1. Solving Rational Equations
Digital Lesson
Solving Rational Equations

2. are rational expressions. For example,
A rational expression is a fraction with polynomials for the numerator and denominator.
are rational expressions.
For example,
If x is replaced by a number making the denominator of a rational expression zero, the value of the rational expression is undefined.
Example: Evaluate for x = –3, 0, and 1. x
 3
undefined
undefined 9
undefined 1
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Rational Expression

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.
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Rational Equation

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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