chapter 6 rational expressions § 6.1 rational functions and multiplying and dividing rational...

Chapter 6

Rational Expressions

§ 6.1

Rational Functions and Multiplying and Dividing


Martin-Gay, Intermediate Algebra, 5ed 33


Q

PRational expressions can be written in the form where P and Q are both polynomials and Q 0.

Examples of Rational Expressions

54

423 2

x

xx22 432

34

yxyx

yx

4

3 2x


To evaluate a rational expression for a particular value(s), substitute the replacement value(s) into the rational expression and simplify the result.

Evaluating Rational Expressions

Example:

Evaluate the following expression for y = 2.

y

y

5

2)

2 2( 25

7

4

7

4


In the previous example, what would happen if we tried to evaluate the rational expression for y = 5?

y

y

5

2 5 25 5 0

3

This expression is undefined!

Evaluating Rational Expressions


We have to be able to determine when a rational expression is undefined.

A rational expression is undefined when the denominator is equal to zero.

The numerator being equal to zero is okay (the rational expression simply equals zero).

Undefined Rational Expressions


Find any real numbers that make the following rational expression undefined.

4515

49 3

x

xx

The expression is undefined when 15x + 45 = 0.

So the expression is undefined when x = 3.

Undefined Rational Expressions

Example:


Simplifying a rational expression means writing it in lowest terms or simplest form.

1PR P R P PQR Q R Q Q

Simplifying Rational Expressions

Fundamental Principle of Rational ExpressionsPQ

For any rational expression and any polynomial R, where R ≠ 0,

or, simply, .PR PQR Q


Simplifying a Rational Expression

1) Completely factor the numerator and denominator of the rational expression.

2) Divide out factors common to the numerator and denominator. (This is the same thing as

“removing the factor of 1.”)

Warning!

Only common FACTORS can be eliminated from the numerator and denominator. Make sure any expression you eliminate is a factor.



Simplify the following expression.

xx

x

5

3572

)5(

)5(7

xx

x

x

7


Example:



20

432

2

xx

xx

)4)(5(

)1)(4(

xx

xx

5

1

x

x


Example:



7

7

y

y 7

)7(1

y

y1


Example:


Multiplying Rational Expressions

The rule for multiplying rational expressions is

QS

PR

S

R

Q

P


as long as Q 0 and S 0.

1) Completely factor each numerator and denominator.

2) Use the rule above and multiply the numerators and denominators.

3) Simplify the product by dividing the numerator and denominator by their common factors.


Multiply the following rational expressions.

12

5

10

63

2 x

x

x

4

1

32252

532

xxx

xxx


Example:


Multiply the following rational expressions.

mnm

m

nm

nm2

2)(

)()(

))((

nmmnm

mnmnm

nm

nm


Example:


Dividing Rational Expressions

The rule for dividing rational expressions is

P R P S PSQ S Q R QR


as long as Q 0 and S 0 and R 0.

To divide by a rational expression, use the rule above and multiply by its reciprocal. Then simplify if possible.


Divide the following rational expression.

25

155

5

)3( 2 xx

155

25

5

)3( 2

x

x

)3(55

55)3)(3(

x

xx3x


Example:


§ 6.2

Adding and Subtracting Rational Expressions


Adding and Subtracting with Common Denominators

The rule for multiplying rational expressions is

If and P RQ Q

Adding or Subtracting Rational Expressions with Common Denominators

are rational expression, then

and P R P R P R P RQ Q Q Q Q Q


Add the following rational expressions.

72

83

72

34

p

p

p

p72

57

p

p

72

8334

p

pp

Adding Rational Expressions

Example:


Subtract the following rational expressions.

2

16

2

8

yy

y

2

168

y

y

2

)2(8

y

y8

Subtracting Rational Expressions

Example:



103

6

103

322 yyyy

y

103

632 yy

y

)2)(5(

)2(3

yy

y

5

3

y

Subtracting Rational Expressions

Example:


To add or subtract rational expressions with unlike denominators, you have to change them to equivalent forms that have the same denominator (a common denominator).

This involves finding the least common denominator of the two original rational expressions.

Least Common Denominators


Finding the Least Common Denominator1) Factor each denominator completely,

2) The LCD is the product of all unique factors each raised to a power equal to the greatest number of times that the factor appears in any factored denominator.



Find the LCD of the following rational expressions.

124

3,

6

1

y

x

y

yy 326 )3(2)3(4124 2 yyy

)3(12)3(32 is LCD theSo 2 yyyy


Example:



2110

24,

34

422

xx

x

xx

)1)(3(342 xxxx

)7)(3(21102 xxxx

7)1)(x3)(x(x is LCD theSo


Example:



12

4,

55

32

2

2 xx

x

x

x

)1)(1(5)1(555 22 xxxx22 )1(12 xxx

21)-1)(x5(x is LCD theSo


Example:



xx 3

2,

3

1

Both of the denominators are already factored.Since each is the opposite of the other, you canuse either x – 3 or 3 – x as the LCD.


Example:


To change rational expressions into equivalent forms, we use the principal that multiplying by 1 (or any form of 1), will give you an equivalent expression.

RQ

RP

R

R

Q

P

Q

P

Q

P

1

Multiplying by 1


Rewrite the rational expression as an equivalent rational expression with the given denominator.

95 729

3

yy

59

3

y

4

4

5 8

8

9

3

y

y

y 9

4

72

24

y

y

Equivalent Expressions

Example:


As stated in the previous section, to add or subtract rational expressions with different denominators, we have to change them to equivalent forms first.

Unlike Denominators


Adding or Subtracting Rational Expressions with Unlike Denominators

1) Find the LCD of the rational expressions.

2) Write each rational expression as an equivalent rational expression whose denominator is the LCD found in Step 1.

3) Add or subtract numerators, and write the result over the denominator.

4) Simplify resulting rational expression, if possible.

Unlike Denominators



aa 6

8

7

15

aa 6

8,

7

15

aa 67

87

76

156

aa 42

56

42

90

a42

146a21

73

Adding with Unlike Denominators

Example:



xx 26

3,

62

5

xx 26

3

62

5

62

3

62

5

xx

62

8

x

)3(2

222

x 3

4

x

Subtracting with Unlike Denominators

Example:



3 and 32

7

x

332

7

x

32

)32(3

32

7

x

x

x

32

96

32

7

x

x

x

32

967

x

x

32

616

x

x

Subtracting with Unlike Denominators

Example:



65,

6

422 xx

x

xx

656

422 xx

x

xx

)2)(3()2)(3(

4

xx

x

xx

)3)(2)(3(

)3(

)3)(2)(3(

)3(4

xxx

xx

xxx

x

)3)(3)(2(

3124 2

xxx

xxx)3)(3)(2(

122

xxx

xx

Adding with Unlike Denominators

Example:


§ 6.3

Simplifying Complex Fractions


Complex Rational Fractions

Complex rational expressions (complex fraction) are rational expressions whose numerator, denominator, or both contain one or more rational expressions.

There are two methods that can be used when simplifying complex fractions.


Simplifying a Complex Fraction (Method 1) 1) Simplify the numerator and the denominator of

the complex fraction so that each is a single fraction.

2) Perform the indicated division by multiplying the numerator of the complex fraction by the reciprocal of the denominator of the complex fraction.

3) Simplify, if possible.



22

22x

x

24

2

24

2x

x

24

24

x

x4 2

2 4x

x 4

4

x

x


Example:


Simplifying a Complex Fraction (Method 2)

1) Multiply the numerator and the denominator of the complex fraction by the LCD of the fractions in both the numerator and denominator.

2) Simplify, if possible.



651321

2

y

y 2

266

yy

2

2

56

46

yy

y


Example:


When you have a rational expression where some of the variables have negative exponents, rewrite the expression using positive exponents.

The resulting expression is a complex fraction, so use either method to simplify the expression.

Simplifying with Negative Exponents


11

22

25 yx

yx

yx

yx1

21

5

1122

yx

yx25

122

22

22

yx

yxyxxy 22 25

1

Simplifying with Negative Exponents

2 2

1 1Simplify .5 2

x yx y

Example:


§ 6.4

Dividing Polynomials: Long Division and Synthetic Division


Dividing a Polynomial by a Monomial


Divide each term in the polynomial by the monomial.

a

aa

3

153612 3 aa

a

a

a

3

15

3

36

3

12 3

aa

5124 2

Example: Divide

, where 0.a b a b cc c c



Example: Divide 4 3 2

210 35 5

5t t t

t

4 3 2

210 35 5

5t t t

t 4 3 2

2 2 210 35 55 5 5

t t tt t t

22 7 1t t


Dividing a polynomial by a polynomial other than a monomial uses a “long division” technique that is similar to the process known as long division in dividing two numbers, which is reviewed on the next slide.

Dividing Polynomials


725643 7256431

4329

6

2585

37

8

6344

32

Divide 43 into 72.

Multiply 1 times 43.

Subtract 43 from 72.

Bring down 5.

Divide 43 into 295.



Bring down 6.

Divide 43 into 376.



Nothing to bring down.32168 .43

We then write our result as



As you can see from the previous example, there is a pattern in the long division technique.

DivideMultiplySubtractBring downThen repeat these steps until you can’t bring down or divide any longer.

We will incorporate this same repeated technique with dividing polynomials.



15232837 2 xxx

x4

xx 1228 2 35 x

5

1535 x

Divide 7x into 28x2.

Multiply 4x times 7x+3.

Subtract 28x2 + 12x from 28x2 – 23x.

Bring down – 15.

Divide 7x into –35x.

Multiply – 5 times 7x+3.

Subtract –35x–15 from –35x–15.

Nothing to bring down.

15

So our answer is 4x – 5.



86472 2 xxx

x2

xx 144 220 x

10

7020 x78

Divide 2x into 4x2.

Multiply 2x times 2x+7.

Subtract 4x2 + 14x from 4x2 – 6x.

Bring down 8.

Divide 2x into –20x.

Multiply -10 times 2x+7.

Subtract –20x–70 from –20x+8.

Nothing to bring down.

8

)72(

78x

x2 10We write our final answer as



Dividing Polynomials Using Long Division

Example: Divide:

21 3 2c c c 2c c c

1. Divide the leading term of the dividend, c2, by the first term of the divisor, x.

c

3. Subtract c2 + c from c2 + 3c – 2.

2. Multiply c by c + 1.2( 1)c c c c

2 c c

2c

2 2( 3 ) ( ) 2c c c c c Continued.

2 3 21

c cc



Example continued:

Bring down the next term to obtain a new polynomial.

21 3 2c c c c

2 c c

24. Repeat the process until the

degree of the remainder is less than the degree of the binomial divisor.

2

2 2c 4 Remainder

( 2)( 1) ( 4)c c 5. Check by verifying that (Quotient)(Divisor) + Remainder = Dividend.

2 3 2 ( 4)c c 2 3 2c c

2 3 2 421 1

c c cc c

2c



Example: Divide: (y2 – 5y + 6) ÷ (y – 2)

y

y2 – 2y

– 3y + 6

– 3

– 3y + 60

22 5 6y y y

No remainder

(y2 – 5y + 6) ÷ (y – 2) = y – 3

Check : (y – 2)(y – 3) = y2 – 5y + 6


Synthetic Division

To find the quotient and remainder when a polynomial of degree 1 or higher is divided by x – c, a shortened version of long division called synthetic division may be used.

21 3 2 x x xx

2( )x x

2x 2

2

( 22 )x 4

Long Division Synthetic Division

Subtraction signs are not necessary.

The original terms in red are not needed because they are the same as the term directly above.

Continued.

21 3 2 x x xx

x2x

2

2 4


Synthetic Division

Continued.

The variables are not necessary.

1 1 3 2x x

1

2

2

2

4

The “boxed” numbers can be aligned horizontally.

21 3 2 x x xx

x2x

2

2 4

1 1 3 2x x

1

2

2

2 4

Continued.


Synthetic Division

The leading coefficient of the dividend can be brought down.

The first two numbers in the last row are the coefficients of the quotient, the last number is the remainder.

To simplify further, the top row can be removed and instead of subtracting, we can change the sign of each entry and add.

1 1 3 2x x

2

2

41

1 2

1 21 1 3 2

2 41

The x + 1 is replaced with a – 1.

Quotient Remainder

Continued.


§ 6.5

Solving Equations Containing Rational

Expressions


Solving Equations

First note that an equation contains an equal sign and an expression does not.

To solve EQUATIONS containing rational expressions, clear the fractions by multiplying both sides of the equation by the LCD of all the fractions.

Then solve as in previous sections.

Note: this works for equations only, not simplifying expressions.


6

71

3

5

x

xx

x 66

71

3

56

xx 7610 x10

73 610

5 1

6

71

30

5

6

71

6

1 true

Solve the following rational equation.Check in the original equation.

Solving Equations

Example:


xxxx 33

1

1

1

2

12

16)1(3

1

1

1

2

116

xx

xxxxxx

2613 xx2633 xx

233 x13 x

Solve the following rational equation.

31x

Solving Equations

Continued.

Example:


21 1 1 13 3 3 3

1 1 1

2 1 3 3

131

1

4

3

2

3

4

3

4

3

4

6 true

Substitute the value for x into the original equation, to check the solution.

So the solution is 31x

Solving Equations

Example continued:



Solving Equations

Continued.

5

1

63

1

107

22

xxxx

x

5235

1

63

1

107

2523

2

xxxxxx

xxx

23523 xxx

63563 xxx66533 xxx

75 x

57x

Example:



Solving Equations

518

1

6521

1

10549

2549

53

true


2

7 2 1 153 6 57 10 7 77 7 5 55 5

18

5

9

518

5

Example continued:



Solving Equations

Continued.

1

2

1

1

xx

111

2

1

111

xx

xxxx

121 xx

221 xx

x3

Example:



Solving Equations

13

2

13

1

4

2

2

1 true

So the solution is x = 3.

Example continued:



Solving Equations

Continued.

aaa

3

2

3

3

9

122

aaaaa

aa

33

3

2

3

3

9

1233

2

aa 323312

aa 263912 aa 26321

a515 a3

Example:



Solving Equations

Since substituting the suggested value of a into the equation produced undefined expressions, the solution is .

212 3 2

39 33 33

0

2

5

3

0

12

Example continued:


§ 6.6

Rational Equations and Problem Solving


Solving Equations for a Specified Variable1) Clear the equation of fractions or rational expressions by

multiplying each side of the equation by the LCD of the denominators in the equation.

2) Use the distributive property to remove grouping symbols such as parentheses.

3) Combine like terms on each side of the equation.

4) Use the addition property of equality to rewrite the equation as an equivalent equation with terms containing the specified variable on one side and all other terms on the other side.

5) Use the distributive property and the multiplication property of equality to get the specified variable alone.

Solving Equations with Multiple Variables


21

111

RRR

2121

21

111RRR

RRRRRR

1221 RRRRRR

2121 RRRRRR

221 RRRRR

RR

RRR

2

21

Solve the following equation for R1

Solving Equations with Multiple Variables

Example:


Finding an Unknown Number

Continued

The quotient of a number and 9 times its reciprocal is 1. Find the number.

Read and reread the problem. If we let

n = the number, then

= the reciprocal of the numbern

1

1.) Understand

Example:


Continued


2.) Translate

The quotient of

a number

n

and 9 times its reciprocal

n

19

is

=

1

1

Example continued:


3.) Solve

Continued


11

9

nn

19

nn

19

n

n

92 n

3,3 n

Example continued:


4.) Interpret


Check: We substitute the values we found from the equation back into the problem. Note that nothing in the problem indicates that we are restricted to positive values.

13

193

133

13

193

133

State: The missing number is 3 or –3.

true true

Example continued:


Ratios and Rates

Ratio is the quotient of two numbers or two quantities.

The units associated with the ratio are important.

The units should match.

If the units do not match, it is called a rate, rather than a ratio.

The ratio of the numbers a and b can also be written as a:b, or .

b

a


Proportion is two ratios (or rates) that are equal to each other.

d

c

b

a

We can rewrite the proportion by multiplying by the LCD, bd.

This simplifies the proportion to ad = bc.

This is commonly referred to as the cross product.

Proportions


Solve the proportion for x.

3

5

2

1

x

x

2513 xx

10533 xx

72 x

27x

Solving Proportions

Continued.

Example:


3

5

23

25

true



3

5

227

127

Solving Proportions

Example continued:


If a 170-pound person weighs approximately 65 pounds on Mars, how much does a 9000-pound satellite weigh?

Marson satellite pound-x

Marson person pound-65

Earthon satellite pound-9000

Earthon person pound-170

000,585659000170 x

pounds 3441170/585000 x

Solving Proportions

Example:


Given the following prices charged for various sizes of picante sauce, find the best buy.

• 10 ounces for $0.99• 16 ounces for $1.69• 30 ounces for $3.29

Solving Proportions

Continued.

Example:


Size Price Unit Price

10 ounces $0.99 $0.99/10 = $0.099

16 ounces $1.69 $1.69/16 = $0.105625

30 ounces $3.29 $3.29/30 $0.10967

The 10 ounce size has the lower unit price, so it is the best buy.

Solving Proportions

Example continued:


Solving a Work Problem

Continued

An experienced roofer can roof a house in 26 hours. A beginner needs 39 hours to do the same job. How long will it take if the two roofers work together?

Read and reread the problem. By using the times for each roofer to complete the job alone, we can figure out their corresponding work rates in portion of the job done per hour.

1.) Understand

Experienced roofer 26 1/26Beginner roofer 39 /39Together t 1/t

Time in hrs Portion job/hr

Example:


Continued


2.) Translate

t

1

39

1

26

1

Since the rate of the two roofers working together would be equal to the sum of the rates of the two roofers working independently,

Example continued:


3.) Solve

Continued


t

1

39

1

26

1

tt

t 781

39

1

26

178

7823 tt

785 t

hours 15.6or 5/78t

Example continued:


4.) Interpret


Check: We substitute the value we found from the proportion calculation back into the problem.

State: The roofers would take 15.6 hours working together to finish the job.

578

1

39

1

26

1

78

5

78

2

78

3 true

Example continued:


Solving a Rate Problem

Continued

The speed of Lazy River’s current is 5 mph. A boat travels 20 miles downstream in the same time as traveling 10 miles upstream. Find the speed of the boat in still water.

Read and reread the problem. By using the formula d=rt, we can rewrite the formula to find that t = d/r.

We note that the rate of the boat downstream would be the rate in still water + the water current and the rate of the boat upstream would be the rate in still water – the water current.

1.) Understand

Down 20 r + 5 20/(r + 5)Up 10 r – 5 10/(r – 5)

Distance rate time = d/r

Example:


Continued


2.) Translate

Since the problem states that the time to travel downstairs was the same as the time to travel upstairs, we get the equation

5

10

5

20

rr

Example continued:


3.) Solve

Continued


5

10

5

20

rr

555

10

5

2055

rr

rrrr

510520 rr

501010020 rr

15010 r

mph 15r

Example continued:


4.) Interpret


Check: We substitute the value we found from the proportion calculation back into the problem.

515

10

515

20

10

10

20

20 true

State: The speed of the boat in still water is 15 mph.

Example continued:


§ 6.7

Variation and Problem Solving


Direct Variation

y varies directly as x, or y is directly proportional to x, if there is a nonzero constant k such that y = kx.

The family of equations of the form y = kx are referred to as direct variation equations.

The number k is called the constant of variation or the constant of proportionality.


If y varies directly as x, find the constant of variation k and the direct variation equation, given that y = 5 when x = 30.

y = kx

5 = k·30

k = 1/6

y = 16

xSo the direct variation equation is

Direct Variation


If y varies directly as x, and y = 48 when x = 6, then find y when x = 15.

y = kx

48 = k·6

8 = k

So the equation is y = 8x.

y = 8·15

y = 120

Direct Variation

Example:


At sea, the distance to the horizon is directly proportional to the square root of the elevation of the observer. If a person who is 36 feet above water can see 7.4 miles, find how far a person 64 feet above the water can see. Round your answer to two decimal places.

Direct Variation

Continued.

Example:


ekd

364.7 k

k64.7

6

4.7k

646

4.7d

7.4 59.2(8) 9.87 miles6 6

d So our equation is

ed6

4.7

We substitute our given value for the elevation into the equation.

Direct Variation

Example continued:


y varies inversely as x, or y is inversely proportional to x, if there is a nonzero constant k such that y = k/x.

The family of equations of the form y = k/x are referred to as inverse variation equations.

The number k is still called the constant of variation or the constant of proportionality.

Inverse Variation


If y varies inversely as x, find the constant of variation k and the inverse variation equation, given that y = 63 when x = 3.

y = k/x

63 = k/3

k = 63·3

k = 189y =

189 x

So the inverse variation equation is

Inverse Variation

Example:


y can vary directly or inversely as powers of x, as well.

y varies directly as a power of x if there is a nonzero constant k and a natural number n such that y = kxn

y varies inversely as a power of x if there is a nonzero constant k and a natural number n such that

nx

ky

Powers of x


The maximum weight that a circular column can hold is inversely proportional to the square of its height.

If an 8-foot column can hold 2 tons, find how much weight a 10-foot column can hold.

Continued.

Powers of x

Example:


2h

kw

6482

2

kk

128k tons28.1100

128

10

1282

w

2

128

hw

So our equation is

We substitute our given value for the height of the column into the equation.

Powers of x

Example continued:


y varies jointly as, or is jointly proportional to two (or more) variables x and z, if there is a nonzero constant k such that y = kxz.

k is STILL the constant of variation or the constant of proportionality.

We can also use combinations of direct, inverse, and joint variation. These variations are referred to as combined variations.

Combined Variation


Write the following statements as equations.• a varies jointly as b and c

a = kbc• P varies jointly as R and the square of S

P = kRS2

• The weight (w) varies jointly with the width (d) and the square of the height (h) and inversely with the length (l)

l

kdhw

2

Combined Variation

Example:

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