Solving Rational Equations Digital Lesson. Copyright © by Houghton Mifflin Company, Inc. All rights...

Solving Rational Equations

Digital Lesson

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2

A rational expression is a fraction with polynomials for the numerator and denominator.

1

3 and ,

3

2 ,

1 2

x

x

xxare 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 1.1

9 and ,

3

2 ,

1 2

x

x

xx

xx

1

3

2

x 1

92

x

x

undefined1 12

1

33

1 0undefined

03

2undefined 9


A rational equation is an equation between rational expressions.

For example, and are rational equations.3

21

xx xx

x

x

x

2

1

1

33

2

4. Check the solutions. 3. Solve the resulting polynomial equation.

2. Clear denominators by multiplying both sides of the equation by the LCM.

1. Find the LCM of the denominators.

To solve a rational equation:


Examples: 1. Solve: .3

1

3

1

x

x

xFind the LCM.

Multiply by LCM = (x – 3).

Solve for x.

LCM = x – 3.

1 = x + 1x = 0

Check. Substitute 0.

Simplify.

3

1

3

1

(0)

(0)(0)

3

1

3

1

True.

2. Solve: .1

21

xx

x – 1 = 2x

Find the LCM.LCM = x(x – 1).

Multiply by LCM.

Simplify.

x = –1 Solve.

1

2

1

xx)1( )1( xxxx


After clearing denominators, a solution of the polynomial equation may make a denominator of the rational equation zero.

Since x2 – 1 = (x – 1)(x + 1),

Since – 1 makes both denominators zero, the rational equation has no solutions.

Example: Solve: .1

1

1

132

xx

x

2x = – 2 x = – 1

3x + 1 = x – 1

Check.

It is critical to check all solutions.

In this case, the value is not a solution of the rational equation.

LCM = (x – 1)(x + 1).

1

1

1

132 xx

x)1)(1( )1)(1( xxxx


Example: Solve: .158

6

3 2

xxx

x

Factor.

Polynomial Equation.

Simplify.

Factor.

The LCM is (x – 3)(x – 5).x2 – 8x + 15 = (x – 3)(x – 5)

x(x – 5) = – 6

x2 – 5x + 6 = 0

(x – 2)(x – 3) = 0

x = 2 or x = 3Check. x = 2 is a solution.Check. x = 3 is not a solution since both sides would be undefined.

158

6

3 2

xxx

xOriginal Equation.

158

6

3 2

xxx

x)5)(3( )5)(3( xxxx


To solve problems involving work, use the formula,

Example: If it takes 5 hours to paint a room, what part of the work is completed after 3 hours?

Three-fifths of the work is completed after three hours.

If one room can be painted in 5 hours then the rate of work is (rooms/hour). The time worked is 3 hours. 5

1

part of work completed = rate of work time worked.

Therefore, part of work completed = rate of work time worked

5

33

5

1part 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.

164

tt

1223 tt

4.25

12125 tt

LCM = 12.

Multiply by 12.

Simplify.

Working together they will paint the room in 2.4 hours.

painter

assistant

rate of work time worked part of work completed

4

1t 4

t

6

1t 6

t


distance = rate time and time = .

To solve problems involving motion, use the formulas,

rate

distance

Examples: 1. If a car travels at 60 miles per hour for 3 hours, what distance has it traveled?

2. How long does it take an airplane to travel 1200 miles flying at a speed of 250 miles per hour?

It takes 4.8 hours for the plane make its trip.

time = = = 4.8.rate

distance

250

1200Since distance = 1200 (mi) and rate = 250 (mi/h),

Since rate = 60 (mi/h) and time = 3 h, then

The car travels 180 miles.distance = rate time = 60 3 = 180.


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.

Let r be the rate of travel (speed) in miles per hour.

300r – 300(r – 10) = r(r – 10)

Trip to client

Trip home

distance rate time

150 rr

150

150 r – 1010

150

r

2

1150

10

150

rrLCM = 2r (r – 10).

Example continued

At what speed was the salesperson driving on the way to the client’s store?

Multiply by LCM.


0 = r2 – 10r – 3000

The salesman drove from home to the client’s store at 60 miles per hour.

The return trip took one-half hour longer.

At 60 mph the time taken to drive the 150 miles from the salesman’s home to the clients store is = 2.5 h.

60

150

At 50 mph (ten miles per hour slower) the time taken to make the return trip of 150 miles is = 3 h.

50

150

r = 60 or – 50

Example continued

Check:

(–50 is irrelevant.)

300r – 300r + 3000 = r2 – 10r

0 = (r – 60)(r + 50)

Solving Rational Equations Digital Lesson. Copyright © by Houghton Mifflin Company, Inc. All rights...

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