Solving Rational Equations 5 5 and Inequalities Warm

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Solving Rational Equations 5 -5 and Inequalities Warm Up Lesson Presentation Lesson Quiz Holt.

Solving Rational Equations 5 -5 and Inequalities Warm Up Lesson Presentation Lesson Quiz Holt. Mc. Dougal Algebra 2 Holt

5 -5 Solving Rational Equations and Inequalities Warm Up Find the least common multiple

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

5 -5 Solving Rational Equations and Inequalities Objective Solve rational equations and inequalities. Holt

5 -5 Solving Rational Equations and Inequalities Objective Solve rational equations and inequalities. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Vocabulary rational equation extraneous solution rational inequality

5 -5 Solving Rational Equations and Inequalities Vocabulary rational equation extraneous solution rational inequality Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities A rational equation is an equation that

5 -5 Solving Rational Equations and Inequalities 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 = d , where r is the average rate r of speed. This equation is a rational equation. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities To solve a rational equation, start by

5 -5 Solving Rational Equations and Inequalities 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Example 1: Solving Rational Equations Solve the

5 -5 Solving Rational Equations and Inequalities Example 1: Solving Rational Equations Solve the equation x – 18 = 3. x x(x) – 18 (x) = 3(x) Multiply each term by the LCD, x. x Simplify. Note that x ≠ 0. x 2 – 18 = 3 x x 2 – 3 x – 18 = 0 (x – 6)(x + 3) = 0 Write in standard form. Factor. x – 6 = 0 or x + 3 = 0 Apply the Zero Product Property. x = 6 or x = – 3 Holt Mc. Dougal Algebra 2 Solve for x.

5 -5 Solving Rational Equations and Inequalities Example 1 Continued 18 Check x –

5 -5 Solving Rational Equations and Inequalities Example 1 Continued 18 Check x – x = 3 6 – 18 6 6– 3 3 Holt Mc. Dougal Algebra 2 3 3 3 18 x– x =3 (– 3) – 18 (– 3) – 3 + 6 3 3

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 1 a Solve

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 1 a Solve the equation 10 = 4 + 2. x 3 10 (3 x) = 4 (3 x) + 2(3 x) x 3 10 x = 12 + 6 x 4 x = 12 x=3 Holt Mc. Dougal Algebra 2 Multiply each term by the LCD, 3 x. Simplify. Note that x ≠ 0. Combine like terms. Solve for x.

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 1 b Solve

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 1 b Solve the equation 6 + 5 = – 7. 4 x 4 6 (4 x) + 5 (4 x) = – 7 (4 x) 4 4 x 24 + 5 x = – 7 x 24 = – 12 x x = – 2 Holt Mc. Dougal Algebra 2 Multiply each term by the LCD, 4 x. Simplify. Note that x ≠ 0. Combine like terms. Solve for x.

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 1 c Solve

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 1 c Solve the equation x = 6 – 1. x x(x) = 6 (x) – 1(x) Multiply each term by the LCD, x. x Simplify. Note that x ≠ 0. x 2 = 6 – x x 2 + x – 6 = 0 (x – 2)(x + 3) = 0 Write in standard form. Factor. x – 2 = 0 or x + 3 = 0 Apply the Zero Product Property. x = 2 or x = – 3 Holt Mc. Dougal Algebra 2 Solve for x.

5 -5 Solving Rational Equations and Inequalities An extraneous solution is a solution of

5 -5 Solving Rational Equations and Inequalities 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Example 2 A: Extraneous Solutions Solve each

5 -5 Solving Rational Equations and Inequalities Example 2 A: Extraneous Solutions Solve each equation. 5 x = 3 x + 4 x– 2 Multiply each term by 5 x 3 x + 4 (x – 2) = (x – 2) x– 2 the LCD, x – 2. x– 2 5 x 3 x + 4 (x – 2) Divide out common (x – 2) = x– 2 factors. 5 x = 3 x + 4 x=2 Simplify. Note that 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Example 2 A Continued Check Substitute 2

5 -5 Solving Rational Equations and Inequalities Example 2 A Continued Check Substitute 2 for x in the original equation. 5 x = 3 x + 4 x– 2 5(2) 3(2) + 4 2– 2 10 0 Holt Mc. Dougal Algebra 2 10 0 Division by 0 is undefined.

5 -5 Solving Rational Equations and Inequalities Example 2 B: Extraneous Solutions Solve each

5 -5 Solving Rational Equations and Inequalities Example 2 B: Extraneous Solutions Solve each equation. 2 x – 5 + x = 11 x– 8 2 Multiply each term by the LCD, 2(x – 8). 2 x – 5 x 2(x – 8) = 11 2(x – 8) + 2(x – 8) x– 8 2 Divide out common factors. 2 x – 5 x 2(x – 8) = 11 2(x – 8) + 2(x – 8) x– 8 2 2(2 x – 5) + x(x – 8) = 11(2) 4 x – 10 + x 2 – 8 x = 22 Holt Mc. Dougal Algebra 2 Simplify. Note that x ≠ 8. Use the Distributive Property.

5 -5 Solving Rational Equations and Inequalities Example 2 B Continued x 2 –

5 -5 Solving Rational Equations and Inequalities Example 2 B Continued x 2 – 4 x – 32 = 0 (x – 8)(x + 4) = 0 Write in standard form. 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Check Example 2 B Continued Write 2

5 -5 Solving Rational Equations and Inequalities Check Example 2 B Continued Write 2 x – 5 + x = x– 8 2 11 x– 8 11 = 0. as 2 x – 5 + x – x– 8 2 Graph the left side of the equation as Y 1. Identify the values of x for which Y 1 = 0. The graph intersects the x-axis only when x = – 4. Therefore, x = – 4 is the only solution. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 2 a 2

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 2 a 2 Solve the equation 216. = x – 16 x– 4 Multiply each term by the LCD, (x – 4)(x +4). 16 (x – 4)(x + 4) = Divide out common factors. 16 (x – 4)(x + 4) = 2 (x – 4 )(x + 4) x– 4 Simplify. Note that x ≠ ± 4. 16 = 2 x + 8 Solve for x. x=4 The solution x = 4 is extraneous because it makes the denominators of the original equation equal to 0. Therefore, the equation has no solution. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 2 b 1

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 2 b 1 x x. = + x– 1 6 Multiply each term by the LCD, 6(x – 1). 1 x x 6(x – 1) = 6(x – 1) + x– 1 6 Solve the equation Divide out common factors. 1 x x 6(x – 1) = 6(x – 1) + x– 1 6 6 = 6 x + x(x – x) Simplify. Note that x ≠ 1. Use the Distributive 6 = 6 x + x 2 – x Property. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 2 b Continued

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 2 b Continued 0 = x 2 + 5 x – 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Example 3: Problem-Solving Application A jet travels

5 -5 Solving Rational Equations and Inequalities 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Example 3 Continued 1 Understand the Problem

5 -5 Solving Rational Equations and Inequalities 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Example 3 Continued 2 Make a Plan

5 -5 Solving Rational Equations and Inequalities Example 3 Continued 2 Make a Plan Let w represent Average the speed of the Distance Speed wind. When the jet (mi) (mi/h) Time (h) 3950 is going east, its East 3950 485 + w speed is equal to 3950 West 3950 485 – w its speed with no 485 – w wind plus w. When the jet is going total time = time east + time west, its speed is 3950 + 3950 equal to its speed 16. 5 = 485 +w 485 – w with no wind minus w. Holt Mc. Dougal Algebra 2

5 -5 3 Solving Rational Equations and Inequalities Solve The LCD is (485 +

5 -5 3 Solving Rational Equations and Inequalities Solve The LCD is (485 + w)(485 – w). 16. 5(485 + w)(485 – w) = + 3950 485 – w 3950 485 + w (485 + w)(485 – w) Simplify. Note that x ≠ ± 485. 16. 5(485 + w)(485 – w) = 3950(485 – w) + 3950 (485 + w) Use the Distributive Property. 3, 881, 212. 5 – 16. 5 w 2 = 1, 915, 750 – 3950 w + 1, 915, 750 + 3950 w 3, 881, 212. 5 – 16. 5 w 2 = 3, 831, 500 Combine like terms. – 16. 5 w 2 = – 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Example 3 Continued 4 Look Back If

5 -5 Solving Rational Equations and Inequalities Example 3 Continued 4 Look Back If the speed of the wind is 55 mi/h, the jet’s speed when going east is 485 + 55 = 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 3 On a

5 -5 Solving Rational Equations and Inequalities 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 3 Continued 1

5 -5 Solving Rational Equations and Inequalities 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 3 Continued 2

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

5 -5 3 Solving Rational Equations and Inequalities Solve 5(2 + c)(2 – c)

5 -5 3 Solving Rational Equations and Inequalities Solve 5(2 + c)(2 – c) = The LCD is (2 – c)(2 + c). 2 2 (2 + c)(2 – c) + (2 + c)(2 – c) 2–c 2+c Simplify. Note that x ≠ ± 2. 5(2 + c)(2 – c) = 2(2 + c) + 2(2 – c) Use the Distributive Property. 20 – 5 c 2 = 4 + 2 c + 4 – 2 c 20 – 5 c 2 = 8 Combine like terms. – 5 c 2 = – 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 3 Continued 4

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 3 Continued 4 Look Back 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 2 + 1. 5 = 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Example 4: Work Application Natalie can finish

5 -5 Solving Rational Equations and Inequalities 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 Natalie’s rate: of the puzzle per hour 8 Renzo’s rate: 1 of the puzzle per hour, where h is the h number of hours needed to finish the puzzle by himself. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Example 4 Continued Natalie’s rate + Renzo’s

5 -5 Solving Rational Equations and Inequalities Example 4 Continued Natalie’s rate + Renzo’s rate hours worked 1 (4. 5) 8 + 1 (4. 5) h = 1 complete puzzle = 1 1 (4. 5)(8 h) + 1 (4. 5)(8 h) = 1(8 h) Multiply by the LCD, 8 h. 8 h 4. 5 h + 36 = 8 h Simplify. 36 = 3. 5 h Solve for h. 10. 3 = h It will take Renzo about 10. 3 hours, or 10 hours 18 minutes to complete a 500 -piece puzzle working by himself. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 4 Julien can

5 -5 Solving Rational Equations and Inequalities 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: 1 of the garden per minute 20 Remy’s rate: 1 of the garden per minute, where m is m the number of minutes needed to mulch the garden by himself. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 4 Continued Julien’s

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 4 Continued Julien’s rate min worked 1 (11) 20 + Remy’s rate min worked + 1 (11) m = 1 complete job = 1 1 (11)(20 m)+ 1 (11)(20 m) = 1(20 m) Multiply by the LCD, 20 m. 20 m 11 m + 220 = 20 m Simplify. 220 = 9 m 24 ≈ m Solve for m. It will take Remy about 24 minutes to mulch the garden working by himself. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities A rational inequality is an inequality that

5 -5 Solving Rational Equations and Inequalities 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Example 5: Using Graphs and Tables to

5 -5 Solving Rational Equations and Inequalities Example 5: Using Graphs and Tables to Solve Rational Equations and Inequalities Solve x x– 6 ≤ 3 by using a graph and a table. Use a graph. On a graphing calculator, Y 1 = x and Y 2 = 3. x– 6 The graph of Y 1 is at or below the graph of Y 2 when x < 6 or when x ≥ 9. Holt Mc. Dougal Algebra 2 (9, 3) Vertical asymptote: x=6

5 -5 Solving Rational Equations and Inequalities Example 5 Continued Use a table. The

5 -5 Solving Rational Equations and Inequalities Example 5 Continued Use a table. The table shows that Y 1 is undefined when x = 6 and that Y 1 ≤ Y 2 when x ≥ 9. The solution of the inequality is x < 6 or x ≥ 9. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 5 a Solve

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 5 a Solve x ≤ 4 by using a graph and a table. x– 3 Use a graph. On a graphing calculator, Y 1 = x and Y 2 = 4. x– 3 The graph of Y 1 is at or below the graph of Y 2 when x < 3 or when x ≥ 4. Holt Mc. Dougal Algebra 2 (4, 4) Vertical asymptote: x=3

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 5 a continued

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 5 a continued Use a table. The table shows that Y 1 is undefined when x = 3 and that Y 1 ≤ Y 2 when x ≥ 4. The solution of the inequality is x < 3 or x ≥ 4. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 5 b Solve

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 5 b Solve 8 x+ 1 = – 2 by using a graph and a table. Use a graph. On a graphing calculator, 8 and Y 2 = – 2. Y 1 = x+ 1 The graph of Y 1 is at or below the graph of Y 2 when x = – 5. Holt Mc. Dougal Algebra 2 (– 5, – 2) Vertical asymptote: x = – 1

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 5 b continued

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 5 b continued Use a table. The table shows that Y 1 is undefined when x = – 1 and that Y 1 ≤ Y 2 when x = – 5. The solution of the inequality is x = – 5. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities You can also solve rational inequalities algebraically.

5 -5 Solving Rational Equations and Inequalities 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Example 6: Solving Rational Inequalities Algebraically Solve

5 -5 Solving Rational Equations and Inequalities Example 6: Solving Rational Inequalities Algebraically Solve 6 x– 8 ≤ 3 algebraically. Case 1 LCD is positive. Step 1 Solve for x. 6 (x – 8) ≤ 3(x – 8) x– 8 6 ≤ 3 x – 24 Multiply by the LCD. Simplify. Note that x ≠ 8. 30 ≤ 3 x Solve for x. 10 ≤ x x ≥ 10 Rewrite with the variable on the left. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Example 6 Continued Solve 6 x– 8

5 -5 Solving Rational Equations and Inequalities Example 6 Continued Solve 6 x– 8 ≤ 3 algebraically. 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. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Example 6: Solving Rational Inequalities Algebraically Solve

5 -5 Solving Rational Equations and Inequalities Example 6: Solving Rational Inequalities Algebraically Solve 6 x– 8 ≤ 3 algebraically. Case 2 LCD is negative. Step 1 Solve for x. 6 (x – 8) ≥ 3(x – 8) x– 8 6 ≥ 3 x – 24 Multiply by the LCD. Reverse the inequality. Simplify. Note that x ≠ 8. 30 ≥ 3 x Solve for x. 10 ≥ x x ≤ 10 Rewrite with the variable on the left. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Example 6 Continued Solve 6 x– 8

5 -5 Solving Rational Equations and Inequalities Example 6 Continued Solve 6 x– 8 ≤ 3 algebraically. Step 2 Consider the sign of the LCD. x– 8>0 LCD is positive. Solve for x. x>8 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 6 x– 8 or x ≥ 10, or {x|x < 8 x ≥ 10}. Holt Mc. Dougal Algebra 2 ≤ 3 is x < 8

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 a Solve

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 a Solve 6 x– 2 ≥ – 4 algebraically. Case 1 LCD is positive. Step 1 Solve for x. 6 (x – 2) ≥ – 4(x – 2) x– 2 6 ≥ – 4 x + 8 – 2 ≥ – 4 x 1 ≤x 2 1 x≥ 2 Holt Mc. Dougal Algebra 2 Multiply by the LCD. Simplify. Note that x ≠ 2. Solve for x. Rewrite with the variable on the left.

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 a Continued

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 a Continued Solve 6 x– 2 ≥ – 4 algebraically. Step 2 Consider the sign of the LCD is positive. x– 2>0 x>2 Solve for x. 1 For Case 1, the solution must satisfy x ≥ 2 and x > 2, which simplifies to x > 2. Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 a Continued

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 a Continued Solve 6 x– 2 ≥ – 4 algebraically. Case 2 LCD is negative. Step 1 Solve for x. 6 (x – 2) ≤ – 4(x – 2) x– 2 6 ≤ – 4 x + 8 – 2 ≤ – 4 x 1 ≥x 2 1 x≤ 2 Holt Mc. Dougal Algebra 2 Multiply by the LCD. Reverse the inequality. Simplify. Note that x ≠ 2. Solve for x. Rewrite with the variable on the left.

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 a Continued

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 a Continued 6 Solve ≥ – 4 algebraically. x– 2 Step 2 Consider the sign of the LCD is negative. x– 2<0 Solve for x. x<2 1 For Case 2, the solution must satisfy x ≤ 2 1 and x < 2, which simplifies to x ≤. 2 The solution set of the original inequality is the union of the solutions to both Case 1 and Case 2. 6 The solution to the inequality x– 2 1 1 2 or x ≤ 2 , or {x| x ≤ x > 2}. 2 Holt Mc. Dougal Algebra 2 ≥ – 4 is x >

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 b Solve

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 b Solve 9 x+ 3 < 6 algebraically. Case 1 LCD is positive. Step 1 Solve for x. 9 (x + 3) < 6(x + 3) x+3 9 < 6 x + 18 – 9 < 6 x 3 – <x 2 3 x>– 2 Holt Mc. Dougal Algebra 2 Multiply by the LCD. Simplify. Note that x ≠ – 3. Solve for x. Rewrite with the variable on the left.

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 b Continued

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 b Continued Solve 9 x+ 3 < 6 algebraically. Step 2 Consider the sign of the LCD is positive. x+3>0 x > – 3 Solve for x. 3 For Case 1, the solution must satisfy x >– 2 3 and x > – 3, which simplifies to x >–. 2 Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 b Continued

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 b Continued Solve 9 x+ 3 < 6 algebraically. Case 2 LCD is negative. Step 1 Solve for x. 9 (x + 3) > 6(x + 3) x+3 9 > 6 x + 18 – 9 > 6 x 3 – >x 2 3 x<– 2 Holt Mc. Dougal Algebra 2 Multiply by the LCD. Reverse the inequality. Simplify. Note that x ≠ – 3. Solve for x. Rewrite with the variable on the left.

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 b Continued

5 -5 Solving Rational Equations and Inequalities Check It Out! Example 6 b Continued 9 Solve < 6 algebraically. x+ 3 Step 2 Consider the sign of the LCD is negative. x+3<0 Solve for x. x < – 3 3 For Case 2, the solution must satisfy x <– 2 and x < – 3, which simplifies to x < – 3. The solution set of the original inequality is the union of the solutions to both Case 1 and Case 2. 9 The solution to the inequality < 6 is x < – 3 x+ 3 3 3 or x > – , or {x| x > – x < – 3}. 2 2 Holt Mc. Dougal Algebra 2

5 -5 Solving Rational Equations and Inequalities Lesson Quiz Solve each equation or inequality.

5 -5 Solving Rational Equations and Inequalities Lesson Quiz Solve each equation or inequality. 1. 2. x+2 = x x– 1 2 6 x = 7 x + 4 x+4 x = – 1 or x = 4 no solution x+2 5 x = – 5 + x = x– 3 5 4 4. ≥ 2 3<x≤ 5 x– 3 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 3. Holt Mc. Dougal Algebra 2