1. Solving Rational Equations and Inequalities
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Algebra 2
Holt Algebra 2

2. Find the least common multiple for each pair.
Warm Up
Find the least common multiple for each pair.
1. 2x2 and 4x2 – 2x
2x2(2x – 1)
2. x + 5 and x2 – x – 30
(x + 5)(x – 6)
Add or subtract. Identify any x-values for which the expression is undefined. 1
x – 2 4x + 3.
5x – 2
4x(x – 2)
–(x – 1) x2 1 x2 x –
x ≠ 0 4.

3. Objective
Solve rational equations and inequalities.

4. Vocabulary
rational equation
extraneous solution
rational inequality

5. Example 3: Problem-Solving Application
A jet travels 3950 mi from Chicago, Illinois, to London, England, and 3950 mi on the return trip. The total flying time is 16.5 h. The return trip takes longer due to winds that generally blow from west to east. If the jet’s average speed with no wind is 485 mi/h, what is the average speed of the wind during the round-trip flight? Round to the nearest mile per hour.

6. Understand the Problem
Example 3 Continued 1
Understand the Problem
The answer will be the average speed of the wind.
List the important information:
The jet spent 16.5 h on the round-trip.
It went 3950 mi east and 3950 mi west.
Its average speed with no wind is 485 mi/h.

7. Example 3 Continued 2 Make a Plan
Let w represent the speed of the wind. When the jet is going east, its speed is equal to its speed with no wind plus w. When the jet is going west, its speed is equal to its speed with no wind minus w.
Distance (mi)
Average Speed (mi/h)
Time (h)
East
3950
485 + w
West
485 – w
485 + w
3950
485 – w
3950
total time =
time east +
time west
16.5
485 + w
3950
485 – w = +

8. Use the Distributive Property.
Solve 3
The LCD is (485 + w)(485 – w).
16.5(485 + w)(485 – w)
485 + w
3950
= (485 + w)(485 – w)
485 – w
(485 + w)(485 – w)
Simplify. Note that x ≠ ±485.
16.5(485 + w)(485 – w) = 3950(485 – w) (485 + w)
Use the Distributive Property.
3,881,212.5 – 16.5w2 = 1,915,750 – 3950w + 1,915, w
3,881,212.5 – 16.5w2 = 3,831,500
Combine like terms.
–16.5w2 = –49,712.5
Solve for w.
w ≈ ± 55
The speed of the wind cannot be negative. Therefore, the average speed of the wind is 55 mi/h.

9. Example 3 Continued
Look Back 4
If the speed of the wind is 55 mi/h, the jet’s speed when going east is = 540 mi/h. It will take the jet approximately 7.3 h to travel 3950 mi east. The jet’s speed when going west is 485 – 55 = 430 mi/h. It will take the jet approximately 9.2 h to travel 3950 mi west. The total trip will take 16.5 h, which is the given time.

10. Check It Out! Example 3
On a river, a kayaker travels 2 mi upstream and 2 mi downstream in a total of 5 h. In still water, the kayaker can travel at an average speed of 2 mi/h. Based on this information, what is the average speed of the current of this river? Round to the nearest tenth.

11. Understand the Problem
Check It Out! Example 3 Continued 1
Understand the Problem
The answer will be the average speed of the current.
List the important information:
The kayaker spent 5 hours kayaking.
She went 2 mi upstream and 2 mi downstream.
Her average speed in still water is 2 mi/h.

12. Check It Out! Example 3 Continued2
Make a Plan
Let c represent the speed of the current. When the kayaker is going upstream, her speed is equal to her speed in still water minus c. When the kayaker is going downstream, her speed is equal to her speed in still water plus c.
Distance (mi)
Average Speed (mi/h)
Time (h) Up 2
2 – c
Down
2 + c
2 – c 2
2 + c 2
total time =
time up- stream +
time down- stream 5
2 – c 2
2 + c = +

13. Use the Distributive Property.
Solve 3
The LCD is (2 – c)(2 + c).
(2 + c)(2 – c)
= (2 + c)(2 – c)
5(2 + c)(2 – c)
2 – c 2
2 + c
Simplify. Note that x ≠ ±2.
5(2 + c)(2 – c) = 2(2 + c) + 2(2 – c)
Use the Distributive Property.
20 – 5c2 = 4 + 2c + 4 – 2c
20 – 5c2 = 8
Combine like terms.
–5c2 = –12
Solve for c.
c ≈ ± 1.5
The speed of the current cannot be negative. Therefore, the average speed of the current is about 1.5 mi/h.

14. Check It Out! Example 3 Continued
Look Back 4
If the speed of the current is about 1.5 mi/h, the kayaker’s speed when going upstream is 2 – 1.5 = 0.5 mi/h. It will take her about 4 h to travel 2 mi upstream. Her speed when going downstream is about = 3.5 mi/h. It will take her 0.5 h to travel 2 mi downstream. The total trip will take about 4.5 hours which is close to the given time of 5 h.

15. Example 4: Work Application
Natalie can finish a 500-piece puzzle in about 8 hours. When Natalie and Renzo work together, they can finish a 500-piece puzzle in about 4.5 hours. About how long will it take Renzo to finish a 500-piece puzzle if he works by himself? 1 8
Natalie’s rate: of the puzzle per hour 1 h
Renzo’s rate: of the puzzle per hour, where h is the number of hours needed to finish the puzzle by himself.

16. Example 4 Continued Natalie’s rate  hours worked Renzo’s rate
1 complete puzzle + = 1 8
(4.5) 1 h
(4.5) + = 1 1 8
(4.5)(8h) + h
(4.5)(8h) = 1(8h)
Multiply by the LCD,8h.
4.5h + 36 = 8h
Simplify.
36 = 3.5h
Solve for h.
10.3 = h
It will take Renzo about 10.3 hours, or 10 hours 17 minutes to complete a 500-piece puzzle working by himself.

17. Check It Out! Example 4
Julien can mulch a garden in 20 minutes. Together Julien and Remy can mulch the same garden in 11 minutes. How long will it take Remy to mulch the garden when working alone?
Julien’s rate: of the garden per minute 1 20
Remy’s rate: of the garden per minute, where m is the number of minutes needed to mulch the garden by himself. 1 m

18. Check It Out! Example 4 Continued
Julien’s rate
 min worked
Remy’s rate
1 complete job + = 1 20
(11) 1 m
(11) + = 1 1 20
(11)(20m)+ m
(11)(20m) = 1(20m)
Multiply by the LCD, 20m.
11m = 20m
Simplify.
220 = 9m
Solve for m.
24 ≈ m
It will take Remy about 24 minutes to mulch the garden working by himself.