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3: Adding and Subtracting Algebraic Fractions

3: Adding and Subtracting Algebraic Fractions

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7.3 • Adding and Subtracting Algebraic Fractions



279



Definition 7.3

If



A

C

and are rational expressions with B ϶ 0, then

B

B



A

C

AϩC

ϩ ϭ

B

B

B



and



C

AϪC

A

Ϫ ϭ

B

B

B



Thus if the denominators of two algebraic fractions are the same, we can add or subtract the

fractions by adding or subtracting the numerators and placing the result over the common

denominator. Here are some examples:

5

7

5ϩ7

12

ϩ ϭ

ϭ

x

x

x

x

8

3

8Ϫ3

5

Ϫ

ϭ

ϭ

xy

xy

xy

xy

14

15

14 ϩ 15

29

ϩ

ϭ

ϭ

2x ϩ 1

2x ϩ 1

2x ϩ 1

2x ϩ 1

3

4

3Ϫ4

Ϫ1

or

Ϫ

ϭ

ϭ

aϪ1

aϪ1

aϪ1

aϪ1



Ϫ



1

aϪ1



In the next examples, notice how we put to use our previous work with simplifying

polynomials.

1x ϩ 32 ϩ 12x Ϫ 32

xϩ3

2x Ϫ 3

3x

ϩ

ϭ

ϭ

4

4

4

4

1x

ϩ

52

Ϫ

1x

ϩ

22

xϩ5

xϩ2

xϩ5ϪxϪ2

3

Ϫ

ϭ

ϭ

ϭ

7

7

7

7

7

13x ϩ 12 ϩ 12x ϩ 32

3x ϩ 1

2x ϩ 3

5x ϩ 4

ϩ

ϭ

ϭ

xy

xy

xy

xy

213n ϩ 12

31n Ϫ 12

213n ϩ 12 Ϫ 31n Ϫ 12

6n ϩ 2 Ϫ 3n ϩ 3

3n ϩ 5

Ϫ

ϭ

ϭ

ϭ

n

n

n

n

n



It may be necessary to simplify the fraction that results from adding or subtracting two

fractions.

14x Ϫ 32 ϩ 12x ϩ 32

4x Ϫ 3

2x ϩ 3

6x

3x

ϩ

ϭ

ϭ

ϭ

8

8

8

8

4

13n

Ϫ

12

Ϫ

1n

Ϫ

52

3n Ϫ 1

nϪ5

3n Ϫ 1 Ϫ n ϩ 5

Ϫ

ϭ

ϭ

12

12

12

12

ϭ



21n ϩ 22

2n ϩ 4

nϩ2

ϭ

ϭ

12

12

6



1Ϫ2x ϩ 32 ϩ 13x Ϫ 12

Ϫ2x ϩ 3

3x Ϫ 1

xϩ2

ϩ 2

ϭ

ϭ 2

2

x Ϫ4

x Ϫ4

x2 Ϫ 4

x Ϫ4

ϭ



xϩ2

1x ϩ 221x Ϫ 22



ϭ



1

xϪ2



Recall that to add or subtract rational numbers with different denominators, we first

change them to equivalent fractions that have a common denominator. In fact, we found that

by using the least common denominator (LCD), our work was easier. Let’s carefully review

the process because it will also work with algebraic fractions in general.



280



Chapter 7 • Algebraic Fractions



Classroom Example

3

5

Add ϩ .

6

7



EXAMPLE 1



Add



3

1

ϩ .

5

4



Solution

By inspection, we see that the LCD is 20. Thus we can change both fractions to equivalent

fractions that have a denominator of 20.

3

1

3 4

1 5

12

5

17

ϩ ϭ a bϩ a bϭ

ϩ

ϭ

5

4

5 4

4 5

20

20

20

Form

of 1



Classroom Example

7

2

Subtract

Ϫ .

12

15



Form

of 1



EXAMPLE 2



Subtract



5

7

Ϫ .

18

24



Solution

If we cannot find the LCD by inspection, then we can use the prime factorization forms.



и3и3

24 ϭ 2 и 2 и 2 и 3

18 ϭ 2



v ¡ LCD ϭ 2



и 2 и 2 и 3 и 3 ϭ 72



5

7

5 4

7 3

20

21

1

Ϫ

ϭ

a bϪ

a bϭ

Ϫ

ϭϪ

18

24

18 4

24 3

72

72

72

Now let’s consider adding and subtracting algebraic fractions with different denominators.



Classroom Example

4m ϩ 3

mϩ1

Add

ϩ

.

5

6



EXAMPLE 3



Add



xϪ2

3x ϩ 1

ϩ

.

4

3



Solution

By inspection, we see that the LCD is 12.

xϪ2

3x ϩ 1

xϪ2

3

3x ϩ 1

4

ϩ

ϭ a

ba b ϩ a

ba b

4

3

4

3

3

4

ϭ

ϭ

ϭ



Classroom Example

xϪ8

xϪ4

Ϫ

.

Subtract

2

10



EXAMPLE 4



31x Ϫ 22



413x ϩ 12



ϩ



12



12



31x Ϫ 22 ϩ 413x ϩ 12

12

3x Ϫ 6 ϩ 12x ϩ 4

15x Ϫ 2

ϭ

12

12



Subtract



nϪ2

nϪ6

Ϫ

.

2

6



Solution

By inspection, we see that the LCD is 6.

nϪ2

nϪ6

nϪ2

3

nϪ6

Ϫ

ϭ a

ba b Ϫ

2

6

2

3

6

ϭ



31n Ϫ 22

6



Ϫ



1n Ϫ 62

6



7.3 • Adding and Subtracting Algebraic Fractions



ϭ



281



31n Ϫ 22 Ϫ 1n Ϫ 62



6

3n Ϫ 6 Ϫ n ϩ 6

ϭ

6

2n

6

n

ϭ

3

ϭ



Don’t forget to simplify!



It does not create any serious difficulties when the denominators contain variables; our

approach remains basically the same.

Classroom Example

2

5

Add

ϩ .

5n

6n



EXAMPLE 5



Add



3

7

ϩ .

4x

3x



Solution

By inspection, we see that the LCD is 12x.

3

7

3 3

7 4

9

28

9 ϩ 28

37

ϩ

ϭ a bϩ a bϭ

ϩ

ϭ

ϭ

4x

3x

4x 3

3x 4

12x

12x

12x

12x



Classroom Example

19

15

Subtract

Ϫ

.

24a

32a



EXAMPLE 6



Subtract



11

5

.

Ϫ

12x

14x



Solution

12x ϭ 2

14x ϭ 2



и 2 и 3 и xv

и7иx



¡ LCD ϭ 2



и 2 и 3 и 7 и x ϭ 84x



11

5

11 7

5 6

Ϫ

ϭ

a bϪ

a b

12x

14x

12x 7

14x 6

77

30

77 Ϫ 30

47

ϭ

Ϫ

ϭ

ϭ

84x

84x

84x

84x



Classroom Example

8

3

Add ϩ

.

x

xϩ5



EXAMPLE 7



Add



4

2

ϩ

.

y

yϪ2



Solution

By inspection, we see that the LCD is y1y Ϫ 22 .

y

2

4

2 yϪ2

4

ϩ

ϭ a



a b

y

y

y

yϪ2

yϪ2

yϪ2

Form

of 1



ϭ

ϭ

ϭ



21y Ϫ 22

y1y Ϫ 22



Form

of 1



ϩ



4y

y1y Ϫ 22



21y Ϫ 22 ϩ 4y

y1y Ϫ 22

2y Ϫ 4 ϩ 4y

6y Ϫ 4

ϭ

y1y Ϫ 22

y1y Ϫ 22



282



Chapter 7 • Algebraic Fractions



Notice the final result in Example 7. The numerator, 6y Ϫ 4, can be factored into 2(3y Ϫ 2).

However, because this produces no common factors with the denominator, the fraction can

6y Ϫ 4

not be simplified. Thus the final answer can be left as

; it is also acceptable to express

y1y Ϫ 22

213y Ϫ 22

it as

.

y1y Ϫ 22

Classroom Example

6

3

Subtract

Ϫ

.

xϩ7

xϪ2



EXAMPLE 8



Subtract



4

7

Ϫ

.

xϩ2

xϩ3



Solution

By inspection, we see that the LCD is 1x ϩ 221x ϩ 32 .

4

7

4

xϩ3

7

xϩ2

Ϫ

ϭ a

ba

bϪ a

ba

b

xϩ2

xϩ3

xϩ2

xϩ3

xϩ3

xϩ2

ϭ

ϭ



41x ϩ 32



1x ϩ 221x ϩ 32



Ϫ



71x ϩ 22



1x ϩ 321x ϩ 22



41x ϩ 32 Ϫ 71x ϩ 22

1x ϩ 221x ϩ 32



ϭ



4x ϩ 12 Ϫ 7x Ϫ 14

1x ϩ 221x ϩ 32



ϭ



Ϫ3x Ϫ 2

1x ϩ 221x ϩ 32



Concept Quiz 7.3

For Problems 1–10, answer true or false.

2x

1

2x ϩ 1

1. The addition problem

is equal to

for all values of x except

ϩ

xϩ4

xϩ4

xϩ4

1

x ϭ Ϫ and x ϭ Ϫ4.

2

2. Any common denominator can be used to add rational expressions, but typically we

use the least common denominator.

10x2z

2x2

and

are equivalent fractions.

3y

15yz

4. The least common multiple of the denominators is always the lowest common

denominator.

3. The fractions



5

3

5. To simplify the expression

we could use 2x Ϫ 1 for the common

ϩ

2x Ϫ 1

1 Ϫ 2x

denominator.

1

5

3

2

, then

ϩ

ϭ

.

2

2x Ϫ 1

1 Ϫ 2x

2x Ϫ 1

Ϫ2

17

3

Ϫ

ϭ

Ϫ4

3

12

4x Ϫ 1

2x ϩ 1

x

ϩ

ϭ

5

6

5

x

3x

5x

5x

Ϫ

ϩ

ϭ

4

2

3

12

2

3

Ϫ5 Ϫ 6x

If x ϶ 0, then

.

Ϫ

Ϫ1ϭ

3x

2x

6x



6. If x ϶

7.

8.

9.

10.



7.3 • Adding and Subtracting Algebraic Fractions



Problem Set 7.3

For Problems 1–34, add or subtract as indicated. Be sure to

express your answers in simplest form. (Objective 1)



39.



y

3y

ϩ

6

4



40.



3y

7y

ϩ

4

5



1.



5

12

ϩ

x

x



2.



17

13

Ϫ

x

x



41.



8x

3x

Ϫ

3

7



42.



5y

3y

Ϫ

6

8



3.



7

5

Ϫ

3x

3x



4.



4

3

ϩ

5x

5x



43.



2x

3x

ϩ

6

5



44.



6x

7x

ϩ

9

12



5.



7

1

ϩ

2n

2n



6.



5

4

ϩ

3n

3n



45.



7n

3n

Ϫ

8

9



46.



8n

7n

Ϫ

10

15



7.



9

13

Ϫ 2

2

4x

4x



8.



12

22

Ϫ 2

2

5x

5x



47.



48.



10.



xϪ2

4

ϩ

x

x



xϪ2

xϩ1

ϩ

5

6



9.



xϩ1

3

ϩ

x

x



xϩ3

xϪ4

ϩ

5

2



49.



xϪ6

xϩ2

ϩ

9

3



50.



xϪ2

xϩ4

ϩ

4

8



11.



3

6

Ϫ

xϪ1

xϪ1



12.



8

10

Ϫ

xϩ4

xϩ4



51.



52.



14.



2x ϩ 3

3

Ϫ

x

x



2n ϩ 3

4n Ϫ 1

ϩ

4

7



13.



xϩ1

1

Ϫ

x

x



3n Ϫ 1 2n ϩ 5

ϩ

3

4



53.



54.



16.



4t Ϫ 1

8t Ϫ 5

ϩ

7

7



5n Ϫ 2

4n ϩ 7

Ϫ

12

6



15.



3t Ϫ 1

2t ϩ 3

ϩ

4

4



4n Ϫ 3

3n ϩ 5

Ϫ

6

18



55.



56.



18.



9a Ϫ 1

4a Ϫ 2

Ϫ

6

6



5x

3x

7x

Ϫ

Ϫ

2

4

6



17.



7a ϩ 2

4a Ϫ 6

Ϫ

3

3



3x

x

5x

ϩ Ϫ

4

6

8



57.



58.



20.



2n Ϫ 5

6n Ϫ 1

ϩ

10

10



4x 5 11x

ϩ Ϫ

3

9

6



19.



4n ϩ 3

6n ϩ 5

ϩ

8

8



x

3

7x

Ϫ

Ϫ

5

10

12



59.



60.



22.



2n Ϫ 6

7n Ϫ 1

Ϫ

5

5



7

5

ϩ

8x

12x



21.



3n Ϫ 7

9n Ϫ 1

Ϫ

6

6



5

1

ϩ

8x

6x



61.



62.



24.



4x ϩ 1

2x ϩ 5

Ϫ

3x

3x



11

8

Ϫ

9y

15y



23.



5x Ϫ 2

8x ϩ 3

Ϫ

7x

7x



5

7

Ϫ

6y

9y



63.



5

11

Ϫ

12x

16x2



64.



4

7

Ϫ 2

9x

6x



65.



3

2

5

Ϫ

ϩ

2x

3x

4x



66.



3

5

10

Ϫ

ϩ

4x

6x

9x



3n



67.



68.



11x Ϫ 9

8x2



3

7

ϩ

x

xϪ5



4

9

ϩ

x

xϪ8



69.



2

3

Ϫ

n

nϪ1



70.



5

7

Ϫ

n

nϩ3



71.



4

6

Ϫ

n

nϩ4



72.



8

3

Ϫ

n

nϪ9



73.



6

12

Ϫ

x

2x ϩ 1



74.



2

6

Ϫ

x

3x Ϫ 2



75.



4

6

ϩ

xϩ4 xϪ3



76.



7

8

ϩ

xϪ2

xϩ1



25.

27.

29.



31x ϩ 22

4x

61n Ϫ 12

3n

213x Ϫ 42

2



7x



ϩ

ϩ

Ϫ



61x Ϫ 12

4x

31n ϩ 22

3n

7xϪ 8

7x2



26.

28.

30.



41x Ϫ 32

5x

21n Ϫ 42

3n



ϩ

ϩ



314x Ϫ 32

8x2



21x ϩ 62

5x

41n ϩ 22



Ϫ



31.



a2

4

Ϫ

aϩ2

aϩ2



32.



n2

16

Ϫ

nϪ4

nϪ4



33.



3x

18

Ϫ

2

1x Ϫ 62

1xϪ62 2



34.



x2 ϩ 5x

4

ϩ

2

1x ϩ 12

1x ϩ 12 2



For Problems 35–80, add or subtract as indicated and express

your answers in simplest form. (Objective 3)

35.



3x

5x

ϩ

8

4



36.



5x

2x

ϩ

3

9



77.



3

9

Ϫ

xϪ2 xϩ1



78.



4

5

Ϫ

xϪ1

xϩ6



37.



7n

4n

Ϫ

12

3



38.



n

7n

Ϫ

6

12



79.



3

4

Ϫ

2x Ϫ 1

3x ϩ 1



80.



6

4

Ϫ

3x Ϫ 4

2x ϩ 3



283



284



Chapter 7 • Algebraic Fractions



Thoughts Into Words

81. Give a step-by-step description of how to do this addition problem:



83. Suppose that your friend does an addition problem as

follows:

51122 ϩ 8172

5

7

60 ϩ 56

116

29

ϩ

ϭ

ϭ

ϭ

ϭ

8

12

81122

96

96

24



3x Ϫ 1

2x ϩ 3

ϩ

6

9

3

3

and

opposites? What should

xϪ2

2Ϫx

3

3

be the result of adding

and

?

xϪ2

2Ϫx



82. Why are



Is this answer correct? What advice would you offer

your friend?



Further Investigations

9

4

ϩ

. Notice

xϪ2

2Ϫx

that the denominators x Ϫ 2 and 2 Ϫ x are opposites; that is,

Ϫ1(2 Ϫ x) = (x Ϫ 2). In such cases, add the fractions as

follows:

Consider the addition problem



9

4

9

4

Ϫ1

ϩ

ϭ

ϩ

a

b

xϪ2

2Ϫx

xϪ2

2 Ϫ x Ϫ1

9

Ϫ4

ϭ

ϩ

xϪ2

xϪ2

ϭ



9 ϩ 1Ϫ42

xϪ2



Form

of 1



5

ϭ

xϪ2



Answers to the Concept Quiz

1. False

2. True

3. True

4. True



7.4



5. True



For Problems 84–89, use this approach to help with the

additions and subtractions.

84.



7

2

Ϫ

xϪ1

1Ϫx



85.



5

1

ϩ

xϪ3

3Ϫx



86.



x

4

ϩ

xϪ4

4Ϫx



87.



Ϫ4

2

ϩ

aϪ1

1Ϫa



88.



2

1

3

ϩ

Ϫ

xϩ3

3Ϫx

x Ϫ9



89.



n

3

Ϫ

2n Ϫ 1

1 Ϫ 2n



2



6. True



7. False



8. False



9. True



10. True



Addition and Subtraction of Algebraic Fractions and

Simplifying Complex Fractions



OBJECTIVES



1



Add or subtract rational expressions for which the denominators can

be factored



2



Simplify complex fractions



In this section, we expand our work with adding and subtracting rational expressions, and we

discuss the process of simplifying complex fractions. Before we begin, however, this seems

like an appropriate time to offer a bit of advice regarding your study of algebra. Success in

algebra depends on having a good understanding of the concepts as well as being able to perform the various computations. As for the computational work, you should adopt a carefully

organized format that shows as many steps as you need in order to minimize the chances of

making careless errors. Don’t be eager to find shortcuts for certain computations before you

have a thorough understanding of the steps involved in the process. This advice is especially

appropriate at the beginning of this section.



7.4 • Addition and Subtraction of Algebraic Fractions and Simplifying Complex Fractions



285



Study Examples 1–4 very carefully. Note that the same basic procedure is followed in

solving each problem:

Step 1 Factor the denominators.

Step 2 Find the LCD.

Step 3 Change each fraction to an equivalent fraction that has the LCD as its denominator.

Step 4 Combine the numerators and place over the LCD.

Step 5 Simplify by performing the addition or subtraction.

Step 6 Look for ways to reduce the resulting fraction.

Classroom Example

4

7

ϩ .

Add 2

z

z ϩ 5z



EXAMPLE 1



Add



3

5

ϩ .

x

x ϩ 2x

2



Solution

1st denominator: x2 ϩ 2x ϭ x1x ϩ 22 v ¡ LCD is x(x ϩ 2)

2nd denominator: x

3

5

3

5 xϩ2

ϩ ϭ

ϩ a

b

x

x xϩ2

x1x ϩ 22

x ϩ 2x

2



This fraction has

the LCD as its

denominator



Classroom Example

1

8

Ϫ

.

Subtract 2

xϪ4

x Ϫ 16



Form

of 1



ϭ



51x ϩ 22

3 ϩ 51x ϩ 22

3

ϩ

ϭ

x1x ϩ 22

x1x ϩ 22

x1x ϩ 22



ϭ



3 ϩ 5x ϩ 10

5x ϩ 13

ϭ

x1x ϩ 22

x1x ϩ 22



EXAMPLE 2



Subtract



4

1

.

Ϫ

x

Ϫ

2

x Ϫ4

2



Solution

x2 Ϫ 4 ϭ 1x ϩ 221x Ϫ 22

xϪ2ϭxϪ2



v ¡ LCD is (x ϩ 2)(x Ϫ 2)



4

1

4

1

xϩ2

Ϫ

ϭ

Ϫ a

ba

b

xϪ2

1x ϩ 221x Ϫ 22

xϪ2

xϩ2

x Ϫ4

2



ϭ

ϭ

ϭ

ϭ



11x ϩ 22

4

Ϫ

1x ϩ 221x Ϫ 22

1x ϩ 221x Ϫ 22

4 Ϫ 11x ϩ 22



1x ϩ 221x Ϫ 22

Ϫx ϩ 2

1x ϩ 221x Ϫ 22

Ϫ11x Ϫ 22



1x ϩ 221x Ϫ 22



ϭϪ



1

xϩ2



ϭ



4ϪxϪ2

1x ϩ 221x Ϫ 22



Note the changing of Ϫx ϩ 2 to Ϫ1(x Ϫ 2)



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