1 / 55

Solving Rational Equations - Lesson Presentation

Learn how to solve rational equations and inequalities, including finding extraneous solutions. Practice with examples and check your solutions thoroughly.

judya
Download Presentation

Solving Rational Equations - Lesson Presentation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Solving Rational Equations and Inequalities 8-5 Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2

  2. –(x – 1) 1 1 1 1 5x – 2 + 4x(x – 2) x – 2 x2 4x x2 x – 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. 3. x ≠ 0 4.

  3. Objective Solve rational equations and inequalities.

  4. Vocabulary rational equation extraneous solution rational inequality

  5. d r 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.

  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. x(x) – (x) = 3(x) 18 18 x x Example 1: Solving Rational Equations Solve the equation x – = 3. 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. 6 – (–3) – 18 18 18 18 (–3) 6 x x Example 1 Continued Check x – = 3 x – = 3 3 3 6 – 3 3 –3 + 6 3 3 3  3 3 

  9. Solve the equation = + 2. (3x) = (3x) + 2(3x) 10 10 4 4 3 x 3 x Check It Out! Example 1a Multiply each term by the LCD, 3x. 10x = 12 + 6x Simplify. Note that x ≠ 0. 4x = 12 Combine like terms. x = 3 Solve for x.

  10. Solve the equation + = – . (4x) + (4x) = – (4x) 7 5 6 5 6 7 4 4 x 4 4 x Check It Out! Example 1b 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. x(x) = (x) – 1(x) 6 6 x x Check It Out! Example 1c Solve the equation x = – 1. 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. = (x – 2) = (x – 2) (x – 2) = (x – 2) 3x + 4 3x + 4 3x + 4 5x 5x 5x x – 2 x – 2 x – 2 x – 2 x – 2 x – 2 Example 2A: Extraneous Solutions Solve each equation. Multiply each term by the LCD, 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. = 3(2)+ 4 3x + 4 5(2) 5x 2 – 2 x – 2 2 – 2 x – 2 10 10 0 0 Example 2A Continued Check Substitute 2 for x in the original equation. Division by 0 is undefined.

  15. x x x 2 2 2 2(x – 8) + 2(x – 8) = 2(x – 8) 2(x – 8) + 2(x – 8) = 2(x – 8) 2x – 5 2x – 5 2x – 5 11 11 11 x – 8 x – 8 x – 8 x – 8 x – 8 x – 8 Example 2B: Extraneous Solutions Solve each equation. + = Multiply each term by the LCD, 2(x – 8). Divide out common factors. 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. x x + = + – = 0. 2 2 2x – 5 2x – 5 11 11 x – 8 x – 8 x – 8 x – 8 Example 2B Continued Check Write as Graph the left side of the equation as Y1. Identify the values of x for which Y1 = 0. The graph intersects the x-axis only when x = –4. Therefore, x = –4 is the only solution.

  18. (x – 4)(x + 4) = (x – 4 )(x + 4) (x – 4)(x + 4) = (x – 4 )(x + 4) 16 16 16 2 2 2 (x – 4)(x + 4) (x – 4)(x + 4) x2 – 16 x – 4 x – 4 x – 4 Check It Out! Example 2a Solve the equation . = Multiply each term by the LCD, (x – 4)(x +4). Divide out common factors. 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.

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

  20. 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.

  21. 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.

  22. 1 Understand the Problem Example 3 Continued 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.

  23. Make a Plan 3950 3950 485 + w 3950 3950 485 – w 16.5 = + 485 – w 485 + w 2 Example 3 Continued 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. total time = time east + time west

  24. 3 Solve 16.5(485 + w)(485 – w) = (485 + w)(485 – w) + (485 + w)(485 – w) 3950 3950 485 + w 485 – w The LCD is (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 – 16.5w2 = 1,915,750 – 3950w + 1,915,750 + 3950w 3,881,212.5 – 16.5w2 = 3,831,500 Combine like terms. Solve for w. –16.5w2 = –49,712.5 w ≈ ± 55 The speed of the wind cannot be negative. Therefore, the average speed of the wind is 55 mi/h.

  25. Look Back 4 Example 3 Continued 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.

  26. 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.

  27. 1 Understand the Problem Check It Out! Example 3 Continued 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.

  28. 2 2 2 – c 2 + c Make a Plan total time = time up- stream + time down- stream 2 2 5 = + 2 – c 2 2 + c Check It Out! Example 3 Continued 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.

  29. 3 2 Solve = (2 + c)(2 – c) + (2 + c)(2 – c) 5(2 + c)(2 – c) 2 – c 2 2 + c The LCD is (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 – 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.

  30. Look Back 4 Check It Out! Example 3 Continued 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.

  31. 1 1 8 h Renzo’s rate: of the puzzle per hour, where h is thenumber of hours needed to finish the puzzle by himself. Natalie’s rate: of the puzzle per hour 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?

  32. Natalie’s rate  hours worked Renzo’s rate  hours worked 1 complete puzzle + = 1 1 1 1 8 h 8 h (4.5)(8h) + (4.5) (4.5) (4.5)(8h) = 1(8h) Example 4 Continued + 1 = 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.

  33. 1 1 20 m Julien’s rate: of the garden per minute Remy’s rate: of the garden per minute, where m is the number of minutes needed to mulch the garden byhimself. 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?

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

  35. 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.

  36. Solve ≤ 3 by using a graph and a table. Use a graph. On a graphing calculator, Y1 = and Y2 = 3. x x x–6 x–6 Example 5: Using Graphs and Tables to Solve Rational Equations and Inequalities (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

  37. 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.

  38. Solve ≥ 4 by using a graph and a table. Use a graph. On a graphing calculator, Y1 = and Y2 = 4. x x x–3 x–3 Check It Out! Example 5a (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

  39. 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.

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

  41. 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.

  42. 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.

  43. Solve ≤ 3 algebraically. 6 (x – 8) ≤ 3(x – 8) x – 8 6 x–8 Example 6: Solving Rational Inequalities Algebraically Case 1 LCD is positive. Step 1 Solve for x. 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

  44. Solve ≤ 3 algebraically. 6 x–8 Example 6 Continued 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.

  45. Solve ≤ 3 algebraically. 6 (x – 8) ≥ 3(x – 8) x – 8 6 x–8 Example 6: Solving Rational Inequalities Algebraically Case 2 LCD is negative. Step 1 Solve for x. 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

  46. Solve ≤ 3 algebraically. 6 6 x–8 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}. Example 6 Continued 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.

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

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

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

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

More Related