1. Solving Rational Equations
Digital Lesson
Solving Rational Equations

2. Objective and Vocabulary
The objective is to be able to solve rational equations.
Lowest Common Denominator (LCD) – The smallest denominator which has all of the original denominators as factors.

3. Solving a Rational Equation
A rational equation, also called a fractional equation, is an equation containing one or more rational expressions. The following is an example of a rational equation:
This denominator would equal zero if x = 0.
This denominator would equal zero if x = 0.
When there is a variable in the denominator, we must avoid any values of the variable that make a denominator zero.
When adding or subtracting rational expressions, we find the LCD and convert fractions to equivalent fractions that have the common denominator.

4. Solving Rational Equations
Solving a Rational Equation
Solving Rational Equations
1) Find the LCM of the denominators.
*Avoid any values of the variable that make a denominator zero.
2) Clear denominators by multiplying both sides of the equation by the LCM.
3) Solve the equation.
4) Check your solution.

5. Solving a Rational Equation
Example
The denominators are 5x, 5, and x. The least common denominator is 5x. We begin by multiplying both sides of the equation by 5x. We will also write the restriction that x cannot equal zero to the right of the equation.
This is the given equation.
Multiply both sides by 5x, the LCD.
Use the distributive property.

6. Solving a Rational Equation
CONTINUED
Divide out common factors in the multiplications.
Multiply.
Subtract.
Add.
Divide.
The proposed solution, 8, is not part of the restriction It should check in the original equation.

7. Solving a Rational Equation
CONTINUED
Check 8: ? ? ? ? ?
true ?
This true statement verifies that the solution is 8 and the solution set is {8}. ?

8. Solving a Rational Equation
EXAMPLE
Solve:
SOLUTION
1) List restrictions on the variable.
This denominator would equal zero if x = 4.
This denominator would equal zero if x = 3.5.
The restrictions are

9. Solving a Rational Equation
CONTINUED
2) Multiply both sides by the LCD. The denominators are x – 4 and 2x – 7. Thus, the LCD is (x – 4)(2x – 7).
This is the given equation.
Multiply both sides by the LCD.
Simplify.

10. Solving a Rational Equation
CONTINUED
3) Solve the resulting equation.
This is the equation cleared of fractions.
Use FOIL on each side.
Subtract from both sides.
Subtract 19x from both sides.
4) Check the proposed solution in the original equation. Notice, there is no proposed solution. And of course, – 7 = – 20 is not a true statement. Therefore, there is no solution to the original rational equation. We say the solution set is , the empty set.

11. Solving a Rational Equation
EXAMPLE
Solve:
SOLUTION
1) List restrictions on the variable. By factoring denominators, it makes it easier to see values that make the denominators zero.
This denominator is zero if x = -4 or x = 2.
This denominator would equal zero if x = -4.
This denominator would equal zero if x = 2.
The restrictions are

12. Solving a Rational Equation
CONTINUED
2) Multiply both sides by the LCD. The factors of the LCD are x + 4 and x – 2. Thus, the LCD is (x + 4)(x – 2).
This is the given equation.
Multiply both sides by the LCD.
Use the distributive property.

13. Solving a Rational Equation
CONTINUED
Simplify.
3) Solve the resulting equation.
This is the equation with cleared fractions.
Use the distributive property.
Combine like terms.
Subtract x from both sides.
Add 5 to both sides.
Divide both sides by 3.

14. Solving a Rational Equation
CONTINUED
4) Check the proposed solutions in the original equation. The proposed solution, 3, is not part of the restriction that Substitute 3 for x, in the given (original) equation. The resulting true statement verifies that 3 is a solution and that {3} is the solution set.

15. Objective #1: Example

16. Objective #1: Example

17. Method 2
The previous example had two or more fractions on one side of the equation. When there is a single fraction on each side of the equation, the equation can be solved as a proportion by cross-multiplying.

18. Objective #1: Example

19. Objective #1: Example

20. Objective #1: Example

21. Objective #1: Example

22. Objective #1: Example

23. Objective #1: Example

24. Objective #1: Example

25. Objective #1: Example

26. Objective #1: Example
CONTINUED

27. Objective #1: Example
CONTINUED

28. Examples: 1. Solve: . LCM = x – 3. 1 = x + 1 x = 0 (0) LCM = x(x – 1).
Find the LCM.
1 = x + 1
Multiply by LCM = (x – 3).
x = 0
Solve for x.
(0)
Check.
Substitute 0.
Simplify.
True.
2. Solve:
LCM = x(x – 1).
Find the LCM.
Multiply by LCM.
x – 1 = 2x
Simplify.
x = –1
Solve.
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Examples: Solve

29. In this case, the value is not a solution of the rational equation.
After clearing denominators, a solution of the polynomial equation may make a denominator of the rational equation zero.
In this case, the value is not a solution of the rational equation.
It is critical to check all solutions.
Example: Solve:
Since x2 – 1 = (x – 1)(x + 1),
LCM = (x – 1)(x + 1).
3x + 1 = x – 1
2x = – 2  x = – 1
Check.
Since – 1 makes both denominators zero, the rational equation has no solutions.
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Example: Solve

30. Example: Solve: . x2 – 8x + 15 = (x – 3)(x – 5) x(x – 5) = – 6
Factor.
The LCM is (x – 3)(x – 5).
Original Equation.
x(x – 5) = – 6
Polynomial Equation.
x2 – 5x + 6 = 0
Simplify.
(x – 2)(x – 3) = 0
Factor.
Check. x = 2 is a solution.
x = 2 or x = 3
Check. x = 3 is not a solution since both sides would be undefined.
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Example: Solve

31. Example: Using Work Formula
To solve problems involving work, use the formula,
part of work completed = rate of work time worked.
Example: If it takes 5 hours to paint a room, what part of the work is completed after 3 hours?
If one room can be painted in 5 hours then the rate of work is (rooms/hour). The time worked is 3 hours.
Therefore, part of work completed = rate of work time worked
part of work completed
Three-fifths of the work is completed after three hours.
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Example: Using Work Formula

32. Let t be the time it takes them to paint the room together.
Example: If a painter can paint a room in 4 hours and her assistant can paint the room in 6 hours, how many hours will it take them to paint the room working together?
Let t be the time it takes them to paint the room together.
painter
assistant
rate of work
time worked
part of work completed t t
LCM = 12.
Multiply by 12.
Simplify.
Working together they will paint the room in 2.4 hours.
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Example: Word Problem

33. Examples: Using Motion Formulas
To solve problems involving motion, use the formulas,
distance = rate  time and time =
Examples: 1. If a car travels at 60 miles per hour for 3 hours, what distance has it traveled?
Since rate = 60 (mi/h) and time = 3 h, then
distance = rate time = = 180.
The car travels 180 miles.
2. How long does it take an airplane to travel miles flying at a speed of 250 miles per hour?
Since distance = 1200 (mi) and rate = 250 (mi/h),
time = = = 4.8.
It takes 4.8 hours for the plane make its trip.
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Examples: Using Motion Formulas

34. Let r be the rate of travel (speed) in miles per hour.
Example: A traveling salesman drives from home to a client’s store 150 miles away. On the return trip he drives 10 miles per hour slower and adds one-half hour in driving time.
At what speed was the salesperson driving on the way to the client’s store?
Let r be the rate of travel (speed) in miles per hour.
Trip to client
Trip home
distance
rate
time
150 r
150
r – 10
LCM = 2r (r – 10).
300r – 300(r – 10) = r(r – 10)
Multiply by LCM.
Example continued
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Example: Word Problem

35. The return trip took one-half hour longer.
Example continued
300r – 300r = r2 – 10r
0 = r2 – 10r – 3000
0 = (r – 60)(r + 50)
r = 60 or – 50
(–50 is irrelevant.)
The salesman drove from home to the client’s store at 60 miles per hour.
Check:
At 60 mph the time taken to drive the 150 miles from the salesman’s home to the clients store is = 2.5 h.
At 50 mph (ten miles per hour slower) the time taken to make the return trip of 150 miles is = 3 h.
The return trip took one-half hour longer.
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Example Continued