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8-5

Solving Rational Equations and Inequalities. 8-5. Warm Up. Lesson Presentation. Lesson Quiz. Holt Algebra 2. –( x – 1). 1. 1. 1. 1. 5 x – 2. +. 4 x ( x – 2). x – 2. x 2. 4 x. x 2. x. –. Warm Up Find the least common multiple for each pair. 1. 2 x 2 and 4 x 2 – 2 x.

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8-5

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  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. 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.

  8. 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.

  9. 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.

  10. 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.

  11. (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.

  12. 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

  13. 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.

  14. 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.

  15. 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.

  16. 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.

  17. 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.

  18. 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.

  19. 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?

  20. 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.

  21. 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.

  22. 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

  23. 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.

  24. 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

  25. 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.

  26. 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.

  27. 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.

  28. 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.

  29. 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.

  30. 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.

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

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

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

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

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