CHAPTER · 6.2 ; Division and Reciprocals. 6.3 : Least Common Multiples and Denominators . 6.4 : Adding Rational Expressions . 6.5 : Subtracting Rational Expressions . 6.6

CHAPTER

6 Rational Expressions and Equations

Slide 2 Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

6.1 Multiplying and Simplifying Rational Expressions 6.2 Division and Reciprocals 6.3 Least Common Multiples and Denominators 6.4 Adding Rational Expressions 6.5 Subtracting Rational Expressions 6.6 Complex Rational Expressions 6.7 Solving Rational Equations 6.8 Applications Using Rational 6.9 Variation and Applications

OBJECTIVES

6.1 Multiplying and Simplifying Rational Expressions


a Find all numbers for which a rational expression is not defined.

b Multiply a rational expression by 1, using an expression such as A/A.

c Simplify rational expressions by factoring the numerator and the denominator and removing factors of 1.

d Multiply rational expressions and simplify.




Rational numbers are quotients of integers.

The following are called rational expressions or fraction expressions. They are quotients, or ratios, of polynomials:




A rational expression is also a division. For example,




When a variable is replaced with a number that produces a denominator equal to zero, the rational expression is not defined.

For example, in the expression when x is replaced with the denominator is 0, and the expression is not defined:

EXAMPLE



1 Find all numbers for which the rational expression is not defined.


EXAMPLE Solution



1


The rational expression is not defined for the replacement numbers 5 and –2.


Multiplying Rational Expressions


We multiply rational expressions in the same way that we multiply fraction notation in arithmetic. To multiply rational expressions, multiply numerators and multiply denominators:




For example,




Any rational expression with the same numerator and denominator is a symbol for 1:


Equivalent Expressions


Expressions that have the same value for all allowable (or meaningful) replacements are called equivalent expressions.

EXAMPLE



2 Multiply.





In algebra, instead of simplifying a fraction like we may need to simplify an expression like To simplify, we can do the reverse of multiplying. We factor the numerator and the denominator and “remove” a factor of 1.

EXAMPLE



5 Simplify:


EXAMPLE



8 Simplify:


EXAMPLE Solution



8





Expressions of the form and are opposites of each other. When either of these binomials is multiplied by –1, the result is the other binomial:




Consider, for example,

At first glance, it appears as though the numerator and the denominator do not have any common factors other than 1. But x – 4 and 4 – x are opposites, or additive inverses, of each other. Thus we can rewrite one as the opposite of the other by factoring out a –1.

EXAMPLE



10 Simplify:


EXAMPLE



11 Multiply and simplify:


EXAMPLE



13 Multiply and simplify:


EXAMPLE Solution



13


CHAPTER




OBJECTIVES

6.2 Division and Reciprocals


a Find the reciprocal of a rational expression. b Divide rational expressions and simplify.


a Find the reciprocal of a rational expression.


Two expressions are reciprocals of each other if their product is 1. The reciprocal of a rational expression is found by interchanging the numerator and the denominator.


Dividing Rational Expressions


EXAMPLE


b Divide rational expressions and simplify.

4 Divide:


EXAMPLE



6 Divide:


EXAMPLE Solution



6


EXAMPLE



8 Divide:


EXAMPLE Solution



8


EXAMPLE Solution



8


CHAPTER




OBJECTIVES

6.3 Least Common Multiples and Denominators


a Find the LCM of several numbers by factoring. b Add fractions, first finding the LCD. c Find the LCM of algebraic expressions by factoring.


Finding LCMs


To find the LCM, use each factor the greatest number of times that it appears in any one factorization.


b Add fractions, first finding the LCD.


Fractions can be added if they have the same denominator. The least common denominator (LCD) of a set of fractions is the least common multiple (LCM) of their denominators.


c Find the LCM of algebraic expressions by factoring.


To find the LCM of two or more algebraic expressions, we factor them. Then we use each factor the greatest number of times that it occurs in any one expression. In Section 6.4, each LCM will become an LCD used to add rational expressions.

EXAMPLE



3 Find the LCM of 12x, 16y, and 8xyz.


EXAMPLE



4 Find the LCM of and


EXAMPLE Solution



4


EXAMPLE



7 Find the LCM of and


EXAMPLE Solution



7


CHAPTER




OBJECTIVES

6.4 Adding Rational Expressions


a Add rational expressions.


Adding Rational Expressions with Like Denominators


To add when the denominators are the same, add the numerators and keep the same denominator. Then simplify if possible.

EXAMPLE



3 Add.



Adding Rational Expressions with Different Denominators


To add rational expressions with different denominators: 1. Find the LCM of the denominators. This is the least

common denominator (LCD). 2. For each rational expression, find an equivalent

expression with the LCD. Multiply by 1 using an expression for 1 made up of factors of the LCD that are missing from the original denominator.

3. Add the numerators. Write the sum over the LCD. 4. Simplify if possible.

EXAMPLE



5 Add:


EXAMPLE



8 Add:


EXAMPLE Solution



8


EXAMPLE Solution



8


EXAMPLE Solution



8


EXAMPLE



10 Add:


EXAMPLE



11 Add:


EXAMPLE Solution



11


EXAMPLE Solution



11


CHAPTER




OBJECTIVES

6.5 Subtracting Rational Expressions


a Subtract rational expressions. b Simplify combined additions and subtractions of

rational expressions.


Subtracting Rational Expressions with Like Denominators


To subtract when the denominators are the same, subtract the numerators and keep the same denominator. Then simplify if possible.

EXAMPLE


a Subtract rational expressions.

2 Subtract:



Subtracting Rational Expressions with Different Denominators


To subtract rational expressions with different denominators: 1. Find the LCM of the denominators. This is the least

common denominator (LCD). 2. For each rational expression, find an equivalent expression

with the LCD. To do so, multiply by 1 using a symbol for 1 made up of factors of the LCD that are missing from the original denominator.

3. Subtract the numerators. Write the difference over the LCD.

4. Simplify if possible.

EXAMPLE



3 Subtract:


EXAMPLE



3 Subtract:


EXAMPLE



6 Subtract:


EXAMPLE Solution



6


EXAMPLE Solution



6


EXAMPLE



7 Subtract:


Factoring 64 – p2, we get (8 – p)(8 + p).Note that the factors 8 – p in the first denominator and p – 8 in the second denominator are opposites.

EXAMPLE Solution



7


EXAMPLE Solution



7


EXAMPLE


b Simplify combined additions and subtractions of rational expressions.

8 Perform the indicated operations and simplify:


EXAMPLE Solution



8


EXAMPLE



9 Perform the indicated operations and simplify:


EXAMPLE Solution



9


EXAMPLE Solution



9


CHAPTER




OBJECTIVES

6.6 Complex Rational Expressions


a Simplify complex rational expressions.




A complex rational expression, or complex fraction expression, is a rational expression that has one or more rational expressions within its numerator or denominator.


Method 1 - Multiplying by the LCM of all Denominators


To simplify a complex rational expression: 1. First, find the LCM of all the denominators of all the

rational expressions occurring within both the numerator and the denominator of the complex rational expression.

2. Then multiply by 1 using LCM/LCM. 3. If possible, simplify by removing a factor of 1.

EXAMPLE



1 Simplify:


EXAMPLE Solution



1


EXAMPLE Solution



1


Multiplying in this manner has the effect of clearing fractions in both the numerator and the denominator of the complex rational expression.

EXAMPLE



3 Simplify:


EXAMPLE Solution



3



Method 2 – Adding in the Numerator and the Denominator


To simplify a complex rational expression: 1. Add or subtract, as necessary, to get a single rational

expression in the numerator. 2. Add or subtract, as necessary, to get a single rational

expression in the denominator. 3. Divide the numerator by the denominator. 4. If possible, simplify by removing a factor of 1.

EXAMPLE



5 Simplify:


EXAMPLE Solution



5


EXAMPLE Solution



5


EXAMPLE



6 Simplify:


EXAMPLE Solution



6


EXAMPLE Solution



6


CHAPTER




OBJECTIVES

6.7 Solving Rational Equations


a Solve rational equations.




A rational, or fraction, equation, is an equation containing one or more rational expressions.


Solving Rational Equations


To solve a rational equation, the first step is to clear the equation of fractions. To do this, multiply all terms on both sides of the equation by the LCM of all the denominators. Then carry out the equation-solving process as we learned it in Chapters 2 and 5.

EXAMPLE



3 Solve:


EXAMPLE



3 Solve:


EXAMPLE



6 Solve:


EXAMPLE



6 Solve:


EXAMPLE



6 Solve:


EXAMPLE



7 Solve:


EXAMPLE Solution



7


EXAMPLE Solution



7


CHAPTER




OBJECTIVES

6.8 Applications Using Rational


a Solve applied problems using rational equations. b Solve proportion problems.

EXAMPLE


a Solve applied problems using rational equations.

1 Sodding a Yard


Charlie’s Lawn Care has two three-person crews who lay sod. Crew A can lay 7 skids of sod in 4 hr, while crew B requires 6 hr to do the same job. How long would it take the two crews working together to lay 7 skids of sod?

EXAMPLE Solution



1


1) Familiarize. It takes crew A 4 hr to do the sodding job alone. Then, in 1 hr, crew A can do 1/4 of the job. It takes crew B 6 hr to do the job alone. Then, in 1 hr, crew B can do 1/6 of the job. Working together, the crews can do

EXAMPLE Solution



1


1) Familiarize. In 2 hr, crew A can do 2(1/4) of the job and crew B can do 2(1/6) of the job. Working together, they can do

EXAMPLE Solution



1


1) Familiarize. In t hr, crew A can do t(1/4) of the job and crew B can do t(1/6) of the job. Working together, they can do of the job

EXAMPLE Solution



1


2) Translate.

where 1 represents the idea that the entire job is completed in time t.

EXAMPLE Solution



1


3) Solve.

EXAMPLE Solution



1


4) Check. The check is let to the students. 5) State. It takes hr for crew A and crew B working together to lay 7 skids of sod.


The Work Principle


Suppose a = the time it takes A to do a job, b = the time it takes B to do the same job, and t = the time it takes them to do the job working together. Then




Problems that deal with distance, speed (or rate), and time are called motion problems. Translation of these problems involves the distance formula, d = r · t, and/or the equivalent formulas r = d/t and t = d/r.


Motion Formulas


The following are the formulas for motion problems:

EXAMPLE



2 Animal Speeds


A zebra can run 15 mph faster than an elephant. A zebra can run 8 mi in the same time that an elephant can run 5 mi. Find the speed of each animal.

EXAMPLE Solution



2


1. Familiarize. We first make a drawing. We let r = the speed of the elephant. Then r + 15 = the speed of the zebra.

EXAMPLE Solution



2


1. Familiarize. We first make a drawing. We let r = the speed of the elephant. Then r + 15 = the speed of the zebra.

EXAMPLE Solution



2


2. Translate.

We know that the animals travel for the same length of time. Thus if we solve each equation for t and set the results equal to each other, we get an equation in terms of r.

EXAMPLE Solution



2


2. Translate.

EXAMPLE Solution



2


3. Solve.

The speed of the elephant is 25 mph, and the speed of the zebra is 25 + 15 or 40 mph.

EXAMPLE Solution



2


4. Check. The check is left to the students. 5. State. The speed of the elephant is 25 mph and the speed of the zebra is 40 mph.


b Solve proportion problems.


A proportion involves ratios. A ratio of two quantities is their quotient. For example, 73% is the ratio of 73 to 100, 73/100. The ratio of two different kinds of measure is called a rate. Suppose an animal travels 720 ft in 2.5 hr. Its rate, or speed, is then


Proportion


An equality of ratios, A/B = C/D, is called a proportion. The numbers within a proportion are said to be proportional to each other.

EXAMPLE



4 Environmental Science


The Fish and Wildlife Division of the Indiana Department of Natural Resources recently completed a study that determined the number of largemouth bass in Lake Monroe, near Bloomington, Indiana. For this project, anglers caught 300 largemouth bass, tagged them, and threw them back into the lake. Later, they caught 85 largemouth bass and found that 15 of them were tagged. Estimate how many largemouth bass are in the lake.

EXAMPLE Solution



4


1. Familiarize. The ratio of the number of largemouth bass tagged to the total number of fish in the lake, F, is 300/F. Of the 85 largemouth bass caught later, 15 fish were tagged. The ratio of fish tagged to fish caught is 15/85. 2. Translate. Assuming that the two ratios are the same, we can translate to a proportion.

EXAMPLE Solution



4


3. Solve.

EXAMPLE Solution



4


4. Check. The check is left to the students. 5. State. We estimate that there are about 1700 largemouth bass in the lake.




Proportions arise in geometry when we are studying similar triangles. If two triangles are similar, then their corresponding angles have the same measure and their corresponding sides are proportional.




To illustrate, if triangle ABC is similar to triangle RST, then angles A and R have the same measure, angles B and S have the same measure, angles C and T have the same measure, and


Similar Triangles


In similar triangles, corresponding angles have the same measure and the lengths of corresponding sides are proportional.

EXAMPLE



6 Similar Triangles


Triangles ABC and XYZ below are similar triangles. Solve for z if a = 8, c = 5, and x = 10.

EXAMPLE Solution



6


CHAPTER




OBJECTIVES

6.9 Variation and Applications


a Find an equation of direct variation given a pair of values of the variables.

b Solve applied problems involving direct variation. c Find an equation of inverse variation given a pair of

values of the variables. d Solve applied problems involving inverse variation. e Find equations of other kinds of variation given values

of the variables.

OBJECTIVES



f Solve applied problems involving other kinds of variation.




A dental hygienist earns $24 per hour. In 1 hr, $24 is earned; in 2 hr, $48 is earned; in 3 hr, $72 is earned; and so on. We plot this information on a graph, using the number of hours as the first coordinate and the amount earned as the second coordinate to form a set of ordered pairs: and so on.




Note that the ratio of the second coordinate to the first is the same number for each point:




Whenever a situation produces pairs of numbers in which the ratio is constant, we say that there is direct variation. Here the amount earned varies directly as the time:

or, using function notation, E(t) = 24t. The equation is an equation of direct variation. The coefficient, 24 in the situation above, is called the variation constant. In this case, it is the rate of change of earnings with respect to time.


Direct Variation


If a situation gives rise to a linear function f(x) = kx, or y = kx, where k is a positive constant, we say that we have direct variation, or that y varies directly as x, or that y is directly proportional to x. The number k is called the variation constant, or constant of proportionality.

EXAMPLE



1 Find the variation constant and an equation of variation in which y varies directly as x, and y = 32 when x = 2.


EXAMPLE



1 Find the variation constant and an equation of variation in which y varies directly as x, and y = 32 when x = 2.


The variation constant, 16, is the rate of change of with respect to x. The equation of variation is y = 16x.



The graph of y = kx, k > 0, always goes through the origin and rises from left to right. Note that as x increases, y increases. The constant k is also the slope of the line.

EXAMPLE


b Solve applied problems involving direct variation.

2 Water from Melting Snow


The number of centimeters W of water produced from melting snow varies directly as S, the number of centimeters of snow. Meteorologists have found that, under certain conditions, 150 cm of snow will melt to 16.8 cm of water. To how many centimeters of water will 200 cm of snow melt?

EXAMPLE Solution



2


EXAMPLE Solution



2



c Find an equation of inverse variation given a pair of values of the variables.


A bus is traveling a distance of 20 mi. At a speed of 5 mph, the trip will take 4 hr; at 20 mph, it will take 1 hr; at 40 mph, it will take 1/2 hr, and so on. We plot this information on a graph, using speed as the first coordinate and time as the second coordinate to determine a set of ordered pairs:




Note that the products of the coordinates are all the same number:

Whenever a situation produces pairs of numbers in which the product is constant, we say that there is inverse variation. Here the time varies inversely as the speed:




The equation is an equation of inverse variation. The coefficient, 20, in the situation above, is called the variation constant. Note that as the first number (speed) increases, the second number (time) decreases.


Inverse Variation


If a situation gives rise to a function f(x) = k/x, or y = k/x, where k is a positive constant, we say that we have inverse variation, or that y varies inversely as x, or that y is inversely proportional to x. The number k is called the variation constant, or constant of proportionality.

EXAMPLE



3 Find the variation constant and an equation of variation in which y varies inversely as x, and y = 32 when x = 0.2.


EXAMPLE



3 Find the variation constant and an equation of variation in which y varies inversely as x, and y = 32 when x = 0.2.




It is helpful to look at the graph of y = k/x, k > 0. The graph is like the one shown at right for positive values of x. Note that as x increases, y decreases.

EXAMPLE


d Solve applied problems involving inverse variation.

4 Musical Pitch


The pitch P of a musical tone varies inversely as its wavelength W. One tone has a pitch of 550 vibrations per second and a wavelength of 1.92 ft. Find the pitch of another tone that has a wavelength of 3.2 ft.

EXAMPLE Solution



4


EXAMPLE Solution



4


Next, we use the equation to find the pitch of a tone that has a wavelength of 3.2 ft:


e Find equations of other kinds of variation given values of the variables.


Consider the equation for the area of a circle, in which A and r are variables and π is a constant:

We say that the area varies directly as the square of the radius.



y varies directly as the nth power of x if there is some positive constant k such that y = kxn.

EXAMPLE



5 Find an equation of variation in which y varies directly as the square of x, and y = 12 when x = 2.





From the law of gravity, we know that the weight W of an object varies inversely as the square of its distance d from the center of the earth:



y varies inversely as the nth power of x if there is some positive constant k such that

EXAMPLE



6 Find an equation of variation in which W varies inversely as the square of d, and W = 3 when d = 5.




Consider the equation for the area of a triangle with height h and base b: A = 1/2bh. We say that the area varies jointly as the height and the base. y varies jointly as x and z if there is some positive constant k such that y = kxz.

EXAMPLE



7 Find an equation of variation in which y varies jointly as x and z, and y = 42 when x = 2 and z = 3.





Different types of variation can be combined. For example, the equation

asserts that y varies jointly as x and the square of z, and inversely as w.

EXAMPLE



8 Find an equation of variation in which y varies jointly as x and z and inversely as the square of w, and y = 105 when x = 3, z = 20, and w = 2.


EXAMPLE Solution



8


EXAMPLE



9 Volume of a Tree


The volume of wood V in a tree varies jointly as the height h and the square of the girth g (girth is distance around). If 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 960 m3 and girth is 2 m?

EXAMPLE Solution



9


The equation of variation is

Therefore, the height of the tree is 75 m.

Documents

CHAPTER · 6.2 ; Division and Reciprocals. 6.3 : Least Common Multiples and Denominators . 6.4 : Adding Rational Expressions . 6.5 : Subtracting Rational Expressions . 6.6