1. Solving Rational Equations and Inequalities
5-5
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. 2x and 4x2 – 2x
2. x + 5 and x2 – x – 30
3. 3x and x2 + 6x + 8
4. 5 and x2 – 5x + 6

3. Objective
Solve rational equations and inequalities.

4. Vocabulary
rational equation
extraneous solution
rational inequality

5. A rational equation is an equation that contains one or more rational expressions. The time t in hours that it takes to travel d miles can be determined by using the equation t = , where r is the average rate of speed. This equation is a rational equation. d r

6. To solve a rational equation, start by multiplying each term of the equation by the least common denominator (LCD) of all of the expressions in the equation. This step eliminates the denominators of the rational expression and results in an equation you can solve by using algebra.

7. Example 1: Solving Rational Equations18 x
Solve the equation x – = 3.
x(x) – (x) = 3(x) 18 x
Multiply each term by the LCD, x.
x2 – 18 = 3x
Simplify. Note that x ≠ 0.
x2 – 3x – 18 = 0
Write in standard form.
(x – 6)(x + 3) = 0
Factor.
x – 6 = 0 or x + 3 = 0
Apply the Zero Product Property.
x = 6 or x = –3
Solve for x.

8. Example 1 Continued 18 x 18 x
Check
x – = 3
x – = 3 18 6
6 – 18
(–3)
(–3) – 3 3
6 – 3 3
–3 + 6 3 3 3  3 3 

9. Check It Out! Example 1a
Solve the equation = 4 x 10 3
(3x) = (3x) + 2(3x) 10 3 4 x
Multiply each term by the LCD, 3x.
10x = x
Simplify. Note that x ≠ 0.
4x = 12
Combine like terms.
x = 3
Solve for x.

10. Check It Out! Example 1b
Solve the equation = – . 5 4 6 x 7
(4x) (4x) = – (4x) 6 x 5 4 7
Multiply each term by the LCD, 4x.
24 + 5x = –7x
Simplify. Note that x ≠ 0.
24 = –12x
Combine like terms.
x = –2
Solve for x.

11. Check It Out! Example 1c 6 x
Solve the equation x = – 1.
x(x) = (x) – 1(x) 6 x
Multiply each term by the LCD, x.
x2 = 6 – x
Simplify. Note that x ≠ 0.
x2 + x – 6 = 0
Write in standard form.
(x – 2)(x + 3) = 0
Factor.
x – 2 = 0 or x + 3 = 0
Apply the Zero Product Property.
x = 2 or x = –3
Solve for x.

12. An extraneous solution is a solution of an equation derived from an original equation that is not a solution of the original equation. When you solve a rational equation, it is possible to get extraneous solutions. These values should be eliminated from the solution set. Always check your solutions by substituting them into the original equation.

13. Example 2A: Extraneous Solutions
Solve each equation. 5x
x – 2
3x + 4 = 5x
x – 2
3x + 4
(x – 2) = (x – 2)
Multiply each term by the LCD, x – 2. 5x
x – 2
3x + 4
(x – 2) = (x – 2)
Divide out common factors.
5x = 3x + 4
Simplify. Note that x ≠ 2.
x = 2
Solve for x.
The solution x = 2 is extraneous because it makes the denominators of the original equation equal to 0. Therefore, the equation has no solution.

14. Example 2A Continued
Check Substitute 2 for x in the original equation. 5x
x – 2
3x + 4 =
5(2)
2 – 2
3(2) + 4
2 – 2 10 10
Division by 0 is undefined.

15. Example 2B: Extraneous Solutions
Solve each equation.
2x – 5
x – 8 x 2 11
x – 8 =
Multiply each term by the LCD, 2(x – 8).
2x – 5
x – 8
2(x – 8) (x – 8) = (x – 8) 11 x 2
Divide out common factors.
2x – 5
x – 8
2(x – 8) (x – 8) = (x – 8) 11 x 2
2(2x – 5) + x(x – 8) = 11(2)
Simplify. Note that x ≠ 8.
Use the Distributive Property.
4x – 10 + x2 – 8x = 22

16. Example 2B Continued
x2 – 4x – 32 = 0
Write in standard form.
(x – 8)(x + 4) = 0
Factor.
x – 8 = 0 or x + 4 = 0
Apply the Zero Product Property.
x = 8 or x = –4
Solve for x.
The solution x = 8 us extraneous because it makes the denominator of the original equation equal to 0. The only solution is x = –4.

17. Check It Out! Example 2a 16
x2 – 16 2
x – 4
Solve the equation =
Multiply each term by the LCD, (x – 4)(x +4). 16
(x – 4)(x + 4) 2
x – 4
(x – 4)(x + 4) = (x – 4 )(x + 4)
Divide out common factors. 16
(x – 4)(x + 4) 2
x – 4
(x – 4)(x + 4) = (x – 4 )(x + 4)
16 = 2x + 8
Simplify. Note that x ≠ ±4.
x = 4
Solve for x.
The solution x = 4 is extraneous because it makes the denominators of the original equation equal to 0. Therefore, the equation has no solution.

18. Check It Out! Example 2b x
x – 1
Solve the equation 1 = 6
Multiply each term by the LCD, 6(x – 1). 1
x – 1
6(x – 1) = (x – 1) (x – 1) x 6
Divide out common factors. 1
x – 1
6(x – 1) = (x – 1) (x – 1) x 6
6 = 6x + x(x – x)
Simplify. Note that x ≠ 1.
Use the Distributive Property.
6 = 6x + x2 – x

19. Check It Out! Example 2b Continued
0 = x2 + 5x – 6
Write in standard form.
0 = (x + 6)(x – 1)
Factor.
x + 6 = 0 or x – 1 = 0
Apply the Zero Product Property.
x = –6 or x = 1
Solve for x.
The solution x = 1 us extraneous because it makes the denominator of the original equation equal to 0. The only solution is x = –6.

20. A rational inequality is an inequality that contains one or more rational expressions. One way to solve rational inequalities is by using graphs and tables.

21. Vertical asymptote: x = 6
Example 5: Using Graphs and Tables to Solve Rational Equations and Inequalities
Solve ≤ 3 by using a graph and a table. x
x – 6 x
x – 6
Use a graph. On a graphing calculator, Y1 = and Y2 = 3.
(9, 3)
The graph of Y1 is at or below the graph of Y2 when x < 6 or when x ≥ 9.
Vertical asymptote: x = 6

22. Example 5 Continued
Use a table. The table shows that Y1 is undefined when x = 6 and that Y1 ≤ Y2 when x ≥ 9.
The solution of the inequality is x < 6 or x ≥ 9.

23. Vertical asymptote: x = 3
Check It Out! Example 5a
Solve ≥ 4 by using a graph and a table. x
x – 3 x
x – 3
Use a graph. On a graphing calculator, Y1 = and Y2 = 4.
(4, 4)
The graph of Y1 is at or below the graph of Y2 when x < 3 or when x ≥ 4.
Vertical asymptote: x = 3

24. Check It Out! Example 5a continued
Use a table. The table shows that Y1 is undefined when x = 3 and that Y1 ≤ Y2 when x ≥ 4.
The solution of the inequality is x < 3 or x ≥ 4.

25. Vertical asymptote: x = –1
Check It Out! Example 5b
Solve = –2 by using a graph and a table. 8
x + 1 8
x + 1
Use a graph. On a graphing calculator, Y1 = and Y2 = –2.
(–5, –2)
The graph of Y1 is at or below the graph of Y2 when x = –5.
Vertical asymptote: x = –1

26. Check It Out! Example 5b continued
Use a table. The table shows that Y1 is undefined when x = –1 and that Y1 ≤ Y2 when x = –5.
The solution of the inequality is x = –5.

27. You can also solve rational inequalities algebraically
You can also solve rational inequalities algebraically. You start by multiplying each term by the least common denominator (LCD) of all the expressions in the inequality. However, you must consider two cases: the LCD is positive or the LCD is negative.

28. Example 6: Solving Rational Inequalities Algebraically
Solve ≤ 3 algebraically. 6
x – 8
Case 1 LCD is positive.
Step 1 Solve for x. 6
x – 8
(x – 8) ≤ 3(x – 8)
Multiply by the LCD.
6 ≤ 3x – 24
Simplify. Note that x ≠ 8.
30 ≤ 3x
Solve for x.
10 ≤ x
Rewrite with the variable on the left.
x ≥ 10

29. Example 6 Continued
Solve ≤ 3 algebraically. 6
x – 8
Step 2 Consider the sign of the LCD.
x – 8 > 0
LCD is positive.
x > 8
Solve for x.
For Case 1, the solution must satisfy x ≥ 10 and x > 8, which simplifies to x ≥ 10.

30. Example 6: Solving Rational Inequalities Algebraically
Solve ≤ 3 algebraically. 6
x – 8
Case 2 LCD is negative.
Step 1 Solve for x. 6
x – 8
(x – 8) ≥ 3(x – 8)
Multiply by the LCD. Reverse the inequality.
6 ≥ 3x – 24
Simplify. Note that x ≠ 8.
30 ≥ 3x
Solve for x.
10 ≥ x
Rewrite with the variable on the left.
x ≤ 10

31. Step 2 Consider the sign of the LCD.
Example 6 Continued
Solve ≤ 3 algebraically. 6
x – 8
Step 2 Consider the sign of the LCD.
x – 8 > 0
LCD is positive.
x > 8
Solve for x.
For Case 2, the solution must satisfy x ≤ 10 and x < 8, which simplifies to x < 8.
The solution set of the original inequality is the union of the solutions to both Case 1 and Case 2. The solution to the inequality ≤ 3 is x < 8 or x ≥ 10, or {x|x < 8  x ≥ 10}. 6
x – 8

32. Check It Out! Example 6a
Solve ≥ –4 algebraically. 6
x – 2
Case 1 LCD is positive.
Step 1 Solve for x. 6
x – 2
(x – 2) ≥ –4(x – 2)
Multiply by the LCD.
6 ≥ –4x + 8
Simplify. Note that x ≠ 2.
–2 ≥ –4x
Solve for x.
≤ x 1 2
Rewrite with the variable on the left.
x ≥ 1 2

33. Check It Out! Example 6a Continued
Solve ≥ –4 algebraically. 6
x – 2
Step 2 Consider the sign of the LCD.
x – 2 > 0
LCD is positive.
x > 2
Solve for x.
For Case 1, the solution must satisfy and x > 2, which simplifies to x > 2.
x ≥ 1 2

34. Check It Out! Example 6a Continued
Solve ≥ –4 algebraically. 6
x – 2
Case 2 LCD is negative.
Step 1 Solve for x. 6
x – 2
(x – 2) ≤ –4(x – 2)
Multiply by the LCD. Reverse the inequality.
6 ≤ –4x + 8
Simplify. Note that x ≠ 2.
–2 ≤ –4x
Solve for x.
≥ x 1 2
Rewrite with the variable on the left.
x ≤ 1 2

35. Check It Out! Example 6a Continued
Solve ≥ –4 algebraically. 6
x – 2
Step 2 Consider the sign of the LCD.
x – 2 < 0
LCD is negative.
x < 2
Solve for x.
For Case 2, the solution must satisfy and x < 2, which simplifies to
x ≤ 1 2
The solution set of the original inequality is the union of the solutions to both Case 1 and Case 2. The solution to the inequality ≥ –4 is x > 2 or x ≤ , or {x|  x > 2}. 6
x – 2
x ≤ 1 2

36. Check It Out! Example 6b
Solve < 6 algebraically. 9
x + 3
Case 1 LCD is positive.
Step 1 Solve for x. 9
x + 3
(x + 3) < 6(x + 3)
Multiply by the LCD.
9 < 6x + 18
Simplify. Note that x ≠ –3.
–9 < 6x
Solve for x.
– < x 3 2
Rewrite with the variable on the left.
x > – 3 2

37. Check It Out! Example 6b Continued
Solve < 6 algebraically. 9
x + 3
Step 2 Consider the sign of the LCD.
x + 3 > 0
LCD is positive.
x > –3
Solve for x.
For Case 1, the solution must satisfy and x > –3, which simplifies to
x >– 3 2

38. Check It Out! Example 6b Continued
Solve < 6 algebraically. 9
x + 3
Case 2 LCD is negative.
Step 1 Solve for x.
Multiply by the LCD. Reverse the inequality. 9
x + 3
(x + 3) > 6(x + 3)
9 > 6x + 18
Simplify. Note that x ≠ –3.
–9 > 6x
Solve for x.
– > x 3 2
Rewrite with the variable on the left.
x < – 3 2

39. Check It Out! Example 6b Continued
Solve < 6 algebraically. 9
x + 3
Step 2 Consider the sign of the LCD.
x + 3 < 0
LCD is negative.
x < –3
Solve for x.
For Case 2, the solution must satisfy and x < –3, which simplifies to x < –3.
x <– 3 2
The solution set of the original inequality is the union of the solutions to both Case 1 and Case 2. The solution to the inequality < 6 is x < –3 or x > – , or {x|  x < –3}.
x > – 3 2 9
x + 3

40. Lesson Quiz
Solve each equation or inequality.
x + 2 x
x – 1 2 = 1.
x = –1 or x = 4 6x
x + 4
7x + 4 = 2.
no solution 3.
x + 2
x – 3 x 5 5
x – 3 =
x = –5 4
x – 3 4.
≥ 2
3 < x ≤ 5
5. A college basketball player has made 58 out of 82 attempted free-throws this season. How many additional free-throws must she make in a row to raise her free-throw percentage to 90%?
158