1. Solving Rational Equations
12-7
Solving Rational Equations
Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 1

2. Warm Up
1. Find the LCM of x, 2x2, and 6.
2. Find the LCM of p2 – 4p and p2 – 16.
Multiply. Simplify your answer. 5.

3. Objectives
Solve rational equations.
Identify extraneous solutions.

4. Vocabulary
rational equation

5. A rational equation is an equation that contains one or more rational expressions. If a rational equation is a proportion, it can be solved using the Cross Product Property.

6.  Example 1: Solving Rational Equations by Using Cross Products
Solve Check your answer.
Check
Use cross products.
5x = (x – 2)(3)
Distribute 3 on the right side.
–1 –1 
5x = 3x – 6
2x = –6
Subtract 3x from both sides.
x = –3

7.  Check It Out! Example 1a Solve . Check your answer. Check
Use cross products.
3n = (n + 4)(1)
Distribute 1 on the right side. 
3n = n + 4
2n = 4
Subtract n from both sides.
n = 2
Divide both sides by 2.

8.  Check It Out! Example 1b Solve . Check your answer. Check
Use cross products. 
4h = (h + 1)(2)
Distribute 2 on the right side.
4h = 2h + 2
2h = 2
Subtract 2h from both sides.
h = 1
Divide both sides by 2.

9.  Check It Out! Example 1c Solve . Check your answer. Check
Use cross products.
21x = (x – 7)(3)
Distribute 3 on the right side.
21x = 3x –21
18x = –21
Subtract 3x from both sides. 
x =
Divide both sides by 18.

10. Some rational equations contain sums or differences of rational expressions. To solve these, you must find the LCD of all the rational expressions in the equation.

11. Example 2A: Solving Rational Equations by Using the LCD
Solve each equation. Check your answer.
Step 1 Find the LCD
2x(x + 1)
Include every factor of the denominator.
Step 2 Multiply both sides by the LCD
Distribute on the left side.

12. Example 2A Continued
Step 3 Simplify and solve.
Divide out common factors.
(2x)(2) +6(x +1) = 5(x +1)
Simplify.
4x + 6x + 6 = 5x + 5
Distribute and multiply.
10x + 6 = 5x + 5
Combine like terms.
Subtract 5x and 6 from both sides.
5x = –1
Divide both sides by 5.

13. Example 2A Continued
Check Verify that your solution is not extraneous. 

14. Example 2B: Solving Rational Equations by Using the LCD
Solve each equation. Check your answer.
Step 1 Find the LCD
(x2)
Include every factor of the denominator.
Step 2 Multiply both sides by the LCD
Distribute on the left side.

15. Example 2B Continued
Step 3 Simplify and solve.
Divide out common factors.
4x – 3 = x2
Simplify.
Subtract x2 from both sides.
–x2 + 4x – 3 = 0
x2 – 4x + 3 = 0
Multiply by – 1.
(x – 3)(x – 1) = 0
Factor.
x = 3, 1
Solve.

16. Example 2B Continued
Check Verify that your solution is not extraneous.  

17. Check It Out! Example 2a
Solve each equation. Check your answer.
Step 1 Find the LCD
a(a +1)
Include every factor of the denominator.
Step 2 Multiply both sides by the LCD
Distribute on the left side.

18. Check It Out! Example 2a Continued
Step 3 Simplify and solve.
Divide out common factors.
3a = 4(a + 1)
Simplify.
3a = 4a + 4
Distribute the 4.
–4 = a
Subtract the 4 and 3a from both sides.

19. Check It Out! Example 2a Continued
Check Verify that your solution is not extraneous. 

20. Check It Out! Example 2b
Solve each equation. Check your answer.
Step 1 Find the LCD
2j(j +2)
Include every factor of the denominator.
Step 2 Multiply both sides by the LCD
Distribute on the left side.

21. Check It Out! Example 2b Continued
Solve each equation. Check your answer.
Divide out common terms.
12j – 10(2j + 4) = 4j + 8
Simplify.
12j – 20j – 40 = 4j + 8
Distribute 10.
–12j = 48
Combine like terms.
j = –4

22. Check It Out! Example 2b Continued
Check Verify that your solution is not extraneous. 

23. Check It Out! Example 2c
Solve each equation. Check your answer.
Step 1 Find the LCD
t(t +3)
Include every factor of the denominator.
Step 2 Multiply both sides by the LCD
Distribute on the right side.

24. Check It Out! Example 2c Continued
Solve each equation. Check your answer.
Divide out common terms.
8t = (t + 3) + t(t + 3)
Simplify.
8t = t t2 + 3t
Distribute t.
0 = t2 – 4t + 3
Combine like terms.
0 = (t – 3)(t – 1)
Factor.
t = 3, 1

25. Check It Out! Example 2c Continued
Check Verify that your solution is not extraneous.  

26. Example 3: Problem-Solving Application
Copy machine A can make 200 copies in 60 minutes. Copy machine B can make 200 copies in 10 minutes. How long will it take both machines working together to make 200 copies?

27. Understand the Problem1
Understand the Problem
The answer will be the number of minutes m machine A and machine B need to print the copies.
List the important information:
Machine A can print the copies in 60 minutes, which is of the job in 1 minute.
Machine B can print the copies in 10 minutes, which is of the job in 1 minute.

28. 2
Make a Plan
The part of the copies that machine A can print plus the part that machine B can print equals the complete job. Machine A’s rate times the number of minutes plus machine B’s rate times the number of minutes will give the complete time to print the copies.
(machine A’s rate) m
(machine B’s rate) +
complete job = m + = 1

29. Solve 3
Multiply both sides by the LCD, 60.
Distribute 60 on the left side.
1m + 6m = 60
7m = 60
Combine like terms.
Divide both sides by 7.
Machine A and Machine B working together can print the copies in a little more than 8.5 minutes.

30. 4
Look Back
Machine A prints of the copies per minute and machine B prints of the copies per minute. So in minutes, machine A prints
of the copies and machine B prints
of the copies. Together, they print

31. Understand the Problem
Check It Out! Example 3
Cindy mows a lawn in 50 minutes. It takes Sara 40 minutes to mow the same lawn. How long will it take them to mow the lawn if they work together? 1
Understand the Problem
The answer will be the number of minutes m Sara and Cindy need to mow the lawn.

32. Understand the Problem1
Understand the Problem
The answer will be the number of minutes m Sara and Cindy need to mow the lawn.
List the important information:
Cindy can mow the lawn in 50 minutes, which is of the job in 1 minute.
Sara can mow the lawn in 40 minutes, which is of the job in 1 minute.

33. 2
Make a Plan
The part of the lawn Cindy can mow plus the part of the lawn Sara can mow equals the complete job. Cindy’s rate times the number of minutes plus Sara’s rate times the number of minutes will give the complete time to mow the lawn.
(Cindy’s rate) m
(Sara’s rate) +
lawn mowed = m + = 1

34. Solve 3
Multiply both sides by the LCD, 200.
5m + 4m = 200
Distribute 200 on the left side.
9m = 200
Combine like terms.
Divide both sides by 9.
Cindy and Sara working together can mow the lawn in a little more than 22.2 minutes.

35. 4
Look Back
Cindy mows of the lawn in 1 minute and Sara mows of the lawn in 1 minute. So, in minutes Cindy mows of the lawn and Sara mows of the lawn. Together they mow lawn.

36. When you multiply each side of an equation by the LCD, you may get an extraneous solution. Recall from Chapter 11 that an extraneous solution is a solution to a resulting equation that is not a solution to the original equation.

37. Extraneous solutions may be introduced by squaring both sides of an equation or by multiplying both sides of an equation by a variable expression.
Helpful Hint

38. Example 4: Extraneous Solutions
Solve Identify any extraneous solutions.
Step 1 Solve.
Use cross products.
Distribute 2 on the left side. Multiply the right side.
2(x2 – 1) = (x + 1)(x – 6)
2x2 – 2 = x2 – 5x – 6
Subtract x2 from both sides. Add 5x and 6 to both sides.
x2 + 5x + 4 = 0
Factor the quadratic expression.
(x + 1)(x + 4) = 0
Use the Zero Product Property.
x = –1 or x = –4
Solve.

39.   Example 4 Continued Solve . Identify any extraneous solutions.
Step 2 Find extraneous solutions.  
Because and
are undefined –1 is
not a solution.
The only solution is – 4, so – 1 is an extraneous solution.

40. Check It Out! Example 4a
Solve. Identify any extraneous solutions.
Step 1 Solve.
Use cross products.
Distribute 3 on the right side. Multiply the left side.
(x – 2)(x – 7) = 3(x – 7)
2x2 – 9x + 14 = 3x – 21
Subtract 3x from both sides. Add 21 to both sides.
X2 – 12x + 35 = 0
Factor the quadratic expression.
(x – 7)(x – 5) = 0
Use the Zero Product Property.
x = 7 or x = 5
Solve.

41. Check It Out! Example 4a Continued
Step 2 Find extraneous solutions.  
Because and
are undefined 7 is
not a solution.
The only solution is 5, so 7 is an extraneous solution.

42. Check It Out! Example 4b
Solve. Identify any extraneous solutions.
Step 1 Solve.
Use cross products.
Distribute 4 on the right side. Multiply the left side.
(x + 1)(x – 3) = 4(x – 2)
x2 – 2x – 3 = 4x – 8
Subtract 4x from both sides. Add 8 to both sides.
X2 – 6x + 5 = 0
Factor the quadratic expression.
(x – 1)(x – 5) = 0
Use the Zero Product Property.
x = 1 or x = 5
Solve.

43. Check It Out! Example 4b Continued
Step 2 Find extraneous solutions.  
1 and 5 are
solutions.
The solutions are 1 and 5, there are no extraneous solutions.

44. Check It Out! Example 4c
Solve. Identify any extraneous solutions.
Step 1 Solve.
Use cross products.
Distribute 6 on the left side. Multiply the right side.
6(x2 + 2x) = 9(x2)
3x2 – 12x = 0
Subtract 9x2 from both sides.
Multiply through with – 1.
3x(x – 4) = 0
Factor the quadratic expression.
3x = 0, or x – 4 = 0
Use the Zero Product Property.
x = 0 or x = 4
Solve.

45. Check It Out! Example 4c Continued
Step 2 Find extraneous solutions.  
Because and
are undefined 0 is
not a solution.
The only solution is 4, so 0 is an extraneous solutions.

46. Lesson Quiz: Part I
Solve each equation. Check your answer. 1. 24
–4, 3 2. 3.
4. Pipe A can fill a tank with water in 4 hours. Pipe B can fill the same tank in 5 hours. How long will it take both pipes working together to fill the tank?

47. Lesson Quiz: Part II
5. Solve
Identify any extraneous solutions.
–5; 3 is extraneous.