Presentation is loading. Please wait.

Presentation is loading. Please wait.

Holt Algebra 1 12-7 Solving Rational Equations Warm Up 1. Find the LCM of x, 2x 2, and 6. 2. Find the LCM of p 2 – 4p and p 2 – 16. Multiply. Simplify.

Similar presentations


Presentation on theme: "Holt Algebra 1 12-7 Solving Rational Equations Warm Up 1. Find the LCM of x, 2x 2, and 6. 2. Find the LCM of p 2 – 4p and p 2 – 16. Multiply. Simplify."— Presentation transcript:

1 Holt Algebra 1 12-7 Solving Rational Equations Warm Up 1. Find the LCM of x, 2x 2, and 6. 2. Find the LCM of p 2 – 4p and p 2 – 16. Multiply. Simplify your answer. 3. 4. 5. 11-8

2 Holt Algebra 1 12-7 Solving Rational Equations Solve rational equations. Identify extraneous solutions. Objectives 11-8

3 Holt Algebra 1 12-7 Solving Rational Equations rational equation Vocabulary 11-8

4 Holt Algebra 1 12-7 Solving Rational Equations 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. 11-8

5 Holt Algebra 1 12-7 Solving Rational Equations Example 1: Solving Rational Equations by Using Cross Products Use cross products. 5x = (x – 2)(3) 5x = 3x – 6 2x = – 6 Solve. Check your answer. x = – 3 Check Distribute 3 on the right side. Subtract 3x from both sides. – 1 – 1 11-8

6 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 1a Solve. Check your answer. Use cross products. Distribute 1 on the right side. Subtract n from both sides. 3n = (n + 4)(1) 3n = n + 4 2n = 4 n = 2 Check Divide both sides by 2. 11-8

7 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 1b Solve. Check your answer. 4h = (h + 1)(2) 4h = 2h + 2 2h = 2 h = 1 Use cross products. Distribute 2 on the right side. Subtract 2h from both sides. Check Divide both sides by 2. 11-8

8 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 1c Solve. Check your answer. 21x = (x – 7)(3) 21x = 3x – 21 18x = – 21 x = Use cross products. Distribute 3 on the right side. Subtract 3x from both sides. Check Divide both sides by 18. 11-8

9 Holt Algebra 1 12-7 Solving Rational Equations 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. 11-8

10 Holt Algebra 1 12-7 Solving Rational Equations 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. 11-8

11 Holt Algebra 1 12-7 Solving Rational Equations Example 2A Continued Step 3 Simplify and solve. (2x)(2) +6(x +1) = 5(x +1) 4x + 6x + 6 = 5x + 5 10x + 6 = 5x + 5 5x = – 1 Divide out common factors. Simplify. Distribute and multiply. Combine like terms. Subtract 5x and 6 from both sides. Divide both sides by 5. 11-8

12 Holt Algebra 1 12-7 Solving Rational Equations Example 2A Continued Check Verify that your solution is not extraneous. 11-8

13 Holt Algebra 1 12-7 Solving Rational Equations Example 2B: Solving Rational Equations by Using the LCD Solve each equation. Check your answer. Step 1 Find the LCD (x 2 ) Include every factor of the denominator. Step 2 Multiply both sides by the LCD Distribute on the left side. 11-8

14 Holt Algebra 1 12-7 Solving Rational Equations Example 2B Continued Step 3 Simplify and solve. Divide out common factors. 4x – 3 = x 2 – x 2 + 4x – 3 = 0 x 2 – 4x + 3 = 0 (x – 3)(x – 1) = 0 x = 3, 1 Simplify. Subtract x 2 from both sides. Factor. Solve. Multiply by – 1. 11-8

15 Holt Algebra 1 12-7 Solving Rational Equations Example 2B Continued Check Verify that your solution is not extraneous. 11-8

16 Holt Algebra 1 12-7 Solving Rational Equations 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. 11-8

17 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 2a Continued Step 3 Simplify and solve. Divide out common factors. 3a = 4(a + 1) 3a = 4a + 4 – 4 = a Simplify. Distribute the 4. Subtract the 4 and 3a from both sides. 11-8

18 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 2a Continued Check Verify that your solution is not extraneous. 11-8

19 Holt Algebra 1 12-7 Solving Rational Equations 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. 11-8

20 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 2b Continued Solve each equation. Check your answer. 12j – 10(2j + 4) = 4j + 8 12j – 20j – 40 = 4j + 8 – 12j = 48 j = – 4 Simplify. Distribute 10. Combine like terms. Divide out common terms. 11-8

21 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 2b Continued Check Verify that your solution is not extraneous. 11-8

22 Holt Algebra 1 12-7 Solving Rational Equations 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. 11-8

23 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 2c Continued Solve each equation. Check your answer. Divide out common terms. 8t = (t + 3) + t(t + 3) 8t = t + 3 + t 2 + 3t 0 = t 2 – 4t + 3 0 = (t – 3)(t – 1) t = 3, 1 Simplify. Distribute t. Combine like terms. Factor. 11-8

24 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 2c Continued Check Verify that your solution is not extraneous. 11-8

25 Holt Algebra 1 12-7 Solving Rational Equations When you multiply each side of an equation by the LCD, you may get an extraneous solution. Recall from previous chapters that an extraneous solution is a solution to a resulting equation that is not a solution to the original equation. 11-8

26 Holt Algebra 1 12-7 Solving Rational Equations Extraneous solutions may be introduced by squaring both sides of an equation or by multiplying both sides of an equation by a variable expression. Helpful Hint 11-8

27 Holt Algebra 1 12-7 Solving Rational Equations Example 4: Extraneous Solutions Solve. Identify any extraneous solutions. Step 1 Solve. 2(x 2 – 1) = (x + 1)(x – 6) 2x 2 – 2 = x 2 – 5x – 6 x 2 + 5x + 4 = 0 (x + 1)(x + 4) = 0 x = – 1 or x = – 4 Use cross products. Distribute 2 on the left side. Multiply the right side. Subtract x 2 from both sides. Add 5x and 6 to both sides. Factor the quadratic expression. Use the Zero Product Property. Solve. 11-8

28 Holt Algebra 1 12-7 Solving Rational Equations Example 4 Continued Solve. Identify any extraneous solutions. Step 2 Find extraneous solutions.  Because and are undefined –1 is not a solution. The only solution is – 4, so – 1 is an extraneous solution. 11-8

29 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 4a Solve. Identify any extraneous solutions. Step 1 Solve. (x – 2)(x – 7) = 3(x – 7) Use cross products. Distribute 3 on the right side. Multiply the left side. 2x 2 – 9x + 14 = 3x – 21 X 2 – 12x + 35 = 0 Subtract 3x from both sides. Add 21 to both sides. (x – 7)(x – 5) = 0 x = 7 or x = 5 Factor the quadratic expression. Use the Zero Product Property. Solve. 11-8

30 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 4a Continued Step 2 Find extraneous solutions. The only solution is 5, so 7 is an extraneous solution. Because and are undefined 7 is not a solution.  11-8

31 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 4b Solve. Identify any extraneous solutions. Step 1 Solve. (x + 1)(x – 3) = 4(x – 2) Use cross products. Distribute 4 on the right side. Multiply the left side. x 2 – 2x – 3 = 4x – 8 X 2 – 6x + 5 = 0 Subtract 4x from both sides. Add 8 to both sides. (x – 1)(x – 5) = 0 x = 1 or x = 5 Factor the quadratic expression. Use the Zero Product Property. Solve. 11-8

32 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 4b Continued Step 2 Find extraneous solutions. The solutions are 1 and 5, there are no extraneous solutions. 1 and 5 are solutions. 11-8

33 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 4c Solve. Identify any extraneous solutions. Step 1 Solve. 6(x 2 + 2x) = 9(x 2 ) Use cross products. Distribute 6 on the left side. Multiply the right side. 3x 2 – 12x = 0 3x(x – 4) = 0 3x = 0, or x – 4 = 0 x = 0 or x = 4 Factor the quadratic expression. Use the Zero Product Property. Solve. Subtract 9x 2 from both sides. Multiply through with – 1. 11-8

34 Holt Algebra 1 12-7 Solving Rational Equations Check It Out! Example 4c Continued Step 2 Find extraneous solutions. The only solution is 4, so 0 is an extraneous solutions. Because and are undefined 0 is not a solution.  11-8

35 Holt Algebra 1 12-7 Solving Rational Equations Lesson Quiz: Part I Solve each equation. Check your answer. 3. 2. 1.24 – 4, 3 11-8

36 Holt Algebra 1 12-7 Solving Rational Equations Lesson Quiz: Part II 4. Solve. Identify any extraneous solutions. – 5; 3 is extraneous. 11-8


Download ppt "Holt Algebra 1 12-7 Solving Rational Equations Warm Up 1. Find the LCM of x, 2x 2, and 6. 2. Find the LCM of p 2 – 4p and p 2 – 16. Multiply. Simplify."

Similar presentations


Ads by Google