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12-7

12-7. Solving Rational Equations. Objectives. Notes. Practice. Holt Algebra 1. Objectives. Solve rational equations. Identify extraneous solutions.

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12-7

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  1. 12-7 Solving Rational Equations Objectives Notes Practice Holt Algebra 1

  2. Objectives Solve rational equations. Identify extraneous solutions.

  3. 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.

  4. 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

  5. Check Check It Out! Example 2 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.

  6. 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.

  7. Example 3: Solving Rational Equations by Using the LCD Solve the 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.

  8. Example 3 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.

  9. Example 3 Continued CheckVerify that your solution is not extraneous. 

  10. Example 4: Solving Rational Equations by Using the LCD Solve the 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.

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

  12.  Example 4 Continued CheckVerify that your solution is not extraneous.

  13. Example 5 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.

  14. Example 5 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

  15.  Example 5 Continued CheckVerify that your solution is not extraneous.

  16. Example 6: 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?

  17. 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:

  18. 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.

  19. 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.

  20. 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

  21. 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.

  22. 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.

  23. Example 7: 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

  24. Because and are undefined –1 is not a solution.   Example 7 Continued Solve . Identify any extraneous solutions. Step 2 Find extraneous solutions. The only solution is – 4, so – 1 is an extraneous solution.

  25. Example 8 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

  26. Because and are undefined 7 is not a solution.   Example 8 Continued Step 2 Find extraneous solutions. The only solution is 5, so 7 is an extraneous solution.

  27. EXIT TICKET: • Solve the equation.

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