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Solve rational equations and inequalities.

Objective. Solve rational equations and inequalities. Vocabulary. rational equation extraneous solution rational inequality. A rational equation is an equation that contains one or more rational expressions.

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Solve rational equations and inequalities.

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  1. Objective Solve rational equations and inequalities.

  2. Vocabulary rational equation extraneous solution rational inequality

  3. A rational equation is an equation that contains one or more rational expressions..

  4. 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.

  5. 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.

  6. 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.

  7. 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.

  8. = (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.

  9. 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

  10. 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.

  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. 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?

  15. 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.

  16. 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?

  17. 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.

  18. 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.

  19. 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

  20. 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.

  21. 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

  22. 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.

  23. 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.

  24. 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

  25. 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.

  26. 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

  27. 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.

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