Ra a E a E a · Solution: Factor the expression, and we get The x divides out to give us Since the x divided out of the numerator and denominator, we would get when x 0. Therefore,

Rational Expressions

and Equations

347

ChaPter 6

Objectives

In this chapter you will learn how to do the following:

1. Simplify rational expressions.

2. Find where a rational expression is undefined.

3. Add, subtract, multiply, and divide rational expressions.

4. Simplify complex fractions.

5. Solve rational equations.

6. Solve applied problems, such as “work” and “distance” problems using rational

equations.

Introduction Problem

While you may not know how to solve the problem below now, you should be able to solve

this introduction problem by the end of section 6.4.

Suppose you are the administrator of the nursing department of a hospital. Your experi-enced nurses can get vital signs in 90 minutes, whereas a less experienced nurse wouldtake 120 minutes. How much time would the two nurses take to get the vital signsfrom all of the patients if both were assigned to the task, and they worked together?Perhaps a more complicated version would be if we had three experienced nurses andtwo less experienced nurses, how much time would it take? Even more complex is thisquestion: In an eight hour shift, if we have three experienced nurses and two trainees,how many times could we expect the patients’ vital signs to be read?

(continues)

6.1 Simplifying rational expressions

You are more likely familiar with the concept that the fraction simplifies to , as the following

picture shows. The shaded regions represent the two fractions, and they are obviously the same size:

The denominators 4 and 8 describe how many parts in which an entire unit is divided. The nu-

merators describe how many of the pieces of that size. In this case, I hope you can see that the

two fractions represent the exact same amount. (The shaded parts are the same size!)

Perhaps better stated, in a generic fashion, is the following principle. It is often used to simplify

fractions.

For example, in simplifying , we cannot cancel the 4x2 in the numerator with

the 4x2 in the denominator, and we can not cancel the 9 in the numerator with the 9 in the de-

nominator because neither of the terms in the numerator or denominator are multiplied together.

We do not have an expression in the form .

34

68

6–8

3–44–3

8–6

The fundamental principle of fractions:a � c—b � c � a�b

4x2 � 94x2 � 12x � 9

a � cb � c

348 ChaPter 6 n Rational�Expressions�and�Equations

In reality, patients on this high-level care optimally have their vital signs read every half

hour. If we have 30 beds filled, how many nurses and trainees should be scheduled to

give optimal care? Do you see the complexity of the problems?

Many problems of this type are standard algebra problems involving a type of problem called

“work” problems. Usually it will require solving equations containing fractions to get rea-

sonable answers.

Our next unit is about rational algebraic expressions, more commonly known as fractions.

You might want to see if tutoring is available at your school.

An expression such as is in the form . Therefore, we can cancel the 4x2 in

the numerator with the 4x2 in the denominator to get .

How can we change an expression such as into an expression in which the

terms are multiplied together? The answer to the question was the title of Chapter 5.

The expression factors into which we can now simplify

to .

Do not go any farther! Do not cancel the 2x in the numerator with the 2x in the denominator.

Do not do it!

If you see an expression such as do NOT cancel the 4 in the numerator with the

4 in the denominator. The right thing to do is to factor the numerator and denominator and seeif there are any factors that cancel out. Remember NEVER CANCEL TERMS! Terms are “added”things.

The expression factors into which simplifies to . This

is as far as it simplifies.

Do not cancel the 2 in the numerator with the 2 in the denominator of to get . Do

not do it!

The x and the 2 in the denominator are not multiplied together; they are added together and

therefore cannot be canceled separately.

Check it with numbers: If simplified to , then we should be able to substitute

any number in for x in both expressions and get the same answer. If we substitute 1 in for x in

, we get . If we substitute 1 in for x in , we get 1. They do not equal, therefore we can

not simplify this expression.

However, did simplify to . Any number that we substitute for x in the

expression will give us the same answer as substituting that number into .

For example, if we substitute 1 in for x in the expression , we get

, which simplifies to or . If we substitute 1 in for x in the expression

, we get , which is . We get the same answer because simplifies to .

4x2

4x2(2x � 9)a � cb � c

1(2x � 9)

4x2 � 94x2 � 12x � 9

4x2 � 94x2 � 12x � 9

(2x � 3) (2x � 3)(2x � 3) (2x � 3)

2x � 32x � 3

2x � 4x2 � 4x � 4

2x � 4x2 � 4x � 4

2(x � 2)(x � 2) (x � 2)

2x � 2

2x � 2

1x

2x � 2

1x

2x � 2

23

1x

2(x � 2)(x � 2) (x � 2)

2x � 2

2x � 2

2(x � 2)(x � 2) (x � 2)

2(x � 2)(x � 2) (x � 2)

2(1 � 2)(1 � 2) (1 � 2)

63 � 3

2x � 2

21 � 2

69

69

23

23

SeCtION 6.1 n Simplifying�Rational�Expressions 349

The rules in algebra are not because someone made them up, it’s because they work with

numbers.

The rule � is even used with reducing to because is , but does

not equal .

You Try These

Simplify the following examples, and check your answers with a neighbor or ask your teacher to

give them to you if possible:

1) 2)

3) 4)

Many look at problems like number 4 in the previous examples and make an incorrect assumption

that you can cancel anything you see in a fraction. A more careful examination of the fundamental

principle would show that, in fact, only factors, not addends, can be cancelled.

The correct way to do number 4 is to first factor the denominator, then cancel the common

factor (x � 2): This example is only valid if x � 2. Let me explain

why.

Division Concerning Zero (Review)

Note that the expression simplifies to 2 as long as x is not 4. If x is 4, we would have an

expression of form which is called indeterminate. Therefore, our simplification of to

2 is only valid if x � 4. Briefly, recall the following:

1. is undefined for n � 0.

2. is 0 for n � 0 .

3. is indeterminate.

a � cb � c

ab

69

23

69

2 � 33 � 3

2 � 33 � 3

23

4050 �21

35 �

x � 2x2 � 5x � 6

�4x2y3

12x2y�

x � 2(x � 2) (x � 3) � 1

x � 3.

2x � 8x � 4

2x � 8x � 4

00,

n0

0n

00


Note: The fraction line in an expression such as is a grouping symbol. There-

fore, is the same as

Following is another unusual type of cancellation:

example 1: Simplify the expression Note that the addends cannot be cancelled.

Solution: We cannot cancel the m in the numerator with an m in the denominator. On closeinspection, if the number �1 was factored out of the numerator (or, for that matter, the de-nominator), then we would have the same factor in the numerator as in the denominator:

In general, any expression of this form can be simplified to �1. It’s only valid if n � m.

Note: If we have can we cancel out the 2s to get ? NO!!!! Check it. is

and definitely does not equal We cannot cancel out addends (things that are added or sub-

tracted) together. We can cancel out things that are multiplied together. For example, does

? Yes. If we can make an expression that has terms added together into an expres-

sion with terms that are multiplied together, we can simplify the expression.

example 2: Simplify the expression and find where the expression is indeterminate.

Solution: Factor the expression so that we can change terms that are added together into terms that are multiplied together. Factoring the numerator and denominator, we get

This simplifies to The (x � 3) was in both the numerator and

denominator. This means when x � 3 � 0, the expression is indeterminate. Solving x � 3 � 0,we get x � �3. Therefore, at x � �3 the expression is indeterminate (not valid).

example 3: Where is the expression undefined or indeterminate?

m � nn � m.

� 1(n � m)n � m Æ

�1(n � m)n � m Æ �1

73

2 � 52 � 1

51

2 � 52 � 1,

51.

7(2)2(3) � 7

3

(x � 2)(x2 � 5x � 6)

.x � 2x2 � 5x � 6

To find where the expression is indeterminate, set any terms that are in both

the numerator and denominator equal to zero, and solve.

x2 � 5x � 64x2 � 9x � 9

(x � 2)(4x � 3) .

(x � 3) (x � 2)(4x � 3) (x � 3) .

5x4x3 � x2 � 3x

x � 2x2 � 5x � 6


Solution: Factor the expression, and we get The x divides out to give us

Since the x divided out of the numerator and denominator, we would get

when x � 0. Therefore, x � 0 is indeterminate (not valid).

In the denominator of the previous example, we have (x � 1)(4x � 3). Set each of these equalto zero, and solve. Setting the x � 1 � 0, we get x � �1; and setting the 4x � 3 � 0, we

would add 3 and divide by 4 to get x � Therefore, at x � �1 and x � , the expression is

undefined (not valid).

example 4: Simplify the rational expression and give a value of x where the sim-

plification is not valid.

Solution: This expression simplifies to The x � 3 divides to 1, and we get

, or simply �(x � 2). However, remember that x � 3 because, if we substitute 3 in

the original expression, we would get which is indeterminate and therefore not valid.

You try this One: Where is the expression undefined or indeterminate?

Solution: It is indeterminate at and undefined at Therefore, is not valid.

example 5: In the previous example, the expression factored into

The (2x � 5) term divided out, leaving us with Since the 2x � 5 � 0

at the expression is indeterminate (not valid) at and since the remaining term in

the denominator, 2x � 5, equals 0 at the expression is undefined (not valid) at

To find where the expression is undefined, set the denominator equal to zero, and solve

after you have cancelled out any terms in the numerator and denominator.

34

34

x2 � 5x � 63 � x

(x � 3) (x � 2)�(x � 3) .

(x � 2)�1

00,

6x2 � 17x � 54x2 � 25

x � �52x � 5

2.x � �52

To find where the expression is 0, set the numerator equal to zero, and solve after

you have cancelled out any terms in the numerator and denominator.

6x2 � 17x � 54x2 � 25

(3x � 1)(2x � 5) .

(3x � 1) (2x � 5)(2x � 5) (2x � 5) .

00

5(x � 1) (4x � 3) .

x � �52 ;x � �5

2 ,

x � 52.x � 5

2,

5xx(x � 1) (4x � 3) .


Setting the remaining numerator, 3x � 1, equal to 0, and solving, we get This tells us

that if we substitute for x in the expression, we will get 0.

It would be very messy to check this, but if we substitute in the original expression or even

the simplified expression, we will get 0. I will check the simplified expression Substi-

tuting for x, we get Let’s check to see

if makes the expression undefined. Substituting for x in we get

We have a nonzero number divided by 0, and that

is undefined. To check the location of the indeterminate value, you would have to substitute the

in the original expression, and you would get which is indeterminate.

(3(�13 ) � 1)

(2(�13 ) � 5)

Æ(�1 � 1)

(�23 � 5)

Æ 0

(�173 )

Æ 0.�13

(3x � 1)(2x � 5) ,5

252

(3( 52) � 1)

(2( 52) � 5)

Æ( 15

2 � 1)(5 � 5) Æ

( 172 )0 .

00,�5

2

�13

(3x � 1)(2x � 5) .

x � �13 .

�13


6.1 exercise Set

Simplify the following expressions by applying the fundamental principal of fractions.

1) 2)

3)

Simplify the following expressions and tell where the expression is invalid (undefined or indeterminate).

4) 5)

6) 7)

8) 9)

10) 11)

12) 13)

14) 15)

16) 17)

18) 19)

20) 21)

4270

6090

35x3y5

42x5y

36x4y10

�48x7y64t2

10t

x � 2x � 2

2t � 2814 � t

4 � t2

t2 � 49 � y2

3 � y

�2 � tt � 2

t2 � 4t2 � 4

y2 � 14y � 49y2 � 49

25 � x2

x2 � 7x � 10

48x2 � 316x2 � 8x � 3

x2 � 10x � 16x2 � 4

36xy � 13x2y � x3yx2y � 9xy

x3 � 4x2 � 3xx2 � 3x

2t2 � 812t2 � 51t � 54

t2 � 5t � 24t2 � 64

x2 � 6x � 72x2 � 36

x2 � 22x � 23x2 � 1


22) 23)

24) 25)

26)

6x3 � 23x2 � 4x6x3 � 25x2 � 4x

x2 � 2x � 8x2 � 9x � 20

16x2 � 44x2 � 1

4x2 � 4 � 10xx2 � 4 � 4x

t4 � 1t2 � 1


6.2 Multiplication and Division of rational expressions

Multiplying Fractions

If I were to ask you what of is, what would be your answer? This picture should tell you the

correct answer.

Even without the blatant hint, you should see that of is . The rule for multiplying fractions is

You should simplify the answer by canceling any like terms from the numerator and denominator.

example:

You Try These #1

Review from arithmetic—check with a neighbor if possible.

1) 2) **

**We need to know how many thirds are in There are 6 thirds in two (why?);

2 more makes 8.

3)

14

12

1–88–1

18

14

12

AB � C

D � A � CB � D

2–3

9—10

3–5

? g g

1 32 9

3 101 5

??

Cancel common factors 2 and 3

35 � 22

337 � 1

3

223.

59 � 3

4


The problems look much harder if we use polynomials for the numerator and denominator, but

they really aren’t. The rule is the same, and we learned how to factor the polynomials in the last

chapter, hopefully. Keep in mind, we can only cancel factors; we cannot cancel terms (added

things) in a fraction.

example: Multiply the fractions and simplify the answer as much as possible.

Solution: The first step is to factor all numerators and denominators.

Now, cancel common factors in the numerator and denominator.

This gives us an answer of

Practice on the following example.

Hopefully, you will get an answer of or something equivalent.

x2 � 4xx � 7 � x2 � 4x � 21

x3 � 16x

x(x � 4)x � 7 �

(x � 7) (x � 3)x(x � 4) (x � 4)

x(x – 4)x + 7 ?

(x + 7)(x – 3)x(x + 4)(x – 4)

x � 3x � 4.

x2 � 2x � 3x2 � 1

� x2 � 5x � 416 � x2

�x � 3x � 4

SeCtION 6.2 n Multiplication�and�Division�of�Rational�Expressions 357

Dividing Fractions

Suppose we had pizzas and wanted to divide them into fourths. The following picture shows

pizzas; count the number of fourths shown.

We can see that the answer should be 10, as there are 10 different fourths in the pizzas.

We’ll check this in a minute.

First, how do we divide 8 by 2? The answer (4, I think!) is the number of 2s in 8. Can you see

that it is also of 8? Multiplying by gives the same result as dividing by 2. This result is gen-

eralized in this rule:

The arrows point out that the divisor has been inverted. Note the operation has changed to

multiplication as well. Let’s test on the pizza problem from before.

example: How many fourths of a pizza are there in two-and-a-half pizzas?

Solution: First, two-and-a-half can be written as five-halves, so what we seek is the answer to

as we expected.

You try the following. They’re a bit more complicated; do not forget to factor first.

212

12

12

A

B�C

D�A

B�D

C

52 � 1

4 � 52 � 4

1 � 10,52 � 1

4.

212

212


You Try These #2

Check your answer with a neighbor, and let your teacher help if your answers disagree.

1) 2) x � 4x2 � 2x � 8

� x2 � 9x � 20x2 � 3x � 10

5t3

6r5 � 20t2

9r3

SeCtION 6.2 n Multiplication�and�Division�of�Rational�Expressions 359

6.2 exercise Set

Perform the following operations, and simplify your answers as much as possible:

1) 2)

3) 4)

5) 6)

7) 8)

9) 10)

11) 12)

13) 14)

15) 16)

17) 18)

19) 20)

21) 22)

23) 24)

25) 26)

27) 28)

x2 � 16x � 3 � x2 � 9

x � 43x � 126x � 30 � x2 � x � 20

x2 � 16

9x2 � 19x2 � 6x � 1

� 3x2 � 11x � 412x2 � 5x � 3

6x(x � 3)3x � 1 �

x2(x � 3)9x2 � 1

2x2 � 11x � 21x2 � 5x � 14

� 4x2 � 9x2 � 6x � 16

12x2 � 5x � 39x2 � 1

� 16x2 � 912x2 � 13x � 3

x2 � 2xy � y2

x2 � x�

x2 � y2

x2 � 1x2 � xy � 12y2

x2 � xy � 20y2 �x2 � 9xy � 20y2

x2 � xy � 30y2

x � 3x2 � 4

� x � 22x2 � x � 15

920 � 16

21

3xy2

4a2b� 2ab

9x2y33a � 15

4a � 2a5a � 25

x2 � 9x2 � 5x � 6

� x2 � 4x � 34 � x2

5xy9t2 �

10x2y12t

a2 � b2

a2 � 2ab � b2 � a � ba � b3x4 � 9x2

2y

2m2 � 8mm2 � 16

m2 � 4m2m2 � 8m

m2 � 16m2 � 4m

x2 � x � 12x2 � 7x � 10

� x2 � 3x � 10x2 � 2x � 8

( x2 � 3x � 4x2 � 9

) ( x � 32x2 � 3x � 2

)

12 � 4x6 � 9

4x � 12x2 � 3x � 2

x2 � 1� x2 � 2x � 3

x2 � x � 6

2x2 � 5x � 12x2 � 2x � 24

� x2 � 9x � 189 � 4x2

3x2 � x � 102x2 � 7x � 4

� 4x2 � 16x2 � 7x � 5

4x � 122x � 10 � x2 � 9

x2 � 252x � 2y

8 �x2(x2 � y2)

24

2x � 116 � x2 � x � 4

4x � 23x2 � 12

x2 � x � 6� x2 � 6x � 9

2x � 4


6.3 Solving rational equations and Inverse Variation

This is a relatively easy section. All that is new is finding a common denominator of fractions,

and that was really discussed in the Review Chapter.

What is the common denominator of the fractions and ? If we were to add these two

fractions, we’d need to convert them both to 36ths to proceed. We get the 36 by considering allpossible factors of 9 and 4, so 3 � 3 � 2 � 2 is the common denominator.

What is the common denominator of and ? The 9 factors as 3 � 3 and, since we already

have a 3, 9 is the least common denominator. Note 3 � 3 � 3 � 27, but 9 is smaller. From thesecond fraction, we need only add prime factors that aren’t in the first fraction’s denominator toget the smallest, known as the LCD or least common denominator.

The basic idea of solving equations with fractions is to transform them into equations without

fractions. This will happen if we multiply by any common denominator, and if we use the least

common denominator, the equation will be less complicated.

example 1: Solve the equation This type of problem was solved in Chapter 2.

Solution: The common denominator of all of these fractions is 2 � 2 � 3 � 3 � 36. The 2 � 2 is

from the denominator 4, the 3 � 3 from 9, and the 3 on the right side adds nothing. We now

proceed to multiply both sides of the equation by 36:

27x � 8x � 12

19x � 12

x �

Suppose we have polynomial denominators. Does it make sense that we would do the same? There

is a slight new twist added to the process, but more on that later. Here’s an example.

example 2: Solve the equation

Solution: Hopefully, you can see the least common denominator is 2(x � 2)(x � 3). What the

next step will do is eliminate all denominators by multiplying by this LCD.

14

79

13

79

3x4 � 2x

9 � 13.

36( 3x4 � 2x

9 � 13)

1219

2x � 2 � 4

x � 3 � 12(x � 2)

SeCtION 6.3 n Solving�Rational�Expressions�and�Inverse�Variation 361


� �

2(x � 3)2 � 2(x � 2)4 � (x � 3)

4(x � 3) � 8(x � 2) � x � 3

4x � 12 � 8x � 16 � x � 3

�4x � 28 � x � 3

�25 � 5x

�5 � x

Caution: Even if a term in the equation has no fraction, it must be multiplied by the LCD.

The following example illustrates this idea:

example 3: Solve .

Solution: It’s not hard to see the LCD is 44. The next step is troublesome for some.

11x � 88 � 8x

�88 � �3x

Note: The LCD 44 also multiplied the 2, even though there was no denominator we needed

to clear.

example 4: Solve the equation .

Solution: Multiply by the common denominator, 12, and we have .

Distribute the 12 through each term to get .

This simplifies to 8x � 48 � x � 2.

Subtracting x from both sides and subtracting 48 from both sides, we get 7x � �46.

Dividing by 7, we get x � .

Now try this one for a real surprise.

2(x � 2) (x � 3) [ 2x � 2 � 4

x � 3 � 12(x � 2) ]

2(x � 2) (x � 3) [ 2x � 2] 2(x � 2) (x � 3) [ 1

2(x � 2) ]

x4 � 2 � 2x

11

44( x4) � 44 � 2 � 44( 2x

11)

x � �88�3 � 88

3

2x3 � 4 � x � 2

12

12[ 2x3 � 4 � x � 2

12 ]

12[ 2x3 ] � 12(4) � 12[ x � 2

12 ]

2(x � 2) (x � 3) [ 4x � 3]

�467

example 5: Solve the equation .

Solution: The x2 � 5x � 4 factors into (x � 4)(x � 1). Next multiply by the LCD (x � 4)(x � 1)

then solve the new equation. You should get x � 4 as a solution. However, 4 is not a solution! You

can check by substituting 4 in the equation. The very first fraction now reads which is unde-

fined. What went wrong?

If you examine very carefully a property we have already been using—the multiplication prop-

erty of equality—you might see the problem.

Multiplication Property of Equality

The equations A � B and A � C � B � C have the same solutions if C � 0.

Do you see it? In our example, we multiplied by the LCD (x � 4)(x � 1) and if x � 4, then

what does (x � 4)(x � 1) equal? See what happened in this example now? Discuss what you

think went wrong with a neighbor, and check with the teacher if neither of you see it.

In fact, the answer to our example here is { }; in other words, there was no solution since our

only solution was not allowed. The method to solve equations with fractions is the following:

1) Simplify the denominators.

2) Multiply the entire equation by the LCD of all of the fractions in the equation.

3) Solve the equation using things you already know how to do (it will not have frac-

tions after step 2).

4) Verify you didn’t accidentally multiply by zero by substituting your answers into

the original equation’s denominators. If any denominator equals zero, exclude

those answers. What’s left is the solution.

Let’s use our step-by-step process to solve the previous equation: � �

.

Step 1: Simplify all the denominators. There is only one denominator that factors. We get

.

Step 2: Multiply by the common denominator. The common denominator is (x � 4)(x � 1).

We get This simplifies to

4(x � 1) � 5(x � 4) � 12.

Step 3: Solve the equation. Distributing the 4 through the first set of parentheses and the 5

through the second set of parentheses, we get 4x � 4 � 5x � 20 � 12. Combining like terms,

we get 9x � 24 � 12. Adding 24, we get 9x � 36. Dividing by 9, we get x � 4.

4x � 4 � 5

x � 1 � 12x2 � 5x � 4

40,

4x � 4 � 5

x � 1 � 12(x � 4) (x � 1)

4(x � 4) (x � 1)x � 4 �

5(x � 4) (x � 1)x � 1 �

12(x � 4) (x � 1)(x � 4) (x � 1) .

4x � 4

5x � 1

12x2 � 5x � 4


Step 4: Verify that our solution does not make any denominator 0. Substituting our

solution of 4 into the original equation, we get �

Notice that this solution gives us division by

zero, which makes the original equation undefined at x � 4. Therefore, x � 4 is not a solution,and since x � 4 was the only possible solution, this equation has no solution. The solutionset is the empty set, { }.

Here are a couple more examples to try:

You Try These #1

1)

2)

example 6: The height of a trapezoid can be found using the formula where h is

the height, A is the area, and b1 and b2 are the lengths of the parallel sides. Solve this equation for A.

Solution: Multiply by the common denominator of and we get

This eliminates the fraction, and we get (b1 � b2)h � 2A. Now

divide by 2, and we get

example 7: Solve the same equation for b2.

Solution: We would still multiply by the common denominator to get (b1 � b2)h � 2A. Atthis point, distribute through the parentheses to get b1h � b2h � 2A. Now subtract b1h, and

we get b2h � 2A � b1h. Now divide by h, and we get

example 8: Solve the equation for n.

Solution: Multiply by the common denominator 5n � 1, and we get

This eliminates the fraction, and we get C(5n � 1) � 2n � w.

Multiply the C through the parentheses, and we get 5Cn � C � 2n � w. Now, since we are solving

4x � 4 � 5

x � 1 � 12x2 � 5x � 4

,

54 � 1 � 12

(4)2 � 5(4) � 4 Æ 4

0 � 53 � 12

0 .

1x � 2 � 4 � 3 � x

x � 2

2xx � 3 � 13x � 3

x2 � 9� x

x � 3

h � 2A(b1 � b2)

,

(b1 � b2) ,

(b1 � b2)h �2A(b1 � b2)(b1 � b2)

.

(b1 � b2)h2 � A

b2 �2A � b1h

h .

44 � 4

C � 2n � w5n � 1

C(5n � 1) �(2n � w) (5n � 1)

(5n � 1) .


for n, get all the terms that have an n in them to one side. Subtracting 2n from both sides and addingC to both sides, we get 5Cn � 2n � w � C. At this point, you need to realize that we can factor an nout of the left side of the equation to get n(5C � 2) � w � C. Our last step is to divide by (5C � 2),

and we get

You Try These #2

1) Solve for y.

2) Solve 3xy � 4x � 6y � 4xy for y.

Variation: Review of Direct Variation

In Chapter 3 we did direct variation. Direct variation is in the form y � kx, where y and x are the

variables and k is some constant.

example 9: Your score on a test varies directly to the amount of time you do homework. Let’s say

that if you study 2 hours you get a 50 on a test. How many hours would it take to get a score of 100?

Solution: The direct variation equation y � kx can be used where y is the score and x is the

number of hours you do homework. Substituting 50 for y and 2 for x in the equation y � kx,

we get 50 � 2k. We can solve this equation to get the constant k � 25. We now have the equa-

tion y � 25x. Now we can use this equation to find the number of hours it takes to get a score

of 100. Substitute 100 for y in the equation y � 25x, and we get 100 � 25x. Solving this, we

get x � 4. Therefore, it would take 4 hours to get a score of 100.

Inverse Variation

A quantity VARIES INVERSELY as another quantity if the product of the two quantities is a con-

stant. For example, if x and y are variables and k is a constant, the fact that x varies inversely as y

is expressed by xy � k or, solving for y, we get

n � w � C5C � 2.

x �2y � 42y � 4

y � kx.



example 10: The number of mistakes you make on a test varies inversely to the number of

hours you do homework. If you study 2 hours, you make 50 mistakes. How many hours would

you need to do homework to reduce the number of mistakes to 10?

Solution: The inverse variation equation can be used where y is the number of mistakesand x is the number of hours you do homework. Substituting 50 for y and 2 for x in the equation

we get Solving this, we get k � 100. We now have the equation

Now we can use this equation to find the number of hours of homework it would take to make

only 10 mistakes. Substitute 10 for y in the equation and we get This is

a rational equation. Multiply by the common denominator x, and we get 10x � 100. Dividingby 10, we get x � 10. It would take 10 hours to reduce the number of mistakes to 10.

example 11: Using the information in the previous example, could you ever do enough home-

work to reduce the number of mistakes to zero?

Solution: We have the equation where y is the number of mistakes. Substituting 0

for y we get Multiplying by the common denominator, x, we get 0 � 100, which isa false statement. This means that this equation has no solution and, therefore, you can nevermake no mistakes.

example 12: The rate (or speed) of doing something is the reciprocal of the time it takes to do

it. We could say that the time it takes to do a job varies inversely to the rate you do the job. Think

about it, the faster you do a job, the less time it takes. If one hose fills a pool in 6 hours, then how

many hours would it take to fill the pool with two hoses, assuming both hoses have the same

water flow as the hose that filled the pool in 6 hours.

Solution: where x is the time it takes to fill the pool and y is the number of hoses. We

get We get k � 6. Substituting 2 for y, we get Multiplying by x, we get 2x � 6

or x � 3. Thus, it would take 3 hours to fill the pool with 2 hoses.

y � kx

y � 100x .50 � k

2.y � kx,

10 � 100x .y � 100

x ,

y � 100x ,

0 � 100x .

y � kx,

2 � 6x .1 � k

6.

6.3 exercise Set

For each of the following, what would be the least common denominator?

1) 2)

3) 4)

5) 6)

Solve each of the equations.

7) 8)

9) 10)

11) Solve for R. 12)

13) 14)

15) 16)

17) 18)

19) Solve for y. 20) Solve for y.

21) Solve for y.

22) The time, t, it takes to drive a distance, d, varies inversely as the rate, r, that you drive.

If it takes 5 hours at a rate of 50 mph, how long will it take at a rate of 60 mph?

( 44x2 � 1

) � ( 34x2 � 4x � 1

)2m2 � 8mm2 � 16

m2 � 4m� 2m2

2x2 � 1

� x2 � 2x � 3x2 � 2x � 1

2x � 2 � 3x � 10

x2 � 4x � 4

�1x2 � 6x � 9

� x � 1x2 � 9

� 1(x � 3)2

16 � 9

4x � 12 � 35x � 15

x � 710 � �3

x � 6x � 1

6 � x9

68 � 6

x � 1x4 � x

6 � 1

2x � 1 � 3

x � �2I � ER � r

x � 2x � 1 � x

x � 2 � 5x � 43(x � 1)

4x � 1 � 5 � x

x � 1 � 2

2x � 1 � 17

24 � 3x � 1

x3x � 16 � 4

x

4x � 2 � 8

x2 � 5x � 6� 3

x � 39

x2 � x � 12� 3

x � 4 � �xx � 3

1x � 1

y � 5zx �

3y � 5y � 5

5y � x � 3

x


23) For a rectangle with fixed area, A, the length varies inversely to the width. If the length

is 10, the width is 15. What would the width be if the length is 5?

24) The amount of force an occupational therapist has to apply to a patient’s arm is in-

versely proportional to the square of the distance from the elbow. If a force of 50 New-

tons is required at a distance of 2 inches, how many inches from the elbow would the

occupational therapist need to apply a force of 2 Newtons?


6.4 applications of rational equations

Some problems from real life that involve equations with fractions as models arise from prob-lems involving work done by two or more workers, or devices. They all can be solved usingthe formula W � r � t (work done � rate � time worked). Also, it is useful to think of the rate

as rate � For example, if it takes you 50 minutes to do your homework, then

you do of your homework every minute, and this is your “homework doing” rate.

Concept check: It takes 30 hours for a water pipe to fill the city pool. What is the rate per hour of

the water pipe? ***

*** Answer:

example 1: One door at a movie theater can allow a full house to empty in 10 minutes. Another

exit door must be installed, since the entire theater must empty in 3 minutes by state law. What

is the most minutes it must take the new door alone to exit a full house if the theater is to be com-

pliant with state law?

Solution: Before tackling this problem, here are a few helpful hints to follow:

1. Make a 3-column chart and title the columns work, rate, and time.

Here’s the chart style.

Now let x be the time it will take the other door to exit a full house.

1time it takes.

150

1—30per hour

Work Rate Time

SeCtION 6.4 n Applications�of�Rational�Expressions 369

2.

3.

4. Usually, we use 1 to represent a complete job, and it will be equal to the sum of the work

done. is the equation we must solve to find the answer. (Note: if the law stated

we needed to empty of the theater in three minutes, we would replace the 1 on the right

side with and get Back to the original multiplying by the

LCD 10x, we get 3x � 30 � 10x, and solving this, we get or minutes.

example 2: One nurse can do assessments on all patients in a section once in 3 hours, but a

student nurse can do assessments in 4 hours. This means checking on every patient exactly once

and taking vital signs.

A. If a nurse and one student nurse work on a shift, how long will it take to assess all of the

patients?

B. If a nurse and two student nurses are working on a shift, how long will it take to assess all of

the patients?

C. If the student nurse needs an extra 20-minute orientation before starting on the assessments,

how long will it take for both a nurse and a student nurse to complete the task?

Work Rate Time

1—10

3

(time required for the job if

both doors are working)

1�x 3

Work Rate Time

3—101—10 3

3�x1�x 3

310 � 3

x � 1

34

310 � 3

x � 1,310 � 3

x � 34. )3

4

x � 427x � 30

7


Solution to part A: Start by letting x be the time it would take to cover the wing if both worked

together. Then we could fill in the table as follows:

Then we can use the fact that work � rate • time to fill out the first column in the table:

The equation we need to solve is and we multiply both sides by 12 to see the so-

lution is hours, or about 1 hour and 43 minutes.

Solution to part B: Here there are two student nurses, each doing of the task. The equation

will be and the solution is

Solution to part C: Here the student nurse will be working 20 minutes (or of an hour) less,

so we will use (The units must match!) The resulting table is given as follows:

Work Rate Time

1�3 x

1�4 x

Work Rate Time

x�31�3 x

x�41�4 x

x3 � x

4 � 1,

x � 127 � 1.714

x4

x � 65 � 1 hours 12 minutes.x

3 � x4 � x

4 � 1,

13

x � 13.

Work Rate Time

x�31�3 x

x � 1�34

1�4 x � 1�3


The equation we need solve is Again we multiply by 12 to get

Distributing the multiplication gives 4x � 3x � 1 � 12 or 7x � 13.

It would take hours or approximately 1 hour 52 minutes.

example 3: A patient is to receive intravenous saline solution from two sources. One source

has a rate of 15 drips per minute, while the other has a rate of 20 drips per minute with the same

size tubing. Assume each drip is mL and the patient must not be over-hydrated; hence we

must stop the drips when 100 mL of the saline solution is in the system.

A. How much time elapses until the nurse should stop the flow of the IV?

B. How much time (from the start of the first source) if the 20-drip source was started 10 minutes

after the other source?

Solution to part A: This time we make the complete job � 100 mL and fill the table as follows:

The equation is Multiply by 3, and we get 300 � 7x or minutes.

x3 �

x�13

4 � 1.

4x � 3( x�13) � 12.

x � 137 � 1.86

115

Work Rate Time (min.)

x 15 drips15 mL

� 1 x

4�3x20 drips15 mL

� 4�3 x

x � 3007 � 43100 � x � 4

3x.


Solution to part B: Fill out the chart as follows: (Note 10 minutes less time operating for the

“fast IV.”)

Here we have and again we multiply by three to get 3x � 4(x � 10) � 300.

Expanding and adding like terms gives us 7x � 340 or or around 49 minutes.

example 4: If we find that a student nurse and a regular nurse together can assess the patients

on a wing in of an hour, as occurred in example 2A, how many times should they be able to

assess the patients on a wing (that is, get vital signs from each patient) on a 12 hour shift?

Solution: Since work � rate • time, and rate � we have

times.

Group Exercise

Complicated Example (but in fact, much like reality): In the intensive care unit, we strive to

only use experienced nurses for staffing. It takes 2.5 hours for an experienced nurse to check vital

signs or assess the patients for all of ICU. An inexperienced nurse (one that hasn’t worked in the

ICU) would take 3 hours. Vital signs must be taken for each patient every fifteen minutes, so

on an eight hour shift, the assessments are made 32 times. Only three experienced nurses are

available for the 3�11 shift. Decide the least number of inexperienced nurses you must have on

the schedule to assess the patients 32 times.

Do not turn the page until you’ve tried to solve it. Hint: We can consider that we need to do

32 “jobs” (32 rounds made) and solve this as a regular work problem.

Work Rate Time

x 1 x

4�3(x � 10) 4�3 x � 10

x � 43(x � 10) � 100,

x � 3407 � 48.57,

127

w � 1127

� 12 Æ1time it takes,

712 � 12

1 Æ 7

Work Rate Time

24—2.51—2.5

8 � 3(3 experienced

nurses, 8 hours)

8x—31�3

8x (x nurses, 8 hours)



The equation is If we multiply both sides by 15 (Do you see why? 7.5 would

also work.), we get 144 � 40x � 480. Subtracting 144 from both sides yields 40x � 336. Di-viding both sides by 40, we have x � 8.4. If we are to insure adequate staffing to meet the“one round every 15 minutes” goal, we must have 9 additional nurses on this shift in the ICU.(Do you see why we cannot “round it off” to 8?)

Note: This one was an extremely complicated one! Test questions will not likely be this hard,

but real world questions may be.

Geometric Method for Solution of Less Complicated Problems

Many problems involving work require that we find how long it takes when two people or things

work together on a job. There is a nice geometric way to solve these (if you have a measuring de-

vice or fairly good graph paper). The following example demonstrates the technique.

example 5: If one hose can siphon a pool dry in 12 hours while another would take 6, how

long would it take if both hoses are working together?

Solution: Solved in the usual way, we would come to the equation The solution

is x � 4 hours. (A chart might be helpful.)

Geometrically, we draw a vertical line of length 12 and a vertical line of length 6 that is parallel

and some small distance apart, both beginning on the same horizontal line segment. Connect

the tops of each segment to the bottoms of the other segment and draw a line segment from

the horizontal to the intersection point. Measure this, and that is the answer. Here’s a sketch:

Hopefully, it will be 4 units as we expected.

Note that the reciprocal of a number is 1 divided by the number, for example, the reciprocal

of 5 is .

242.5 � 8x

3 � 32.

x12 � x

6 � 1.

612

measure this

15

example 6: A number added to 10 times its reciprocal is equal to 7. Find the number.

Solution: Let n be the number. Then is the reciprocal and n � 10 � 7. Multiplying both

sides by the least common denominator n, we get n2 � 10 � 7n . Subtracting 7n from bothsides, our equivalent equation is n2 − 7n � 10 � 0. This factors as (n − 2)(n − 5) � 0.

By the zero factor property n � 2 � 0 or n � 5 � 0, we see that there are two solutions forthis problem, n � 2 an� n � 5.

Check that 2 � 10 � � 7 and also 5 � 10 � � 7.

example 7: (Distance � rate × time) fA truck drives 150 miles on flatland in the same time that it drives 100 miles through mountains.

The average flatland speed is 20 mph faster than the mountain speed. What is the truck’s speed in

the flatlands?

Solution: Let r � average speed in flatland. Recall d � r * t, and we can fill out the table below.

Note that the time is distance divided by rate because if d � r × t, then t � .

From the problem, the times are “the same,” which translates into “�”. The equation we must

solve is

� .

Multiplying both sides by the least common denominator r(r � 20), we get

150(r � 20) � 100r.

Distributing on the left side, we see

150r � 3000 � 100r.

Subtracting 150r from both sides, we now have

�3000 � �50r.

Finally, dividing by �50, we see the solution:

r � 60 mph.

Since r represents the speed of the truck in the flatlands, this means the speed of the truck in

the flatlands is 60 mph and the speed of the truck in the mountains is r � 20 or 40 mph.

15

12

dr

Distance RateDistance

Time�————Rate

Flatland 150 r 150——r

Mountains 100 r � 20 100——r � 20

100r � 20

150r

1n

1n


example 8: (Distance � rate � time) kA search helicopter can carry enough fuel for a 5 hour flight. The wind is blowing out to sea at 30

mph, and the airspeed of the helicopter is 120 mph. How far out to sea can the helicopter fly

before it must return?

Solution: Let d be the distance out to sea that the helicopter can fly. Then we construct a table,

the top line is for flying out, the bottom is flying back.

The total time for the round trip is 5 hours, so

� � 5.

Multiplying both sides by the LCD 450, we get

450 � 450 � 5.

which simplifies to 3d � 5d � 2250, and adding like terms 8d � 2250. Now dividing by 8, we

see that the helicopter can fly 281.25 miles before returning.

example 9: A cyclist bikes at a constant speed for 25 miles. He then returns home at the same

speed but takes a different route. His return trip takes one hour longer and is 30 miles. Find his

speed. bSolution: Set up the following chart using the equation Distance � Rate � Time.

Solving the equation 25 � r � t for t, we get t � , and similarly, solving 30 � r � t yields

t � . We know the time returning is one hour longer, so � � 1 . Multiplying bothsides by the LCD, r, gives 30 − 25 � r or r � 5.

Distance RateDistance


Flying out d 150 d——150

Flying in d 90 d——90

d90

d150

( d150 � d

90)

Distance Rate (speed)Distance


Leaving home 25 miles r 25——r

Returning home 30 miles r 30——r

25r

25r

30r

30r


6.4 exercise Set

1) If it takes one hose 6 hours to fill a pool, and we are able to use a neighbor’s hose (which would

fill the pool at the same rate), how long would it take to fill the pool with both hoses?

2) If we must wait an hour to start using the neighbor’s hose, since they are watering their

tomatoes, how much time (from when we began filling the pool with our hose) will it take?

3) It is possible to use the “geometric method” to solve the following problem: One hose fills a

tank in 6 hours, but if a second hose is added, the tank can be filled in 4 hours. How long

would it take the second hose to fill the tank alone? Describe how you would do this, and then

carry your idea out to get the solution using the geometric method. Check your answer by

solving it the algebraic way. Discuss disadvantages of the geometric way.

4) One experienced nurse can assess the patients in a ward on a shift in 20 minutes, and if an in-

experienced nurse joins, they can do assessments in 12 minutes. How long would it take the

inexperienced nurse to assess the ward?

5) One IV is giving saline at a rate of 100 mL every 30 minutes while another is giving saline at

a rate of 100 mL every 20 minutes. If the flow of the IVs must stop before the patient receives

more than 100 mL of saline, how much time will elapse?

6) One IV is giving saline at a rate of 100 mL every 30 minutes while another is giving saline at

a rate of 100 mL every 20 minutes. If the flow of the IVs must stop before the patient receives

more than 400 mL of saline, how much time will elapse?

12


Note that there is more than one way to solve the GROUP problems.

(Group problem 1)

7) If a second IV (giving 100 mL every 30 minutes) is started 20 minutes after another IV

(giving 100 mL every 20 minutes), how many more minutes will pass until the patient

has received 400 mL?

(Group problem 2)

8) If the second IV (giving100 mL every 30 minutes) is started 10 minutes after the first

(giving 100 mL every 20 minutes), how many more minutes will pass until the patient

has received 400 mL?

(Group problem 3)

9) If an IV giving 100 mL every 20 minutes is started 10 minutes after an IV giving 100 mL

every 30 minutes, how many more minutes will pass until the patient has received 400 mL?

10) One nurse’s assistant can scrub and prepare an operating room in 40 minutes. How

long would it take three nurse’s assistants working together?

Note: Exam questions from this section will not be as tough as the GROUP problems!

11) Bill can mow his entire lawn in 2 hours. If his son helps him, they can get it done in an

hour and fifteen minutes. How long would it take his son to mow the grass alone?

12) In a large building, one ventilation system can clear the air in 7.5 hours. If a second

ventilation system is added, the air is cleared in 3 hours. How long would it take the

second system to clear the air alone?

13) A pontoon boat at full throttle can travel 15 miles upriver in the same time that it can

travel 25 miles downriver. If the speed of the river’s current is 2 mph, what is the full

throttle speed of the boat?


14) At the Seattle sewage treatment plant, one pipe can fill a settling tank in 6 hours while

another can fill the same tank in 9 hours. A drain can empty a full tank in 4 hours.

a) If both pipes are opened, how long does it take to fill the settling tank?

b) If both pipes are opened at midnight, what time is the settling tank filled?

c) If both pipes and the drain are opened at midnight, how many hours will it take to

fill the settling tank?

d) If both pipes are opened at midnight, and the drain is opened at 1 a.m., at what time

is the settling tank filled?

e) If the pipes are opened at midnight and the drain is opened at 1 a.m. but closed

again at 2 a.m., at what time is the settling tank filled?

15) A car travels 400 miles on level terrain in the same amount of time it travels 160 miles

on mountainous terrain. If the rate of the car is 30 miles per hour less in the mountains

than on level ground, find its rate in the mountains.

16) A boat moves 9 kilometers upstream in the same amount of time it moves 17 kilome-

ters downstream. If the rate of the current is 6 kilometers per hour, find the rate of the

boat in still water.

17) Jim can run 5 miles per hour on level ground on a non-windy day. One windy day, he

runs 14 miles with the wind, and in the same amount of time runs 5 miles against the

wind. What is the rate of the wind?

18) Five divided by the sum of a number and 5, minus the quotient of 3 and the difference

of the number and 5 is equal to 6 times the reciprocal of the difference of the number

squared and 25. What is the number?


6.5 adding and Subtracting rational expressions, Part 1

We learned in an earlier chapter how to add and subtract fractions. Fractions are a simple type of

rational expression where the numerator and denominator are numbers. The broader definition

of rational expressions includes fractions whose numerator and denominator can be polynomials.

The rules for adding and subtracting are identical; therefore, it might be a good idea to refresh

how you add and subtract fractions with the following examples:

You Try These

Add or subtract as indicated. Check with a neighbor when you are finished.

1) 2)

3) 4)

5) 6)

Problems 1, 2, and 3 were not as complicated since the fractions added or subtracted had the

same denominator. The rule for adding rational expressions is the same and is listed below:

The rule for subtracting is similar. There is a space for you to list that rule below:

In the examples below, note that they all have the same denominator or nearly so.

17 � 3

717 � 3

7

310 � 4

558 � 3

8

712 � 2

15310 � 4

5

AC � B

C � A � BC


example 1: Simplify

Solution: This simplifies to

example 2: Simplify

Solution: is the same as or, more simply,

This simplifies to When we factored �1 from the denominator of

the second fraction, we realized that each fraction had the same denominator.

example 3: Simplify

Solution: Since both fractions have the same denominator, we can add or subtract the numer-

ators to get Factoring the numerator gives us The factor x � 4 can be

cancelled from the numerator and denominator of the fraction using the fundamental principalof fractions from section 1 of this chapter. Recall from that section that the simplification isonly valid for x � 4. (Why?) This simplifies to 2.

example 4: Simplify

Solution: Since both fractions have the same denominator, we can add or subtract the numer-

ators to get Now factor the denominator to get Now, as long

as x � 3, the expression can be simplified to

example 5: Simplify Note: The fraction line is a group-

ing symbol like parentheses. This expression is the same as �

It is useful to write the expression this way so you do not forget to distribute the negative signthrough both terms.

Solution: Since we have a common denominator, we can write this expression as

Notice that I put parentheses around the (6x � 2) so I would re-

member that the subtraction sign affects both the 6x and the 2. Distribute through parentheses,

and we get Combine like terms to get Now, factor to

2xx � 3 � 5

x � 3.

2x � 5x � 3 .

2xx � 3 � 5

3 � x.

2xx � 3 � 5

�(x � 3)2x

x � 3 � 53 � x

2x � 5x � 3 .2x

x � 3 � 5x � 3.

2xx � 4 � 8

x � 4.

2(x � 4)x � 4 .2x � 8

x � 4 .

xx2 � x � 6

� 3x2 � x � 6

.

(x � 3)(x � 3) (x � 2) .x � 3

x2 � x � 6.

1x � 2.

3(3x � 2)6x2 � 13x � 28

� 6x � 26x2 � 13x � 28

.

3(3x � 2)6x2 � 13x � 28

(6x � 2)6x2 � 13x � 28

.

3(3x � 2) � (6x � 2)6x2 � 13x � 28

.

3x � 46x2 � 13x � 28.

9x � 6 � 6x � 26x2 � 13x � 28

.

SeCtION 6.5 n Adding�and�Subtracting�Rational�Expressions,�Part�1 381

get because these x values would make the de-

nominator equal to 0.

Practice the problems in the following 6.5 Exercise Set, which already have a common denomi-

nator, before going on to the problems in which we need to get the least common denominator.

(3x � 4)(3x � 4) (2x � 7) Æ 1

2x � 7; x � �43 , 72


6.5 exercise Set

Add or subtract as indicated, simplifying your answer whenever possible.

1) 2)

3) 4)

5) 6)

7) 8)

9) 10)

11) 12)

2y2 � 1

�y � 1y2 � 1

x2x2 � 8

� 22x2 � 8

3x2

x � 4 � 48x � 4

25x � 7

5x

xx2 � 9

� 3 � 2xx2 � 9

xx2 � 9

� 3x2 � 9

x � 4x2 � 3x � 4

� x � 1x2 � 3x � 4

3x � 42x2 � 7x � 15

� x � 12x2 � 7x � 15

4(x � 2y )9x2 � 16y2 �

x � 4y9x2 � 16y2

3x � 46x2 � 23x � 4

� 3 � 3x6x2 � 23x � 4

x � 46x2 � 13x � 8

�4(x � 3)

6x2 � 13x � 82x � 9

x2 � 6x � 8� x � 3

x2 � 6x � 8


6.6 adding and Subtracting rational expressions, Part 2

Adding and Subtracting Rational Expressions When the Denominators Are Different

Recall at the beginning of this section you were asked to add You cannot add these since

the denominators must be the same number. That’s an easy fix. We can convert to if we

multiply by 1 in the form Note yields Now the problem is really adding

with the result giving The following example is similar.

example 1: Add First, factoring, we get We

cannot add the fractions in the current form since they do not have the same denominator. We

can correct this by multiplying by 1 in the form the problem becomes

or Before continuing, note that we

multiplied by 1 in such a way to include the missing factor (x � 2) in the denominator. Now that

we have a common denominator, the addition can take place. The result is

simplifying to . It turns out that the numerator is one of the factors of the denominator,

since

More on the Process with Different Denominators

Recall at the beginning of this section that you also did the problem Here the least com-

mon denominator (LCD) was a large part of the problem. The LCD is found by factoring the num-bers 12 and 15. The next step is compiling the product of the prime factors as follows:

12: 22 � 3

15: 3 � 5

LCD: 22 � 3 � 5 or 60

310 � 4

5.

810

45

810.2

2 � 45

22.4

51110.3

10 � 810,

5x � 6(x � 2) (x � 2) � 4

(x � 2) .5x � 6x2 � 4

� 4x � 2.

x � 2x � 2;4

x � 2

5x � 6x2 � 4

�4(x � 2)x2 � 4

.5x � 6(x � 2) (x � 2) �

4(x � 2)(x � 2) (x � 2)

5x � 6 � 4(x � 2)x2 � 4

,

x � 2x2 � 4

x � 2x2 � 4

Æ(x � 2)

(x � 2) (x � 2) Æ 1x � 2.

712 � 2

15.


Each fraction can then be renamed as a fraction with the LCD as the denominator by multi-

plying it by 1. See that is missing the factor 5 in its denominator? Multiply it by to get

See that is missing the factor 4 in its denominator? Multiply it by to get The problem

can now be written , so the answer is simplifying to

The next example is much like the problem we just did.

example 2: Subtract the fractions

Solution: The factored form of the problem is The LCD can

be seen to be x(x � 2)(x � 3). Multiplying the fraction by and the fraction

by we get Now that the denomi-

nators are the same, we can write the solution as which simplifies to

You try the following problems.

You Try These

1) 2)

3560.5

57

128

60.44

215

920.27

60,3560 � 8

60712 � 2

15

xx2 � 5x � 6

� 2x2 � 2x

.

x(x � 2) (x � 3) � 2

x(x � 2) .xx

x(x � 2) (x � 3)

x2

x(x � 2) (x � 3) �2(x � 3)

x(x � 2) (x � 3) .x � 3x � 3,2

x(x � 2)

x2 � 2(x � 3)x(x � 2) (x � 3) ,

x2 � 2x � 6x(x � 2) (x � 3) .

6aa2 � a � 6

� 7aa2 � 7a � 10

3xx2 � 7x � 10

� 2x2 � 5x


6.6 exercise Set

Add or subtract the algebraic fractions as indicated.

1) 2)

3) 4)

5) 6)

7) 8)

9)

Note that problems 7 and 8 yield a “quick method” of simplification if the denominators are“relatively prime.” (That is, they have no common factors). Roughly said, this formula,

works as problems 7 and 8 show.

10) 11)

12) 13)

14) 15)

16) 17)

18)

19)

20) 21)

x2x � 5 � x � 4

2x � 353x � 2

9x2

x � 2x � 3 � x2 � 2x

x2 � 92y � 5y � 2 �

4y � 12y2 � y � 6

7x2 � 5x � 6

� 2x2 � 1

x � 3x � 2 � x

x2 � 4

ab � c

dab � c

d

2x2 � 4

� 3x � 2

ab�c

d Æ ad�bcbd ,

2x � 9x2 � 6x � 8

� x � 3x2 � 2x � 8

�4x � 1 � 2x

1 � x2

3x � 84x2 � 4x � 1

� x � 34x2 � 1

2x � 3x2 � 3x � 4

� 3x � 2x2 � 16

6x � y �

4 � yy � x

5x � 22x2 � 9x � 9

� x � 76x2 � 13x � 6

x � 2x2 � 25

� 3xx2 � 15x � 50

62x2 � 5x � 2

� 52x2 � 3x � 2

3x2

x2 � 4x� 2x � 1

x2 � 10x � 24

2x(20x2 � 5)

� 1(2x � 1) � 2

(12x2 � 12x � 3)

x � 2 � 3x � 5(x � 1)�12

6x2 � 22x � 12� 5x

9x2 � 4


6.7 Complex rational expressions

Do not let the word “complex” make this sound difficult, as it is not. Before discussing the concept,

examine the following problem.

example 1: If you go on a round trip in an ambulance, and the trip is x miles, you drive there

averaging 60 mph and return averaging 40 mph. What was your average speed for the trip? Hint:

it is not 50 mph.

Solution: from d � r � t, we get so the two parts of the trip have elapsed time and

The distance traveled for the entire trip is x � x or 2x miles. If we let r be the average rate

for the trip, then from d � r � t, we get and solving this for r by di-

viding both sides by we get the expression A simplified form

of the left side of this equation will give us the average rate. Note that the expression is a frac-tion that contains fractions. Hence, it is called a complex fraction. This section of the chapterwill discuss how such expressions can be simplified.

Method 1: (Best used when the denominator contains the sum or difference of fractions.)

Multiply both the numerator and denominator of the complex fraction by the least common

denominator (LCD) of all of the “inner” fractions. This is a legitimate operation since we are

really multiplying by 1 in a disguised form.

Method 2: (Best used when the denominator is a single fraction.) If the denominator is a single

fraction (or has been made into one by addition), then the rule for dividing by a fraction applies:

“invert the denominator and multiply it by the numerator.” (See section 2 of this chapter).

Solving Example 1 Using Method 1: The expression can be simplified by multiply-

ing the numerator and denominator by 120: The

average speed for the round trip in the example above is 48 miles per hour.

Solving Example 1 Using Method 2: The expression can be simplified by adding

the denominator, getting a result of , and then inverting the denominator

and multiplying getting the same result.

Try an appropriate method on each of these examples.

x60t � d

r ,

x40.

2x � r � ( x60 � x

40) ;

2xx

60 � x40

� r.( x60 � x

40) ,

2xx

60 � x40

2xx

60 � x40

� 120120 Æ 240x

2x � 3x Æ 240x5x Æ 48.

2xx

60 � x40

2x2x

120 � 3x120

Æ 2x5x

120

2x � 1205x Æ 48,

SeCtION 6.7 n Complex�Rational�Expressions 387

You Try These

1) 2)

3) 4)

5)

example: Simplify the following complex fraction: .

Solution: Rewrite as .

Multiply the numerator and denominator by the LCD, x2y2 and we get

.

This simplifies to .

9x3

6x2

13 � 1

256 � 2

3

x � 1x

x2 � 1

ab � 2

ab � 4b

a

1x � 11 � x2

x

x�2 � y�2

y�2 � x�2

1x2 � 1

y2

1y2 � 1

x2

x�2 � y�2

y�2 � x�2

[1x2 � 1

y2

1y2 � 1

x2 ] [x2y2

1x2y2

1]

y2 � x2

x2 � y2


6.7 exercise Set

Simplify the following complex fractions.

1) 2)

3) 4)

5) 6)

7) 8)

9)

10) In electronics, the total resistance, R, in a circuit containing parallel resistors r1 and r2

(see picture below) is given by the formula Simplify the right side.

x2 � 1x

(x � 1)2

x

x � 11 � 1

x

1 � 5x

1 � 25x2

1x � 2 � 1

x2

tt � 3

t2

3

54 � 1

638 � 5

12

1x � h � 1

xh

x�1 � 2�1

x�2 � 2�2

1 � 12

2 � 12

R � 11r1

� 1r2

.

r1

r2

SeCtION 6.7 n Complex�Rational�Expressions 389

6.8 Graphs of rational Functions

Any function that has integer exponents is a rational function. Really, linear functions are a type

of rational function. However, in this section we will be studying rational functions that have a

variable to a positive exponent in the denominator.

example 1a: Graph the function f(x) � .

Solution: We will plot some points to get an idea of the shape of this graph.

These points are plotted below.

1x


x y or f(x)

�10 f(�10) � Æ or �0.11�10

�110

�3 f(�3) � Æ or �0.31�3

�13

�2 f(�2) � Æ or �0.51�2

�12

�1 f(�1) � Æ or �11�1

�11

0 f(0) � Æ undefined10

1 f(1) � Æ 111

2 f(2) � Æ 0.512

3 f(3) � Æ 0.313

10 f(10) � Æ 0.1110

We can sketch the graph on each side of the y-axis.

Notice that the farther the graph goes to the right the closer the graph gets to the y-axis. For

example, if we evaluate the function f(x) � at x � 100, we would get a y-value of or

0.01. If we evaluate the function at x � 1,000, we would get a y-value of or 0.001. The

same is true on the left side of the y-axis. If we evaluate the function at x � �100, we would

get a y-value of or �0.01. If we evaluate the function at x � �1,000, we would get a

y-value of or �0.001.

As the x values approach infinity (get larger, farther to the right), the y values are getting closer

to 0.

As the x values approach negative infinity (farther to the left), the y values are getting closer

to 0.

example 1b: Does the graph of the previous function f(x) � appear to have a horizontal as-

ymptote and if so, what is the horizontal asymptote?

Solution: Notice that the graph seems to level out on the far left and far right. This is a good

indication that this graph has a horizontal asymptote. The larger the values we put in for x,

the closer the x values get to the x-axis or y � 0. (Note: The graph of y � 0 is the x-axis).

1100

1x

11,000

1�100

1�1,000

horizontal asymptote

A horizontal asymptote is a horizontal line that the graph gets closer to as

the x values approach infinity or negative infinity.

1x

SeCtION 6.8 n Graphs�of�Rational�Expressions 391

example 1c: What is happening to the function f(x) � as the x values get close to 0?

Solution: If we evaluate the function at x � 0, we get f(0) � which is undefined. This means

that there is no y-value for the x-value of 0. Therefore the left side of the graph will not connect

to the right side of the graph. To get an idea of what the graph looks like near x � 0, we will

need to plot points close to 0.

Notice on the right side of the graph, as the x values get closer to 0 the y values are getting larger.

We can see that on the left side of the graph, as the x values get closer to 0 the y values are getting

larger in the negative direction.

1x

10


x y or f(x)

�12 f � Æ 1 � or �2(�1

2 ) 2�1

1�12

�14 f � Æ 1 � or �4(�1

4 ) 4�1

1�14

�110 f � �10(�1

10 )0 f(0) � Æ undefined1

0

110 f � 10( 1

10)14 f � 4( 1

4)12 f � 2( 1

2)

If we would evaluate the function f(x) � for positive values getting closer and closer to 0, the

y values would get larger and larger without bound. For example, f � Æ 1,000.

If we would evaluate the function f(x) � for negative values getting closer and closer to 0, they values would get larger and larger without bound in the negative direction.

For example, f � Æ �1,000.

The function f(x) � has a vertical asymptote at x � 0. (Note: x � 0 is the y-axis.)

example 1d: What is the domain of f(x) � ?

Solution: The only value for x that does not evaluate to a “real” number is 0. This is because

of the fact that f(0) � and division by 0 is undefined. Therefore the domain is all Real num-

bers except 0. We could also write the domain as all x values such that x � 0. In set notationwe would write {x | x � 0}.

example 1e: What is the range of f(x) � ?

Solution: When we are checking on the range, we need to see how high and low the graph

goes and if there is any y value that the function does not pass through. Looking at the graph,

we can see the y values approach negative infinity as the x values get closer to 0 from the left.

Vertical asymptote

A vertical asymptote is a vertical line that the graph gets closer to in the following manner:

As the x values get closer to a particular value,

the y values approach infinity or negative infinity.

1x

11

1,000( 1

1,000)

1x

1�1

1,000( �1

1,000)

1x

Domain

The domain is the set of all x values (input values) that result in y-values (output values)

that are real numbers.

range

The range is the set of all y values (output values) that

are possible for a particular function.

1x

10

1x


We can also see that the y values approach positive infinity as the x values get closer to 0 from

the right. That would make us think that the range is all real numbers. However, there is a

horizontal asymptote at y � 0 and the graph never passes through the x-axis (y � 0). There-

fore, the range is all real numbers except 0. We could write the range as all y values such that

y � 0. In set notation we would write {y|y � 0}.

You Try These

1) For the function f(x) � � 5 determine the following: the location of any hori-

zontal asymptote, location of any vertical asymptote, and the domain and range and

graph the function.

2) For the function f(x) � − 2 determine the following: the location of any horizon-

tal asymptote, location of any vertical asymptote, and the domain and range and graph

the function.

3) What is the location of the horizontal asymptote, location of the vertical asymptote, and

the domain and range of functions in the form f(x) � � � k?

4) For the function f(x) � � 5 determine the following: the location of any hori-

zontal asymptote, location of any vertical asymptote, and the domain and range andgraph the function.

3x � 2

�2x � 4

ax � h

1(x � 3)2


6.8 exercise Set

For problems 1 through 8 give the location of any horizontal or vertical asymptote and give the

domain and range of the function.

1) f(x) � � 3 2) f(x) � � 5

3) f(x) � � 5 4) f(x) � � 1

5) f(x) � 6) f(x) � � 5

7) f(x) � 8) f(x) � � 2

9) The cost in millions of dollars for the federal government to seize p percent of a certain ille-

gal drug as it enters the country is c(p) � where p is the percent and c is the cost.

a) Find the cost of seizing 45%. (Use 45 not 0.45 for your value of p.)

b) What percentage of illegal drugs would be removed if 36 million dollars could be allo-

cated to the cost of removing illegal drugs?

c) Find the cost of seizing 75%.

d) What percentage of illegal drugs would be removed if 150 million dollars could be allo-

cated to the cost of removing illegal drugs?

e) Find the cost of seizing 90%.

f) Find the cost of seizing 95%.

g) Find the cost of seizing 99%.

h) Find the cost of seizing 99.9%.

i) According to this model, would it be possible to seize 100% of the drug?

2x � 1

5x � 4

13x � 10

�12x � 1

�12x � 2

1x � 100

�3(x � 1)2

2(x � 1)2

45p100 � p


j) Does this function have a vertical asymptote? If so, what is its location?

k) What is the meaning of the vertical asymptote in terms of this problem?

10) The cost in dollars of removing p percent of the air pollutants in the stack emission of a

utility company that burns coal to generate electricity is c(p) � .

a) Find the cost of removing 50% of air pollutants.

b) What percentage of air pollutants would be removed if $70,000 could be allocated to the

cost of removing air pollutants?

c) What percentage of air pollutants would be removed if $1,000,000 could be allocated to

the cost of removing air pollutants?

d) Find the cost of removing 90% of air pollutants.

e) Find the cost of removing 95% of air pollutants.

f) Find the cost of removing 99% of air pollutants.

g) Find the cost of removing 99.9% of air pollutants.

h) According to this model, would it be possible to remove 100% of the pollutants?

Explain.

i) Does this function have a vertical asymptote? If so, what is its location?

j) What is the meaning of the vertical asymptote in terms of this problem?

65,000p100 � p


review Problems for Chapter 6

Simplify each expression.

1) 2)

3)

Multiply or divide as indicated in the following:

4) � 5) �

Add or subtract as indicated in the following:

6) 7)

Solve the equations.

8) � − 9) � �

Solve each problem.

10) It takes one hose 20 hours to fill a fish pond, while it takes another hose 30 hours.

a) How long would it take if both hoses are used?

b) If the 20 hour hose starts at midnight and the 30 hour hose starts at 1:00 a.m., at what

time is the pond full?

11) The intensity of light (measured in lumens) varies inversely as the square of the distance

from the light source. If a light bulb is measured at 10 lumens when it is 2 feet from the

light source, how many lumens will be measured if it is 9 feet from the light source?

x3 � 4xx2 � 3x � 2

2x � 6x2 � 3x

x � 4x2 � 5x � 4

x2 � 10x � 16x2 � x � 6

x2 � 5x � 24x2 � x � 12

25yz2

18x3y43x2

50y3z

3x2 � 2x � 8

� 2x2 � 3x � 2

9x � 76x2 � 3x � 4

6x2

y � 3y � 1

2y � 164y � 4

y2y � 2

12

2x � 1 � 1

x � 2

reVIeW PrOBLeMS FOr ChaPter 6 397

Simplify the complex fractions.

12) 13)

14)

15) If an ambulance drives to a distant hospital at 100 mph and then returns at 60 mph, what is

the average speed of the round trip?

�5x2

15�x4

34 � 1

423 � 1

2

1x2 � 1

� 2x2 � 2x � 1

3x � 1 � 4

x � 1


Documents

Ra a E a E a · Solution: Factor the expression, and we get The x divides out to give us Since the x divided out of the numerator and denominator, we would get when x 0. Therefore,