1. Bell Ringer
1. What is the restriction on rational expressions which results in extraneous solutions?
2. When solving equations with fractions, what do you do to “get rid of the fraction”?

2. Solving Rational Equations
Monday, October 24, 2016

3. 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.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
Rational Equation

4. Other Methods
If you have two rational expressions set equal, you can cross multiply to solve.
If there is more than one term, sometimes you can combine then cross multiply.
If the expressions have the same denominator, just cross it out and solve.

5. 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.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
Examples: Solve

6. 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.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
Example: Solve

7. 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.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
Example: Solve

8. 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.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
Example: Using Work Formula

9. 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.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
Example: Word Problem

10. 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.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
Examples: Using Motion Formulas

11. 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
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
Example: Word Problem

12. 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.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
Example Continued

13. Assignments Classwork: Rational Equations (1-15)
Homework: Rational Equations (1-10)

14. Exit Ticket When solving rational equations…
1. When can you cross out the denominators?
2. When can you cross multiply?
3. If you can’t do #1 or #2, what do you do?
4. Why do you need to check your solutions?