1. Rational Equations
Technical Definition: An equation that contains a rational expression
Practical Definition: An equation that has a variable in a denominator
Example:

2. A rational equation is an equation between rational expressions.
For example, and are rational equations.
To solve a rational equation:
1. Find the LCM of the denominators.
2. Clear denominators by multiplying both sides of the equation by the LCM.
3. Solve the resulting polynomial equation.
4. Check the solutions.
Rational Equation

3. Solving Rational Equations
Find “restricted values” for the equation by setting every denominator that contains a variable equal to zero and solving
Find the LCD of all the fractions and multiply both sides of equation by the LCD to eliminate fractions
Solve the resulting equation to find apparent solutions
Solutions are all apparent solutions that are not restricted

4. 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.
Example: Solve

5. Example

6. Check. x = 3 is not a solution since both sides would be undefined.
Example: Solve:
x2 – 8x + 15 = (x – 3)(x – 5)
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.
Example: Solve

7. This checks in the original equation, so the solution is 7.

8. Applications of Rational Expressions
Word problems that translate to rational expressions are handled the same as all other word problems
On the next slide we give an example of such a problem

9. Example
When three more than a number is divided by twice the number, the result is the same as the original number. Find all numbers that satisfy these conditions. .

11. 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.
Example: Using Work Formula

12. painter assistant rate of work time worked part of work completed
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.
Example: Word Problem

13. Tom can mow a lawn in 4 hours. Perry can mow the same lawn in 5 hours
Tom can mow a lawn in 4 hours. Perry can mow the same lawn in 5 hours. How long would it take both of them, working together with two lawn mowers, to mow the lawn?
UNDERSTAND the problem
Question: How long will it take the two of them to mow the lawn together?
Data: Tom takes 4 hours to mow the lawn. Perry takes 5 hours to mow the lawn.
Tom can do 1/4 of the job in one hour
Perry can do 1/5 of the job in one hour

14. Develop and carryout a PLAN
Tom can mow a lawn in 4 hours. Perry can mow the same lawn in 5 hours. How long would it take both of them, working together with two lawn mowers, to mow the lawn?
Develop and carryout a PLAN
Let t represent the total number of hours it takes them working together. Then they can mow 1/t of it in 1 hour.
Translate to an equation.
Tom can do 1/4 of the job in one hour
Together they can do 1/t of the job in one hour
Perry can do 1/5 of the job in one hour

15. Tom can mow a lawn in 4 hours. Perry can mow the same lawn in 5 hours
Tom can mow a lawn in 4 hours. Perry can mow the same lawn in 5 hours. How long would it take both of them, working together with two lawn mowers, to mow the lawn?

16. At a factory, smokestack A pollutes the air twice as fast as smokestack B.When the stacks operate together, they yield a certain amount of pollution in 15 hours. Find the time it would take each to yield that same amount of pollution operating alone.
1/x is the fraction of the pollution produced by A in 1 hour.
1/2x is the fraction of the pollution produced by B in 1 hour.
1/15 is the fraction of the total pollution produced by A and B in 1 hour.

17. 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.
Examples: Using Motion Formulas

18. Trip to client Trip home distance rate time 150 r 150 r – 10
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
Example: Word Problem

19. (–50) is irrelevant.) Example continued 300r – 300r + 3000 = r2 – 10r
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.
Example Continued

20. An airplane flies 1062 km with the wind
An airplane flies 1062 km with the wind. In the same amount of time it can fly 738 km against the wind. The speed of the plane in still air is 200 km/h. Find the speed of the wind.

21. r = 36 km/h

24. Rational
Inequalities