Copyright © 2011 Pearson Education, Inc. Rational Expressions and Equations CHAPTER 7.1Simplifying...

Copyright © 2011 Pearson Education, Inc.

Rational Expressions and Equations

CHAPTER

7.1 Simplifying Rational Expressions7.2 Multiplying and Dividing Rational Expressions7.3 Adding and Subtracting Rational Expressions with the

Same Denominator7.4 Adding and Subtracting Rational Expressions with

Different Denominators7.5 Complex Rational Expressions7.6 Solving Equations Containing Rational Expressions7.7 Applications with Rational Expressions, Including

Variation

77


Simplifying Rational Expressions7.17.1

1. Evaluate rational expressions.2. Find numbers that cause a rational expression to be

undefined.3. Simplify rational expressions containing only monomials.4. Simplify rational expressions containing multiterm

polynomials.

3Copyright © 2011 Pearson Education, Inc.

Rational expression: An expression that can be

written in the form , where P and Q are

polynomials and Q 0.

P

Q

Some rational expressions are

5

2

3

18

x

x 2

7

9

x

x

2 2 15

4 20

x x

x


Example 1

Evaluate the expression when

a. x = 2 b. x = –1

7 9,

1

x

x

Solution

a. 7 9

1

x

x

7

1

2

2

9

14 9

3

5 2 or 1

3 3

b. 7 9

1

x

x

7 91

1 1

7 9

0

16

0

which is undefined


Finding Values That Make a Rational Expression UndefinedTo determine the value(s) that make a rational expression undefined, 1. Write an equation that has the denominator set equal to zero. 2. Solve the equation.


Example 2aFind every value for the variable that makes the expression undefined.4

6 5

z

z

Solution 6 5 0z 6 5z

The original expression is undefined if z is replaced by 5/6.

Set the denominator equal to 0.

Add 5 to both sides.

Divide both sides by 6.5

6z


Example 2bFind every value for the variable that makes the expression undefined.

3 2

6

5 4y y y

Solution 3 25 4 0y y y

2 5 4 0y y y 4 1 0y y y

0 or 4 0 or 1 0y y y 4 1y y

The original expression is undefined if y is replaced by 0, –4, or –1.

Set the denominator equal to 0.

Factor out the monomial GCF, y.

Use the zero factor theorem.


Simplifying Rational ExpressionsTo simplify a rational expression to lowest terms:

1. Factor the numerator and denominator completely. 2. Divide out all the common factors in the numerator

and denominator.3. Multiply the remaining factors in the numerator and

the remaining factors in the denominator.

1, where , , and

1

are polynomials and and are not 0.

PR P PP Q

QR Q Q

R Q R


Example 3

Simplify

Solution

7

3

4.

24

x

x

2 2

2 2 2 3

x x x x x x x

x x x

4

6

x

Write the numerator and denominator in factored form, then eliminate the common factors. Multiply the remaining factors.

7

3

4

24

x

x


Example 4 Simplify.a. b.

Solution

a.

b.

3

5

6

42

a

a

2 2

2 4

24

72

xy z

x y

3

5

6

42

a

a

2 3

2 3 7

a a a

a a a a a

2

1

7a

2 2

2 4

24

72

xy z

x y

2 2 2 3

2 2 2 3 3

x y y z z

x x y y y y

2

23

z

xy

Write the numerator and denominator in factored form, then eliminate the common factors. Multiply the remaining factors.


Example 5

Simplify .

8

2 4

xy

x

Solution

2 2 2

2 2

x y

x

Write the numerator and denominator in factored form, then divide out the common factor, which is 2.

8

2 4

xy

x

4

2

xy

x


Example 6a

Simplify.

2

2

6 12

7 14

x x

x x

Solution

6 2

7 2

x x

x x

6 2

7 2

x x

x x

6

7

Write the numerator and denominator in factored form, then divide out the common factors, x and x + 2.

2

2

6 12

7 14

x x

x x


Example 6b

Simplify.

2

2

25

2 15

y

y y

Solution

5 5

3 5

y y

y y

Factor the numerator and denominator completely. Then divide out the common factor, y – 5.

2

2

25

2 15

y

y y

5

3

y

y


Example 6c

Simplify.

2

2

3 9 12

6 30 24

x x

x x

SolutionFactor out the GCF.

2

2

3 9 12

6 30 24

x x

x x

2

2

3 3 4

6 5 4

x x

x x

3 4 1

6 4 1

x x

x x

3 4 1

2 3 4 1

x x

x x

4

2 4

x

x

Factor the polynomial factors.

Divide out common factors.


Example 7

Simplify.

2

2

4 4 48

3 3 18

x x

x x

Solution

4 4 4 16 or

3 2 3 6

x x

x x

4 4 3

3 2 3

x x

x x

2

2

4 4 48

3 3 18

x x

x x


Example 8

Simplify.

22 21

6 2

x x

x

Solution

2 7 3

2 3

x x

x

22 21

6 2

x x

x

2 7 1 3

2 3

x x

x

2 7 2 7 or

2 2

x x


Simplify.

a) 5

b)

c)

d)

2

2

6 4

2 2

n n

n n

2

2

3 2n n

n n

23 2

1

n n

n

3 2

1

n

n

7.1


Simplify.

a) –10x

b)

c)

d)

2

2

10 25

25

x x

x

10

5

x

x

5

5

x

x

5

5

x

x

7.1


Multiplying and Dividing Rational Expressions7.27.2

1. Multiply rational expressions.2. Divide rational expressions.3. Convert units of measurement using dimensional

analysis.


, where , , , P R PR

P Q RQ S QS

Multiplying Rational ExpressionsTo multiply rational expressions,

1. Factor each numerator and denominator completely.

2. Divide out any numerator factor with any matching denominator factor.

3. Multiply numerator by numerator and denominator by denominator.

4. Simplify as needed.

and are polynomials and 0 and 0.S Q S


Example 1

Multiply. 26 25

5 18

a ab

b

Solution 26 25

5 18

a ab

b

2 3 5 5

5 2 3 3

a a b b

b

5

1 3

a a b

25

3

a b


Example 2a

Solution

1 1

2 x x

Write numerators and denominators in factored form.

Multiply the remaining numerator factors and denominator factors.

3

5 3 12

2 8 15

x x

x x

3 45

2 4 5 3

xx

x x x x

2

1

2x


Example 2b

Multiply. 220 4 3

9 2 10

x x

x x

Solution

2 1

3 1

x



4 5 3

3 3 2 5

x x x

x x

2 2 1 5 3

3 3 2 5

x x x

x x

2

3

x


2 1 2 2

2 2 5 2 1 1

x x

x x

Example 2c

Multiply. 2

2

2 4

20 2 3 1

x x

x x

Solution2

2

2 4

20 2 3 1

x x

x x

2 2

or 5 2 1 10 5

x x

x x




Dividing Rational Expressions

, where , , , and

are polynomials and 0, 0, and 0.

P R P SP Q R S

Q S Q R

Q R S


Dividing Rational ExpressionsTo divide rational expressions,

1. Write an equivalent multiplication statement with the reciprocal of the divisor.

2. Factor each numerator and denominator completely. (Steps 1 and 2 are interchangeable.)

3. Divide out any numerator factor with any matching denominator factor.

4. Multiply numerator by numerator and denominator by denominator.

5. Simplify as needed.


Example 3

Divide . 2

3

6 20

35 5

x y y

z z

Solution

2

2

3

70

x

z

Write an equivalent multiplication statement.

Divide out common factors, and multiply remaining factors.

2

3

6 20

35 5

x y y

z z

2

3

6 5

35 20

x y z

z y

3 2 5

5 7 5 2 2

x x y z

z z z y


2 1 3 3 1 2

2 1 2 3 1 2 1

x x x x

x x x x

Example 4

Divide. 2 2

2 2

2 7 3 6 5 1

2 3 2 3 5 2

x x x x

x x x x

Solution 2 2

2 2

2 7 3 6 5 1

2 3 2 3 5 2

x x x x

x x x x

2 2

2 2

2 7 3 3 5 2

2 3 2 6 5 1

x x x x

x x x x

3

2 1

x

x

Write an equivalent multiplication statement.

Divide out common factors, and multiply remaining factors.


Using Dimensional Analysis to Convert Between Units of MeasurementTo convert units using dimensional analysis, multiply the given measurement by conversion factors so that the undesired units divide out, leaving the desired units.


Example 5a

Convert 300 ounces to pounds

Solution

300 oz. 1 lb.

1 16 oz. 18.75 lb.300 oz.

300 lb.

16


Example 5b

Convert 17 yards to inches.

Solution

17 yd. 3 ft. 12 in.

1 1 yd. 1 ft. 612 inches17 yards


Example 5c

Convert 25 miles per hour to feet per second.

Solution

25 mi. 5280ft. 1 hr. 1min .

1 hr. 1 mi. 60 min. 60sec.

25 miles

1 hour36.6 ft./sec.


Multiply.

a)

b)

c)

d)

2

2

6 9 2

4 3

m m m

m m

3

2

m

m

3

2

m

m

3

2

m

m

3

2

m

m

7.2


Divide.

a)

b)

c)

d)

2 5 6 2

6 6

n n n

n n

3n

3

2

n

n

3 2

2

n n

n

3 2

2

n n

n

7.2


Adding and Subtracting Rational Expressions with the Same Denominator7.37.3

1. Add or subtract rational expressions with the same denominator.


Adding or Subtracting Rational Expressions (Same Denominator)To add or subtract rational expressions that have the same denominator:1. Add or subtract the numerators and keep the same denominator. 2. Simplify to lowest terms. (Remember to factor the

numerators and denominators completely in order to simplify).


Example 1

Add.

Solution

7

18 18

y y

7

18 18

y y 7

18

y y

8

18

y

2 2 2

3 3 2

y

4

9

y

Since the rational expressions have the same denominator, we add numerators and keep the same denominator.

Factor.

Divide out the common factor, 2.


Example 2a

Subtract.

Solution

18

7 7

z

z z

Warning: Although it may be tempting to do so, we cannot divide out the z’s because they are terms, not factors.

18

7

z

z

18

7 7

z

z z


Example 2b

Subtract.

Solution

2 16

4 4

x

x x

4 4

4

x x

x

2 16

4 4

x

x x

2 16

4

x

x

4x Divide out the common factor, x – 4.

Note: The numerator can be factored, so we may be able to simplify.


Example 2c

Subtract.

Solution

7 11

12 12

x

7 11

12

x

7 11

12 12

x

7 11

12

x


2 3

5

b b

b b

Example 3

Add.

Solution

2 2

2 2

7 4

5 5

b b b b

b b b b

Combine like terms in the numerator.

2 2

2 2

7 4

5 5

b b b b

b b b b

2 2

2

7 4

5

b b b b

b b

2

2

2 3

5

b b

b b

2 3

5

b

b

Factor the numerator and the denominator.

Divide out the common factor, b.


Example 4

Add.

Solution

2 2

3 3

5 6 3 2

2 8 2 8

x x x x

x x x x

Combine like terms in the numerator.

Factor the numerator and the denominator.

2 2

3 3

5 6 3 2

2 8 2 8

x x x x

x x x x

2 2

3

5 6 3 2

2 8

x x x x

x x

2

3

2 8 8

2 8

x x

x x

2 2 2

2 2 2

x x

x x x


continued

2

2 2 or

2 2

x x

x x x x

2 2 2

2 2 2

x x

x x x

Divide out the common factors, 2 and x + 2.


Example 5a

Subtract.

Solution

2 23 2 7 8

3 1 3 1

x x x x

x x

2 23 2 7 8

3 1

x x x x

x

22 1

3 1

x x

x

2 23 2 7 8

3 1 3 1

x x x x

x x

2 23 2 7 8

3 1

x x x x

x

Note: To write an equivalent addition, change the operation symbol from a minus sign to a plus sign and change all the signs in the subtrahend (second) polynomial.


Example 5b

Subtract.

Solution

3

2 2

5 5 5 22

2 6 2 6

x x x

x x x x

3

2

5 5 5 22

2 6

x x x

x x

Note: This numerator is the difference of cubes. 3

2

5 5 5 22

2 6

x x x

x x

3

2

27

2 6

x

x x

3

2 2

5 5 5 22

2 6 2 6

x x x

x x x x


continued

Note: This numerator is the difference of cubes.

23 3 9

2 3

x x x

x x

3

2

27

2 6

x

x x

2 3 9

2

x x

x


Add.

a)

b)

c)

d)

2 2

2 5 7

3 4 3 4

x x

x x x x

2

3 12

3 4

x

x x

3 4

1 4

x

x x

2

1 4

x

x x

3

1x

7.3


Subtract.

a)

b)

c)

d)

2 2

1 5 3

2 1 2 1

a a

a a a a

2

2 6

1

a

a

2

2 3

1

a

a

2

4

1a

4

1a

7.3


Adding and Subtracting Rational Expressions with Different Denominators7.47.4

1. Find the LCD of two or more rational expressions.2. Given two rational expressions, write equivalent rational

expressions with their LCD.3. Add or subtract rational expressions with different

denominators.


Remember that when adding or subtracting fractions with different denominators, we must first find a common denominator. It is helpful to use the least common denominator (LCD), which is the smallest number that is evenly divisible by all the denominators.

38 2 212 2 3

3LCD 2 3 = 24


Finding the LCDTo find the LCD of two or more rational expressions,1. Find the prime factorization of each denominator. 2. Write a product that contains each unique prime

factor the greatest number of times that factor occurs in any factorization. Or if you prefer to use exponents, write the product that contains each unique prime factor raised to the greatest exponent that occurs on that factor in any factorization.

3. Simplify the product found in step 2.


Example 1

Find the LCD of

Solution We first factor the denominators 12y2 and 8y3 by writing their prime factorizations.

2 3

5 2 and .

12 8y y

2 2 212 2 3 y y 3 3 38 2 y y

The unique factors are 2, 3, and y. To generate the LCD, include 2, 3, and y the greatest number of times each appears in any of the factorizations.


continued

Note: We can compare exponents in the prime factorizations to create the LCD. If two factorizations have the same prime factors, we write that prime factor in the LCD with the greater of the two exponents.

3 3 3LCD = 2 3 24y y

The greatest number of times that 2 appears is three times (in 23 • y3).

The greatest number of times that 3 appears is once (in 22 • 3 • y2).

The greatest number of times that y appears is three times (in 23 • y3).


Example 2

Find the LCD.

Solution Factor the denominators x2 – 25 and 2x – 10.

2

8 3 and

25 2 10x x

2 25 5 5x x x 2 10 2 5x x

The unique factors are 2, (x + 5), and (x – 5). The greatest number of times that 2 appears is once.The greatest number of times that (x + 5) appears is once.The greatest number of times that (x – 5) appears is once.

LCD = 2 5 5x x


Example 3

Write as equivalent rational expressions with their LCD.

Solution The LCD is 24x2. For each rational expression, we multiply both the numerator and denominator by an appropriate factor so the denominator becomes 24x2.

2

7 5 and

8 12x x

7

8x

7 3

8 3

x

x x

2

21

24

x

x

2

5

12x 2

5 2

12 2x

2

10

24x


Example 4

Write as equivalent rational expressions with their LCD.

Solution In the previous example, we found the LCD to be .

2

8 3 and

25 2 10x x

2 5 5x x

2

8

25x 8

5 5x x

8 2

5 5 2x x

16

2 5 5x x

Write the denominator in factored form.

Note: Another way to determine the appropriate factor is to think of it as the factor in the LCD that is missing from the original denominator.

Multiply the numerator and the denominator by the same factor, 2, to get the LCD, 2(x + 5)(x – 5).


continued

3

2 10x 3

2 5x

3 5

2 5 5

x

x x

3 5

2 5 5

x

x x

Write the denominator in factored form.

Multiply the numerator and the denominator by the same factor, (x + 5), to get the LCD, 2(x + 5)(x – 5).


Adding or Subtracting Rational Expressions with Different DenominatorsTo add or subtract rational expressions with different denominators,1. Find the LCD. 2. Write each rational expression as an equivalent expression with the LCD.3. Add or subtract the numerators and keep the LCD.4. Simplify.


Example 5

Add.

Solution The LCD is 24x2.

2

3 5 1 +

8 6

x x

x x

2

3 5 1 +

8 6

x x

x x

2

3 3 5 1 4 +

8 3 6 4

x x x

x x x

Write equivalent rational

expressions with the LCD, 24x2.

2

2 2

3 9 20 4 +

24 24

x x x

x x

2

2

3 9 20 4

24

x x x

x

2

2

3 29 4

24

x x

x

Add numerators.

Note: Remember that to add polynomials, we combine like terms.


Example 6

Add .

Solution

2 2

6 4 +

2 4 3x x x x

2 2

6 4 +

2 4 3x x x x

6 4

+ 2 1 3 1x x x x

6 3 4 2 +

2 1 3 2 3 1

x x

x x x x x x


continued

6 3 4 2 +

2 1 3 2 3 1

x x

x x x x x x

6 18 4 8

+ 2 1 3 2 3 1

x x

x x x x x x

6 18 4 8

2 1 3

x x

x x x

10 10

2 1 3

x

x x x

10 1

2 1 3

x

x x x

10

2 3x x


Example 7

Add.

Solution

7 3 +

6 6

x

x x

Since x – 6 and 6 – x are additive inverses, we obtain the LCD by multiplying the numerator and denominator of one of the rational expressions by –1. We chose the second rational expression.

7 3 +

6 6

x

x x

3 17

+ 6 6 1

x

x x

7 3 +

6 6

x

x x

3 7

6

x

x


Add.

a)

b)

c)

d)

5 8

4 3 12n n

23

3 12n 13

3 12n

13 4

3 12

n

n

5

3 12n

7.4


Subtract.

a)

b)

c)

d)

5 2 3

5 5

x x

x x

3 1

5

x

x

7 3

5

x

x

7 3

5

x

x

7 3

5

x

x

7.4


Complex Rational Expressions7.57.5

1. Simplify complex rational expressions.


Complex rational expression: A rational expression that contains rational expressions in the numerator or denominator.

Examples:

3456

3

2

xx

y7

2tt

2 14

13

nn

n


Simplifying Complex Rational ExpressionsTo simplify a complex rational expression, use one of the following methods.Method 11. Simplify the numerator and denominator if needed.2. Rewrite as a horizontal division problem. Method 21. Multiply the numerator and denominator of the complex rational expression by their LCD. 2. Simplify.


Example 1 —Method 1

Simplify.

Solution Write the numerator fractions as equivalent fractions with their LCD, 12, and write the denominator fractions with their LCD, 24.

3 54 65 18 6

3 54 65 18 6

3(3) 5(2)4(3) 6(2)5(3) 1(4)8(3) 6(4)

9 1012 1215 424 24

19121124


continued

19 11

12 24

19 24

12 11

38

11

1

2

19121124


Example 1—Method 2

Simplify.

Solution Multiply the numerator and denominator by 24.

3 54 65 18 6

3 54 65 18 6

243 54 65 1

26

48

3 24 5 244 1 6 15 24 1 248 1 6 1

1

46

13

1

4

1

18 20

15 4

38

11


Example 1 Simplify.

Solution Write the numerator and denominator as equivalent fractions. (Method 1)

12

23

y

y

12

23

y

y

2( ) 1(1)1( ) (1)3( ) 2(1)1( ) (1)

yy yyy y

2 1

3 2

yy

yy

2 1 3 2

y y

y y

2 1

3 2

y y

y y2 1

3 2

y

y

2 1

3 2

yy yyy y


Example 1Simplify.

Solution Multiply the numerator and denominator by the LCD of all the rational expressions. (Method 2)

32

11

xx

xx

32

11

xx

xx

32

11

2

2

xx

xx

x

x

2 3 22 1 1

1 2 1 21 1 1

x x xx

x x xx

2 6

2 2( 1)

x

x x

2 6

2 2 2

x

x x

2 26 6 or

4 2 2(2 1)

x x

x x


Simplify.

a)

b)

c)

d)

abcd

ab

cd

ac

bdad

bc

1ad

bc

7.5


Simplify.

a)

b)

c) a + 1

d) a – 1

1

11

aa

a

1a

a

1a

a

7.5


Solving Equations Containing Rational Expressions7.67.6

1. Solve equations containing rational expressions.


Example 1Solve.

Solution

3

6 4 8

x x

3

6 4 8

x x

24 243

6 4 8

x x

324 2

62

84

44

x x

4 6 9 x x2 9 x2 9

2 2

x

9

2x

Multiply both sides by 24.

Distribute and divide out common factors.

Subtract 6x from both sides.

Divide both sides by 2.

Simplify both sides.

4

1 1

6

1

3


Example 2Solve.

Solution

2 5 1

4x x

4 42 5 1

4x x

x x

Multiply both sides by 4x.

Distribute and divide out common factors.

Simplify both sides.

2 5 1

4x x

42 5 1

44 4

x xx x x

8 20 x

12x


Extraneous Solution: An apparent solution that does not solve its equation.

Solving Equations Containing Rational Expressions To solve an equation that contains rational expressions, 1. Eliminate the rational expressions by multiplying both sides of the equation by their LCD. 2. Solve the equation using the methods we learned in Chapters 2 (linear equations) and 7 (quadratic equations).3. Check your solution(s) in the original equation. Discard any extraneous solutions.


Example 3Solve.

Solution If x = 6, then the equation is undefined, so the solution cannot be 6.

2

6 6

x

x

2

6 6

x

x

6( 6) 6( 6)2

6 6

xx

xx

6 ( 6)2 x x

6 2 12 x x

4 12x

3x

Divide out common factors.

Distribute 2 to clear the parentheses.

Subtract 2x on both sides.

Divide both sides by 4.


Example 4

Solve.Solution

2 1 6

3 1 3 1

x

x x x

(3 1) (2 1 6

3 33

1 11)

x x xx

x x xx

(3 1) (3 1) (2 1 6

3 1 3 13 1)

x

x xx x x x

xx x

2 3 1 6 ( ) x x x x22 3 1 6 x x x

26 1 0x x (3 1)(2 1) 0 x x

3 1 0 or 2 1 0 x x 1 1 or

3 2 x x

In the original equation, when x = -1/3 it is undefined, so the only solution is x = ½.


Example 5

Solve.

Solution

2

2 11

7 10 21 3

a a

a a a a

2 11

7 ( 7)( 3) 3

a a

a a a a

2 1( 7)( 3) 1 ( 7)( 3)

7 ( 7)( 3) 3

a aa a a a

a a a a

2 ( 3) ( 7)( 3) 1 ( 7) a a a a a a

2 2 22 6 ( 10 21) 1 7 a a a a a a

2 2 22 6 10 21 1 7 a a a a a a2 24 21 7 1 a a a a


continued

2 24 21 7 1 a a a a

11 21 1 a

11 22 a

11 22

11 11

a

2a

2

2 11 .

7 10 21 3

a a

a a a a

This solves the original equation


Solve.

a) 2

b) 5

c) 7

d) 10

5 7

20 2

a

a

7.6


Solve.

a) 2

b) 1

c) 2 and 5

d) 3 and 4

2

2 10 1 2 10

1 1 1

a a a

a a a

7.6


Applications with Rational Expressions, Including Variation7.77.7

1. Use tables to solve problems with two unknowns involving rational expressions.

2. Solve problems involving direct variation.3. Solve problems involving inverse variation.4. Solve problems involving joint variation.5. Solve problems involving combined variation.


Tables are helpful in problems involving two or more people working together to complete a task.For each person involved, the rate of work, time at work, and amount of the task completed are related as follows:

Since the people are working together, the sum of their individual amounts of the task completed equals the whole task.

Person’s rate of work

Time at work

Amount of the task completed by that person

=

Amount completed by one person

Amount completed by the other person

Whole task+ =


Example 1Louis and Rebecca own a painting business. Louis can paint an average size room in 5 hours. Rebecca can paint the same room in 3 hours. How long would it take them to paint the same room working together?

UnderstandLouis paints at a rate of 1 room in 5 hours, or 1/5 of a room per hour.Rebecca paints at a rate of 1 room in 3 hours, or 1/3 of a room per hour.


continued

Plan and Execute Louis’ amount completed + Rebecca’s amount completed = 1 room

Category Rate of Work

Time at Work

Amount of Task Completed

Louis t t

Rebecca t t

1513

1513

1 11

5 3t t


continued 1 1

15 3

t t

1 115 15(1)

5 3t t

1 1

15 15 155 3

t t

3 5 15t t 8 15t

15

8t

3

1

5

1


continued

AnswerWorking together, it takes Louis and Rebecca or 1 7/8 hours to paint an average size room.

Check

158

1 15 1 151

5 8 3 8

15 151

40 24

45 751

120 120

1201

120


Example 2A sports car travels 15 mph faster than a loaded truck on the freeway. In the same time that the sports car travels 156 miles, the truck travels 120 miles. Find the speed of each vehicle.

UnderstandUse a table to organize the information.

Vehicle Distance Rate Time

Sports car 156 miles r + 15

Truck 120 miles r

156

15r 120

r


continuedPlan and Execute

156 12015 15

15r r r r

r r

156 120

15r r

156 120 15r r

156 120 1800r r

36 1800r

50r


continued

AnswerThe truck travels at 50 mph while the sports car travels at r + 15 or 65 mph.

Check 156 120

15r r

156 120

65 50

2.4 2.4


Direct variation: Two variables, y and x, are in direct variation if y = kx, where k is a constant.

In words, direct variation is written as “y varies directly as x” or “y is directly proportional to x” and these phrases translate to y = kx.


Example 3Suppose y varies directly as x. When y = 12, x = 6. Find y when x = 9.Solution Replace y with 12 and x with 6, then solve for k.

y = kx12 = k 62 = k

Replace k with 2 in y = kx so that we have y = 2x.y = 2xy = 2(9) = 18


Example 4Nigel’s paycheck varies directly as the number of hours worked. For 15 hours of work, the pay is $188.25. Find the pay for 27 hours of work.Understand Translating “paycheck varies directly as the number of hours worked,” we write p = kh, where p represents the paycheck and h represents the hours.Plan Use p = kh and replace p with 188.25 and h with 15, in order to find the value of k.


continuedExecute 188.25 = k 15

12.55 = k

Replace k with 12.55. p = kh p = 12.55h

When hours = 27: p = 12.55(27) p = 338.85

Answer Working 27 hours will earn a paycheck of $338.85.


Inverse variation: Two variables, y and x, are in inverse variation if where k is a constant.

In words, inverse variation is written as “y varies inversely as x” or “y is inversely proportional to x, and these phrases translate to .”

,k

yx

,k

yx


Example 5If the temperature is constant, the pressure of a gas in a container varies inversely as the volume of the container. If the pressure is 15 pounds per square foot in a container with 4 cubic feet, what is the pressure in a container with 1.5 cubic feet?Understand Because the pressure and volume vary inversely, we can write

where P represents the pressure and V represents the volume.

kP

V


continuedPlan Use pressure is 15 when volume is 4 to determine the value of k.Execute

Answer The pressure is 40 pounds per square foot when the volume is 1.5 cubic feet.

kP

V

154

k

60 k

Replacing k with 60, solve for P when V is 1.5.

60P

V

60

1.5P 40


Joint variation: If y varies jointly as x and z, then y = kxz, where k is the constant of variation.


Solution

Suppose a varies jointly with b and c. If a = 72 when b = 3 and c = 8, find a when b = 4 and c = 2.

We have a = kbc, so

72 3 8k

3 kThe equation of variation is a = 3bc.

a = 3bc

a = 3(4)(2) = 24

Example 6


Example 7 The volume of wood V in a tree varies jointly as the height h and the square of the girth g (girth is the distance around). The volume of a redwood tree is 216 m3 when the height is 30 m and the girth is 1.5 m, what is the height of a tree whose volume is 1700 m3 and girth is 2.5 m?Solution Find k using the first set of data.

V = khg2

216 = k 30 1.52

3.2 = k


continued

The equation of variation is V = 3.2hg2. We substitute the second set of data into the equation.Volume is 1700 m3 and girth is 2.5 m.

V = 3.2hg2

1700 = 3.2 h 2.52

1700 = 20h85 = h

The height of the tree is 85 m.


Danielle can paint a 4800-square foot house in 8 days, and Paige can do the same job in only 6 days. How long will it take them to paint a 4800-square foot house if they work together?

a) 14 days

b) 7 days

c)

d)

33 days

72

3 days5

7.7


The distance a car travels varies directly with the amount of gas it carries. On the highway, a van travels 112 miles using 7 gallons of gas. How many gallons are required to travel 544 miles?

a) 15 gallons

b) 16 gallons

c) 32 gallons

d) 34 gallons

7.7


If y varies inversely as x and y = 8 when x = 10, what is y when x is 16?

a) 2

b) 5

c) 10

d) 16

7.7

Copyright © 2011 Pearson Education, Inc. Rational Expressions and Equations CHAPTER 7.1Simplifying...

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Transcript of Copyright © 2011 Pearson Education, Inc. Rational Expressions and Equations CHAPTER 7.1Simplifying...