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Solving Rational Equations with Cross Products and LCD

This lesson presentation provides step-by-step instructions and examples for solving rational equations using cross products and the least common denominator (LCD). It covers topics such as finding the LCD, multiplying both sides by the LCD, and simplifying the equation. Additionally, it emphasizes the importance of checking for extraneous solutions.

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Solving Rational Equations with Cross Products and LCD

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  1. Preview Warm Up California Standards Lesson Presentation

  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. California Standards Preparation for 15.0 Students apply algebraic techniques to solve rate problems, work problems, and percent mixture problems.

  4. Vocabulary rational equation extraneous solutions

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

  7. –1 –1 Additional Example 1 Continued Check Substitute –3 for x in the original equation.

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

  9. Check  Check It Out! Example 1a Continued Substitute 2 for n in the original equation.

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

  11. Check  Check It Out! Example 1b Continued Substitute 1 for h in the original equation.

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

  13. Substitute for x in the original equation. Check It Out! Example 1c Continued Check 

  14. Some rational equations contain sums or differences of rational expressions. To solve these, you must find a common denominator for all the rational expressions in the equation.

  15. Additional 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.

  16. Additional 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.

  17. Additional Example 2A Continued Check 

  18. Additional 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.

  19. Additional 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 or 1 Solve.

  20.  Additional Example 2B Continued Check

  21. 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.

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

  23. Check It Out! Example 2a Continued Check

  24. 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.

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

  26. Check It Out! Example 2b Continued Check

  27. When you multiply each side of an equation by the LCD, you may get an extraneous solution. An extraneous solution is a solution to a resulting equation that is not a solution to the original equation. Because of extraneous solutions, it is especially important to check your answers.

  28. Additional Example 3: Extraneous Solutions Solve . Check your answer. 2(x2– 1) = (x + 1)(x– 6) Use cross products. Distribute 2 on the left side. Multiply the right side. 2x2– 2 = x2– 5x– 6 x2 + 5x + 4 = 0 Subtract x2 from both sides. Add 5x and 6 to both sides. (x + 1)(x + 4) = 0 Factor the quadratic expression. Use the Zero Product Property. x = –1 or x = –4 Solve.

  29. Because and are undefined, –1 is not a solution.   Additional Example 3 Continued Solve . Check your answer. Check –1 is an extraneous solution. The only solution is –4.

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

  31. Because and are undefined, 7 is not a solution.   Check It Out! Example 3a Continued Check 7 is an extraneous solution. The only solution is 5.

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

  33.  Check It Out! Example 3b Continued Check 1 and 5 are solutions. There are no extraneous solutions. The solutions are 1 and 5.

  34. Subtract 9x2 from both sides. Multiply through with – 1. Check It Out! Example 3c Solve. Identify any extraneous solutions. 6(x2 + 2x) = 9(x2) Use cross products. 6x2 + 12x) = 9x2 Distribute 6 on the left side. Multiply the right side. 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

  35. Because and are undefined, 0 is not a solution.   Check It Out! Example 3c Continued Check 0 is an extraneous solution. The only solution is 4.

  36. Lesson Quiz Solve each equation. Check your answer. –4, 3 1. 24 2. 3. 4. Solve . Check your answer. –5

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