 9
– =
(2) 17 ? 2 3 14 9
+ – = 17 ? 2 3 14 9
+ – = 17 ? 1
The LCD is 9. 6 9 14
+ – = 17 ? 6 9 = ? 

16. Check It Out: Example 2A
Solve.
+ = – 3n 4 5 4 1 4
Multiply both sides by 4 to clear fractions, and then solve.
( ) ( ) 5 4 –1 3n
= 4
( ) ( ) ( ) 3n 4 5 –1
= 4
Distributive Property.
3n + 5 = –1

17. Check It Out: Example 2A Continued
– 5 – Subtract 5 from both sides.
3n = –6 3n 3 –6 =
Divide both sides by 3.
n = –2

18. Check It Out: Example 2B
Solve.
+ – = x 3 5x 9 13 1 3
The LCD is 9.
( ) x 3 1 5x 9 13
– = 9( )
Multiply both sides by 9.
9( ) + 9( )– 9( ) = 9( ) 5x 9 x 3 13 1
Distributive Property.
5x + 3x – 13 = 3
8x – 13 = 3 Combine like terms.

19. Check It Out: Example 2B Continued
8x – 13 = Combine like terms.
Add 13 to both sides.
8x = 16 = 8x 8 16
Divide both sides by 8.
x = 2

20. Check It Out: Example 2B Continued3 5x 9 13 1 3
+ – = 1 3
Substitute 2 for x.
5(2) 9
– =
(2) 13 ? 1 3 10 9
+ – = 2 13 ?
The LCD is 9. 3 9 10
+ – = 6 13 ? 3 9 = ? 

21. Additional Example 3: Travel Application
On Monday, David rides his bicycle m miles in 2 hours. On Tuesday, he rides three times as far in 5 hours. If his average speed for the two days is 12 mi/h, how far did he ride on Monday? Round your answer to the nearest tenth of a mile.
David’s average speed is his total distance for the two days divided by the total time.
Total distance
Total time =
average speed

22. Additional Example 3 Continued
2 + 5
= 12
m + 3m
Substitute m + 3m for total distance and for total time. 7
= 12 4m
Simplify.
= 7(12) 7 4m
Multiply both sides by 7.
4m = 84
84 4
4m 4 =
Divide both sides by 4.
m = 21
David rode 21.0 miles.

23. Check It Out! Example 3
On Saturday, Penelope rode her scooter m miles in 3 hours. On Sunday, she rides twice as far in 7 hours. If her average speed for two days is 20 mi/h, how far did she ride on Saturday? Round your answer to the nearest tenth of a mile.
Penelope’s average speed is her total distance for the two days divided by the total time.
Total distance
Total time =
average speed

24. Check It Out! Example 3 Continued
3 + 7
= 20
m + 2m
Substitute m + 2m for total distance and for total time. 10
= 20 3m
Simplify.
= 10(20) 10 3m
Multiply both sides by 10.
3m = 200
3m 3 =
Divide both sides by 3.
m  66.67
Penelope rode approximately 66.7 miles.

25. Lesson Quiz for Student Response Systems
Lesson Quizzes
Standard Lesson Quiz
Lesson Quiz for Student Response Systems 25

26. Lesson Quiz Solve. 1. 6x + 3x – x + 9 = 33 2. 8(x + 2) + 5 = 29 x = 3=
5. Linda is paid double her normal hourly rate for each hour she works over 40 hours in a week. Last week she worked 52 hours and earned $544. What is her hourly rate?
x = 3
x = 1 5 8 x 8 33 8
x = 28
x = 1 9 16 25 21
– = 6x 7 2x 21
$8.50

27. Lesson Quiz for Student Response Systems
1. Solve 4p p = 88
A. p = 5
B. p = 7
C. p = 75
D. p = 91 27

28. Lesson Quiz for Student Response Systems
2. Solve 4(x + 3) + 5 = 109
A. x = 4
B. x = 23
C. x = 26
D. x = 101 28

29. Lesson Quiz for Student Response Systems
3. Solve
A. x = 5
B. x = 6
C. x = 21
D. x = 101 29