1 / 47

12-7

12-7. Solving Rational Equations. Warm Up. Lesson Presentation. Lesson Quiz. Holt Algebra 1. Warm Up 1. Find the LCM of x , 2 x 2 , and 6. 2. Find the LCM of p 2 – 4 p and p 2 – 16. Multiply. Simplify your answer. 3. 4. . 5. Objectives.

unity
Download Presentation

12-7

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. 12-7 Solving Rational Equations Warm Up Lesson Presentation Lesson Quiz Holt Algebra 1

  2. Warm Up • 1. Find the LCM of x, 2x2, and 6. • 2. Find the LCM of p2 – 4p and p2 – 16. • Multiply. Simplify your answer. • 3. 4. 5.

  3. Objectives Solve rational equations. Identify extraneous solutions.

  4. Vocabulary rational equation

  5. A rational equation is an equation that contains one or more rational expressions. If a rational equation is a proportion, it can be solved using the Cross Product Property.

  6. Solve . Check your answer.  –1 –1 Example 1: Solving Rational Equations by Using Cross Products Check Use cross products. 5x = (x– 2)(3) Distribute 3 on the right side. 5x = 3x– 6 2x = –6 Subtract 3x from both sides. x = –3

  7. Solve . Check your answer. Check  Check It Out! Example 1a Use cross products. 3n = (n + 4)(1) Distribute 1 on the right side. 3n = n + 4 Subtract n from both sides. 2n = 4 n = 2 Divide both sides by 2.

  8. Check  Check It Out! Example 1b Solve . Check your answer. Use cross products. 4h = (h + 1)(2) Distribute 2 on the right side. 4h = 2h + 2 2h = 2 Subtract 2h from both sides. h = 1 Divide both sides by 2.

  9. Check Check It Out! Example 1c Solve . Check your answer. Use cross products. 21x = (x– 7)(3) Distribute 3 on the right side. 21x = 3x–21 Subtract 3x from both sides. 18x = –21  x = Divide both sides by 18.

  10. Some rational equations contain sums or differences of rational expressions. To solve these, you must find the LCD of all the rational expressions in the equation.

  11. Example 2A: Solving Rational Equations by Using the LCD Solve each equation. Check your answer. Step 1 Find the LCD 2x(x + 1) Include every factor of the denominator. Step 2 Multiply both sides by the LCD Distribute on the left side.

  12. Example 2A Continued Step 3 Simplify and solve. Divide out common factors. (2x)(2) +6(x +1) = 5(x +1) Simplify. 4x + 6x + 6 = 5x + 5 Distribute and multiply. 10x + 6 = 5x + 5 Combine like terms. Subtract 5x and 6 from both sides. 5x = –1 Divide both sides by 5.

  13. Example 2A Continued CheckVerify that your solution is not extraneous. 

  14. Example 2B: Solving Rational Equations by Using the LCD Solve each equation. Check your answer. Step 1 Find the LCD (x2) Include every factor of the denominator. Step 2 Multiply both sides by the LCD Distribute on the left side.

  15. Example 2B Continued Step 3 Simplify and solve. Divide out common factors. 4x– 3 = x2 Simplify. Subtract x2 from both sides. –x2 + 4x– 3 = 0 x2– 4x + 3 = 0 Multiply by – 1. (x– 3)(x– 1) = 0 Factor. x = 3, 1 Solve.

  16.  Example 2B Continued CheckVerify that your solution is not extraneous.

  17. Check It Out! Example 2a Solve each equation. Check your answer. Step 1 Find the LCD a(a +1) Include every factor of the denominator. Step 2 Multiply both sides by the LCD Distribute on the left side.

  18. Check It Out! Example 2a Continued Step 3 Simplify and solve. Divide out common factors. 3a = 4(a + 1) Simplify. 3a = 4a + 4 Distribute the 4. –4 = a Subtract the 4 and 3a from both sides.

  19. Check It Out! Example 2a Continued CheckVerify that your solution is not extraneous.

  20. Check It Out! Example 2b Solve each equation. Check your answer. Step 1 Find the LCD 2j(j +2) Include every factor of the denominator. Step 2 Multiply both sides by the LCD Distribute on the left side.

  21. Check It Out! Example 2b Continued Solve each equation. Check your answer. Divide out common terms. 12j –10(2j + 4) = 4j + 8 Simplify. 12j– 20j– 40 = 4j + 8 Distribute 10. –12j = 48 Combine like terms. j = –4

  22. Check It Out! Example 2b Continued CheckVerify that your solution is not extraneous.

  23. Check It Out! Example 2c Solve each equation. Check your answer. Step 1 Find the LCD t(t +3) Include every factor of the denominator. Step 2 Multiply both sides by the LCD Distribute on the right side.

  24. Check It Out! Example 2c Continued Solve each equation. Check your answer. Divide out common terms. 8t = (t + 3) + t(t + 3) Simplify. 8t = t + 3 + t2 + 3t Distribute t. 0 = t2– 4t + 3 Combine like terms. 0 = (t– 3)(t– 1) Factor. t = 3, 1

  25.  Check It Out! Example 2c Continued CheckVerify that your solution is not extraneous.

  26. Example 3: Problem-Solving Application Copy machine A can make 200 copies in 60 minutes. Copy machine B can make 200 copies in 10 minutes. How long will it take both machines working together to make 200 copies?

  27. 1 Understand the Problem • Machine A can print the copies in 60 minutes, which is of the job in 1 minute. • Machine B can print the copies in 10 minutes, which is of the job in 1 minute. The answer will be the number of minutes m machine A and machine B need to print the copies. List the important information:

  28. Make a Plan + = complete job (machine A’s rate) (machine B’s rate) m m 2 = + m m 1 The part of the copies that machine A can print plus the part that machine B can print equals the complete job. Machine A’s rate times the number of minutes plus machine B’s rate times the number of minutes will give the complete time to print the copies.

  29. 3 Solve Multiply both sides by the LCD, 60. Distribute 60 on the left side. 1m + 6m = 60 Combine like terms. 7m = 60 Divide both sides by 7. Machine A and Machine B working together can print the copies in a little more than 8.5 minutes.

  30. Machine A prints of the copies per minute and machine B prints of the copies per minute. So in minutes, machine A prints of the copies and machine B prints of the copies. Together, they print 4 Look Back

  31. 1 Understand the Problem Check It Out! Example 3 Cindy mows a lawn in 50 minutes. It takes Sara 40 minutes to mow the same lawn. How long will it take them to mow the lawn if they work together? The answer will be the number of minutes m Sara and Cindy need to mow the lawn.

  32. 1 Understand the Problem • Cindy can mow the lawn in 50 minutes, which is of the job in 1 minute. • Sara can mow the lawn in 40 minutes, which is of the job in 1 minute. The answer will be the number of minutes m Sara and Cindy need to mow the lawn. List the important information:

  33. Make a Plan + = lawn mowed (Cindy’s rate) (Sara’s rate) m m 2 m + m = 1 The part of the lawn Cindy can mow plus the part of the lawn Sara can mow equals the complete job. Cindy’s rate times the number of minutes plus Sara’s rate times the number of minutes will give the complete time to mow the lawn.

  34. 3 Solve Multiply both sides by the LCD, 200. Distribute 200 on the left side. 5m + 4m = 200 9m = 200 Combine like terms. Divide both sides by 9. Cindy and Sara working together can mow the lawn in a little more than 22.2 minutes.

  35. 4 Look Back Cindy mows of the lawn in 1 minute and Sara mows of the lawn in 1 minute. So, in minutes Cindy mows of the lawn and Sara mows of the lawn. Together they mow lawn.

  36. When you multiply each side of an equation by the LCD, you may get an extraneous solution. Recall from Chapter 11 that an extraneous solution is a solution to a resulting equation that is not a solution to the original equation.

  37. Helpful Hint Extraneous solutions may be introduced by squaring both sides of an equation or by multiplying both sides of an equation by a variable expression.

  38. Example 4: Extraneous Solutions Solve . Identify any extraneous solutions. Step 1 Solve. Use cross products. Distribute 2 on the left side. Multiply the right side. 2(x2– 1) = (x + 1)(x– 6) 2x2– 2 = x2– 5x– 6 Subtract x2 from both sides. Add 5x and 6 to both sides. x2 + 5x + 4 = 0 Factor the quadratic expression. (x + 1)(x + 4) = 0 Use the Zero Product Property. Solve. x = –1 or x = –4

  39. Because and are undefined –1 is not a solution.   Example 4 Continued Solve . Identify any extraneous solutions. Step 2 Find extraneous solutions. The only solution is – 4, so – 1 is an extraneous solution.

  40. Check It Out! Example 4a Solve. Identify any extraneous solutions. Step 1 Solve. Use cross products. Distribute 3 on the right side. Multiply the left side. (x– 2)(x– 7) = 3(x– 7) 2x2– 9x + 14 = 3x– 21 Subtract 3x from both sides. Add 21 to both sides. X2– 12x + 35 = 0 Factor the quadratic expression. (x– 7)(x –5) = 0 Use the Zero Product Property. Solve. x = 7 or x = 5

  41. Because and are undefined 7 is not a solution.   Check It Out! Example 4a Continued Step 2 Find extraneous solutions. The only solution is 5, so 7 is an extraneous solution.

  42. Check It Out! Example 4b Solve. Identify any extraneous solutions. Step 1 Solve. Use cross products. Distribute 4 on the right side. Multiply the left side. (x + 1)(x– 3) = 4(x– 2) x2– 2x– 3 = 4x– 8 Subtract 4x from both sides. Add 8 to both sides. X2– 6x + 5 = 0 Factor the quadratic expression. (x– 1)(x –5) = 0 Use the Zero Product Property. Solve. x = 1 or x = 5

  43.  Check It Out! Example 4b Continued Step 2 Find extraneous solutions. 1 and 5 are solutions. The solutions are 1 and 5, there are no extraneous solutions.

  44. Subtract 9x2 from both sides. Multiply through with – 1. Check It Out! Example 4c Solve. Identify any extraneous solutions. Step 1 Solve. Use cross products. Distribute 6 on the left side. Multiply the right side. 6(x2 + 2x) = 9(x2) 3x2– 12x= 0 3x(x– 4) = 0 Factor the quadratic expression. 3x = 0, or x –4 = 0 Use the Zero Product Property. Solve. x = 0 or x = 4

  45. Because and are undefined 0 is not a solution.   Check It Out! Example 4c Continued Step 2 Find extraneous solutions. The only solution is 4, so 0 is an extraneous solutions.

  46. Lesson Quiz: Part I Solve each equation. Check your answer. –4, 3 1. 24 2. 3. 4. Pipe A can fill a tank with water in 4 hours. Pipe B can fill the same tank in 5 hours. How long will it take both pipes working together to fill the tank?

  47. Lesson Quiz: Part II 5. Solve . Identify any extraneous solutions. –5; 3 is extraneous.

More Related