1. Splash Screen

2. Five-Minute Check (over Lesson 2–8) CCSS Then/Now New Vocabulary
Example 1: Real-World Example: Mixture Problem
Example 2: Real-World Example: Percent Mixture Problem
Example 3: Real-World Example: Speed of One Vehicle
Example 4: Real-World Example: Speed of Two Vehicles
Lesson Menu

3. Solve 6r + t = r – 1 for r. A. B. C. D.
5-Minute Check 1

4. Solve 4c – d = 4a – 2c + 1 for c. A. B. C. D.
5-Minute Check 2

5. for h. A. B. C. D.
5-Minute Check 3

6. A. 4.6 cm
B. 5.8 cm
C. 6.2 cm
D. 6.4 cm
5-Minute Check 4

7. A. 236.25 miles B. 472.5 miles C. 945 miles D. 1285.75 miles
5-Minute Check 5

8. Mathematical Practices
Content Standards
A.REI.1 Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
A.REI.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
Mathematical Practices
1 Make sense of problems and persevere in solving them.
4 Model with mathematics.
Common Core State Standards © Copyright National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved.
CCSS

9. You translated sentences into equations.
Solve mixture problems.
Solve uniform motion problems.
Then/Now

10. uniform motion problem rate problem
weighted average
mixture problem
uniform motion problem
rate problem
Vocabulary

11. Let w = the number of pounds of cheaper cat food. Make a table.
Mixture Problem
PETS Mandisha feeds her cat gourmet cat food that costs $1.75 per pound. She combines it with cheaper food that costs $0.50 per pound. How many pounds of cheaper food should Mandisha buy to go with 5 pounds of gourmet food, if she wants the average price to be $1.00 per pound?
Let w = the number of pounds of cheaper cat food. Make a table.
Example 1

12. Price of gourmet cat food price of cheaper cat food
Mixture Problem
Write and solve an equation using the information in the table.
Price of gourmet cat food
plus
price of cheaper cat food
equals
price of mixed cat food.
w = (5 + w)
w = 1.00(5 + w) Original equation
w = 5 + 1w Distributive Property
w – 0.5w = 5 + 1w – 0.5w Subtract 0.5w from each side.
8.75 = w Simplify.
Example 1

13. 8.75 – 5 = 5 + 0.5w – 5 Subtract 5 from each side.
Mixture Problem
8.75 – 5 = w – 5 Subtract 5 from each side.
3.75 = 0.5w Simplify.
Divide each side by 0.5.
7.5 = w Simplify.
Answer: Mandisha should buy 7.5 pounds of cheaper cat food to be mixed with the 5 pounds of gourmet cat food so that the average price is $1.00 per pound of cat food.
Example 1

14. Cheryl bought 3 ounces of glass beads that cost $1. 79 an ounce
Cheryl bought 3 ounces of glass beads that cost $1.79 an ounce. The seed beads cost $0.99 an ounce. How many ounces of seed beads can she buy if she only wants the beads to be $1.29 an ounce for her craft project?
A. 3.1 ounces
B. 5 ounces
C. 4.6 ounces
D. 2 ounces
Example 1

15. Percent Mixture Problem
AUTO MAINTENANCE A car’s radiator should contain a solution of 50% antifreeze. Bae has 2 gallons of a 35% antifreeze. How many gallons of 100% antifreeze should Bae add to his solution to produce a solution of 50% antifreeze?
Let g = the number of gallons of 100% antifreeze to be added. Make a table.
Example 2

16. Write and solve an equation using the information in the table.
Percent Mixture Problem
Write and solve an equation using the information in the table.
Amount of antifreeze in 35% solution
plus
amount of antifreeze in 100% solution
equals
amount of antifreeze in 50% solution.
0.35(2) (g) = (2 + g)
0.35(2) + 1.0(g) = 0.50(2 + g) Original equation
g = g Distributive Property
g – 0.50g = g – 0.50g Subtract 0.50g from each side.
Example 2

17. 0.70 + 0.50g – 0.70 = 1 – 0.70 Subtract 0.70 from each side.
Percent Mixture Problem
g = 1 Simplify.
g – 0.70 = 1 – 0.70 Subtract 0.70 from each side.
0.50g = 0.30 Simplify.
Divide each side by 0.50.
g = 0.6 Simplify.
Answer: Bae should add 0.6 gallon of 100% antifreeze to produce a 50% solution.
Example 2

18. A recipe calls for mixed nuts with 50% peanuts
A recipe calls for mixed nuts with 50% peanuts pound of 15% peanuts has already been used. How many pounds of 75% peanuts needs to be added to obtain the required 50% mix?
A. B.
C. D. lb
Example 2

19. Speed of One Vehicle
AIR TRAVEL Nita took a non-stop flight to visit her grandmother. The 750-mile trip took three hours and 45 minutes. Because of bad weather, the return trip took four hours and 45 minutes. What was her average speed for the round trip?
Understand We know that Nita did not travel the same amount of time on each portion of her trip. So, we will need to find the weighted average of the plane’s speed. We are asked to find the average speed for both portions of the trip.
Example 3

20. Speed of One Vehicle
Plan First, find the rate of the going portion, and then the return portion of the trip. Because the rate is in miles per hour, convert 3 hours and 45 minutes to hours and 4 hours 45 minutes to hours.
Going
Formula for rate
Example 3

21. Return Formula for rate
Speed of One Vehicle
Return
Formula for rate
Because we are looking for a weighted average, we cannot just average the speeds. We need to find the weighted average for the round trip.
Example 3

22. Speed of One Vehicle
Solve
Substitution
Simplify.
Example 3

23. Answer: The average speed was about 176 miles per hour.
Speed of One Vehicle
Answer: The average speed was about 176 miles per hour.
Check The solution of 176 miles per hour is between the going portion rate 200 miles per hour, and the return rate, miles per hour. So, the answer is reasonable.
Example 3

24. In the morning, when traffic is light, it takes 30 minutes to get to work. The trip is 15 miles through towns. In the afternoon, when traffic is a little heavier, it takes 45 minutes. What is the average speed for the round trip?
A. 24 miles per hour
B. 30 miles per hour
C. 15 miles per hour
D. 45 miles per hour
Example 3

25. Speeds of Two Vehicles
RESCUE A railroad switching operator has discovered that two trains are heading toward each other on the same track. Currently, the trains are 53 miles apart. One train is traveling at 75 miles per hour and the other 40 miles per hour. The faster train will require 5 miles to stop safely, and the slower train will require 3 miles to stop safely. About how many minutes does the operator have to warn the train engineers to stop their trains?
Step 1 Draw a diagram.
53 miles apart
Takes 5 miles to stop
Takes 3 miles to stop
53 – (5 + 3) = 45 miles
Example 4

26. Distance traveled by fast train distance traveled by other train
Speeds of Two Vehicles
Step 2 Let m = the number of hours that the operator has to warn the train engineers to stop their trains safely. Make a table.
Step 3 Write and solve an equation using the information in the table.
Distance traveled by fast train
plus
distance traveled by other train
equals
45 miles.
75m m =
Example 4

27. 75m + 40m = 45 Original equation
Speeds of Two Vehicles
75m + 40m = 45 Original equation
115m = 45 Simplify.
Divide each side by 115.
m ≈ 0.39 Round to the nearest hundredth.
0.39 × 60 = 23.4 Convert to minutes by multiplying by 60.
Answer: The operator has about 23 minutes to warn the engineers.
Example 4

28. Two students left the school on their bicycles at the same time, one heading north and the other heading south. The student heading north travels 15 miles per hour, and the one heading south travels at 17 miles per hour. After about how many minutes will they be 7.5 miles apart?
A. 17 minutes
B. 15 minutes
C. 14 minutes
D. 30 minutes
Example 4

29. End of the Lesson