1. Rational Expressions Topic 4: Solving Rational Expressions

2. I can determine the non-permissible values for the variable in a rational equation. I can determine, algebraically, the solution to a rational equation, and explain the strategy used to solve the equation. I can explain why a value obtained in solving a rational equation may not be a solution of the equation. I can solve a contextual problem that involves a rational equation.

3. Before You Start… Quadratic equations can be solved in three ways: By factoring (sum and product, decomposition, or box method) By graphing By using the quadratic formula

4. Example Solve the following polynomial equations, if possible. Solving Quadratic Equations a) Is there a common factor? No. 3 terms. Factor using decomposition. -10 Factors of -10 1 ×-10-1 × 10 2 ×-5-2 × 5 We are looking for factors that have a product of -10 and a sum of 3. -2 and 5 work. Therefore, we break the b-value up into -2x and 5x. Notice that the value of the equation has not changed. Now we find that there is a GCF of x to be pulled out of the first two terms and a GCF of 5 to be pulled out of the last two terms. Pull out the common binomial factor (x-2). Solve. Remember that if either of the factors =0, the equation is =0.

5. Example Solving Quadratic Equations b) 12Factors of 12 1 ×12-1 × -12 2 × 6-2 × -6 3 × 4-3 × -4 We are looking for factors that have a product of 12 and a sum of -7. -3 and -4 work. Therefore, we break the b-value up into -3x and -4x. Notice that the value of the equation has not changed. Now we find that there is a GCF of x to be pulled out of the first two terms and a GCF of -4 to be pulled out of the last two terms. Pull out the common binomial factor (x-3). Is there a common factor? Yes. 3 terms. Factor using decomposition. Solve. Remember that if either of the factors =0, the equation is =0.

6. Example Solving Quadratic Equations c) Is there a common factor? No. 3 terms. Factor using decomposition. -12Factors of -12 -1 ×121 × -12 -2 × 62 × -6 -3 × 43 × -4 We are looking for factors that have a product of -12 and a sum of -1. 3 and -4 work. Therefore, we break the b-value up into -4x and 3x. Notice that the value of the equation has not changed. Now we find that there is a GCF of 2x to be pulled out of the first two terms and a GCF of 3 to be pulled out of the last two terms. Pull out the common binomial factor (x-2). Solve. Remember that if either of the factors =0, the equation is =0.

7. Example Solving Quadratic Equations d) Is there a common factor? No. 3 terms. Factor using decomposition. 30 Factors of 30 1 ×30-1 × -30 2 × 15-2 × -15 3 × 10-3 × -10 5 × 6-5 × -6 We are looking for factors that have a product of 30 and a sum of 17. 2 and 15 work. Therefore, we break the b-value up into 2x and 15x. Notice that the value of the equation has not changed. Now we find that there is a GCF of x to be pulled out of the first two terms and a GCF of 5 to be pulled out of the last two terms. Pull out the common binomial factor (3x+2). Solve. Remember that if either of the factors =0, the equation is =0.

8. Example Solving Quadratic Equations e) -4 Factors of -4 1 ×-4-1 × 4 2 × -2-2 × 2 We are looking for factors that have a product of -4 and a sum of 2. None of these work so this equation is not factorable. We must use the Quadratic Formula!! Identify your variables. In this equation: a = 1, b = 2, c = -4 Substitute your a, b, and c values into the quadratic formula. Make sure to put brackets around each of your substitutions. When you enter this in your calculator, a bracket will automatically be inserted under the square root. Make sure to close it.

9. Example Solving Quadratic Equations e) (Continued) You have to enter this in your calculator twice. Once with the positive and once with the negative. Enter the numerator. Press enter. Then divide by the denominator (in brackets) like below. x = 1.24, -3.24

10. Information A rational equation is an equation that involves one or more rational expressions. A rational equation can be solved algebraically by: identifying the non-permissible values factoring all denominators to determine the LCD multiplying each term by the lowest common denominator (LCD) solving the resulting equation for the variable

11. Information On some occasions, you will not be able to solve a resulting quadratic equation by factoring. In cases like this, you can use the quadratic formula.

12. Information Recall from Topic 2, that: a non-permissible value is a value for the variable that makes the denominator of a rational expression equal to 0. a restriction is the non-permissible value stated at the end of a question along with the answer. an inadmissible value is a solution that is not valid in the context of the word problem. An extraneous root is a solution to an equation that is not permissible in the original equation.

13. Example 1 Solving a rational equation x ≠ 2, x ≠ 1

14. Example 1 c) Verify your solution using substitution. Solving a rational equation In order to verify a solution, take your answer and plug it into the original equation. If the two sides are equal after simplifying, your solution it correct. The solution x = -3 is correct!

15. Example 2 For each of the following rational equations, state the non-permissible values, solve the equation, and verify your solution. a) Solving rational equations NPV’s: x ≠ 0 LCD: 3x Multiply all terms by the LCD.

16. Example 2 b) Solving rational equations NPV’s: x ≠ 0 x ≠ -6 LCD: x(x+6) Multiply all terms by the LCD.

17. Example 2 c) Solving rational equations NPV’s: x ≠ -1, x ≠ 0 LCD: 6x(x+1) Now you need to factor the trinomial. Multiply all terms by the LCD.

18. Example 3 Solve the following equations. a) Solving rational equations with extraneous roots NPV’s: x ≠ 3, x ≠ 0 LCD: x(x-3) This is not a permissible solution, since 3 is one of the NPV’s. No Solution!

19. Example 3 b) Solving rational equations with extraneous roots NPV’s: a ≠ 2, a ≠ -2 LCD: (a+2)(a - 2) No Solution! (2 is a non permissible value!) The second denominator can be factored first to make things easier!

20. Example 4 Solving a problem that involves a rational equation The salt concentration (c) is 3.75. Put this into the equation in place of c. NPV’s: t ≠ -25 LCD: 25+t

21. Example 4 b) Verify the solution. Solving a problem that involves a rational equation

22. Example 5 Dog sled races are a tradition in many northern Canadian communities. In one race, the distance was 210 km. The first place dog team’s average speed was 5 km/h faster than the average speed of the second place dog team. The second place dog team crossed the finish line 1 hour after the first place dog team. Determine the average speed of the first place and second place dog teams. Speed-Distance-Time Problem

23. Example 5 Step 1: Define the variables. Let x represent the average speed of the first place dog team. The average speed of the second place dog team is represented as ___________ Speed-Distance-Time Problem x - 5

24. Example 5 Step 2: Create a rational equation to find x. Organize the information in a distance-speed-time table. a)Use the information in the problem to fill in first two columns of the table below. b)Fill in the third column using only the information in first two columns and the relationship between distance, speed and time. Speed-Distance-Time Problem 210 x - 5

25. Example 5 c) Use the third column and the information in the problem to create a rational equation to find x. Use the relationship: (time difference) = (time of slower) – (time of faster) Speed-Distance-Time Problem The second team finished one hour earlier than the second.

26. Example 5 Step 3: Solve the equation Speed-Distance-Time Problem NPV’s: x ≠ 0, 5 LCD: x(x-5) Multiply all terms by the lowest common denominator.

27. Example 5 Step 4: Answer the question. What was the average speed of the first place dog team and second place dog team? Speed-Distance-Time Problem The first place dog team went (on average) 35km/h. The second place dog team went (on average) 5 km/h slower: 30 km/h. Remember:

28. Example 6 When they work together, Stuart and Lucy can deliver flyers to all the homes in their neighbourhood in 20 min. When Lucy works alone, she can deliver the flyers in 9 minutes less time than Stuart can when he works alone. When Stuart works alone, how long does he take to deliver the flyers? a) Define a variable to represent what it is you are looking for. b) Using the variable from part a), define an expression that represents the time that Lucy takes to deliver the flyers when she works alone. Solving a rational expressions with a inadmissible solution Let x represent how long it takes for Stuart to deliver the flyers on his own. Lucy would take x-9 to deliver the flyers on her own.

29. c) Organize the information in a table. Example 6 If it takes 34 minutes to complete something, you do 1/34 of the work in one minute. Following this logic…

30. d) Write an equation that relates the three rates from part b). State the restrictions on the variable. Example 6

31. e) When Stuart works alone, how long does he take to deliver the flyers? Example 6 LCD: 20x(x-9) 4 does not make sense in the context of the problem.

32. Example 6 f) Verify that your solutions is correct. Determining the probability Put 45 in for x to make sure that it works in the equation.

33. Need to Know A rational equation is an equation that involves one or more rational expressions. An extraneous root is a solution to an equation that is not permissible in the original equation. An inadmissible solution is a solution that is permissible in the original equation, but is not valid in the context of the word problem. A rational equation can be solved algebraically by:  identifying the non-permissible values  multiplying each term by the lowest common denominator (LCD) to eliminate the fractions from the equation  solving the resulting equation for the variable, identifying any extraneous roots.

34. Need to Know

35. Each solution to a rational equation can be verified by substituting it into the original equation. If the left side equals the right side, then it is a valid solution to a rational equation in a non-contextual problem. If a root of a rational equation is a non-permissible value of the rational expressions in the original equation, it is an extraneous root. When solving a contextual problem, it is important to check for solutions that are valid solutions to the equation but are inadmissible in the context of the problem. You’re ready! Try the homework from this section.