7
7.1 Introduction to Rational Expressions 7.2 Multiplication and Division of Rational Expressions 7.3 Addition and Subtraction with Like Denominators 7.4 Addition and Subtraction with Unlike Denominators 7.5 Complex Fractions 7.6 Rational Equations and Formulas 7.7 Proportions and Variation
Rational Expressions
There is nothing wrong with making mistakes. Just don't respond with encores.
—ANONYMOUS
ne of the most significant problems facing the U.S. transportation system is chronic highway congestion. According to a newly released “Highway Statistics,” Americans drove about 3 trillion miles in 2009. Our ability to keep traffic moving smoothly and safely is key to keeping our economy strong, and traffic congestion costs motorists more than $87 billion annually in wasted time and fuel. A 2007 study found that, collectively, Americans spend as many as 4.2 billion hours stuck in traffic each year. If the amount of traffic doubles on a highway, the time spent waiting in traffic may more than double. At times, only a slight increase in the traffic rate can result in a dramatic increase in the time spent waiting. (See Example 6 in Section 7.1.) Mathematics can describe this effect by using rational expressions. In this chapter we introduce rational expressions and some of their applications.
O
ISBN 1-256-49082-2
Source: Randy James, “America: Still Stuck in Traffic.” Time.com, July 9, 2009; U.S. Department of Transportation, 2011.
Have you ever been moving smoothly in traffic, only to come to a sudden halt? Mathematics shows that in certain conditions, if the number of cars on a road increases even slightly, then the movement of traffic can slow dramatically. To understand why this occurs, we will consider how rational expressions can be used to model traffic flow. (See Example 6.)
Basic Concepts
Recall that a rational number is any number that can be expressed as a ratio of two integers p 0. In this chapter, we discuss rational expressions, which can be written as q , where q the ratio of two polynomials. Because examples of polynomials include 3, 2 x, x 2
NEW VOCABULARY n n n n Rational expression Lowest terms Vertical asymptote Probability
4, and x 3 x2 3 x3 x2
1,
it follows that examples of rational expressions include 3 , 2x 2x x
2
4
,
4
, and
1 . 4
RATIONAL EXPRESSION
P A rational expression can be written as Q, where P and Q are polynomials. A rational expression is defined whenever Q 0.
We can evaluate polynomials for different values of a variable. For example, for x = 2 the polynomial x 2 - 3x + 1 evaluates to (2)2 - 3(2) + 1 = - 1. Rational expressions can be evaluated similarly.
EXAMPLE 1
Evaluating rational expressions
If possible, evaluate each expression for the given value of the variable. y2 1 (a) x = 2 (b) y = -4 x + 1 2y - 1 2 - x 5z + 8 (c) 2 z = 1 (d) x = -3 x - 2 z - 2z + 1
Solution
1 1 (a) If x = 2, then x + 1 = 2 + 1 = 1. 3
(b) If y =
4, then 2y - 1 = 2( 4) - 1 = - 16. 9
ISBN 1-256-49082-2
y2
(
4)2
5(1) + 8 + (c) If z = 1, then z2 5z 2z 8 1 = 12 - 2(1) + 1 , or 13, which is undefined because division 0 + by 0 is not possible.
READING CHECK
• When is a rational expression undefined?
UNDEFINED EXPRESSIONS Division by 0 is undefined. As a result, rational expressions are different from polynomials because they are undefined whenever their denominators are 0. For example, the expression in Example 1(c) is undefined when z = 1.
EXAMPLE 2
Determining when a rational expression is undefined
Find all values of the variable for which each expression is undefined. 4 1 1 - 6r 4t (a) (d) 2 (c) 2 (b) x t - 3 r - 4 x + 1
Solution (a) A rational expression is undefined when its denominator is 0. Thus 1 is undefined when x x = 0. (b) The expression t 4t 3 is undefined when its denominator, t 3, is 0, or when t = 3.
1 (c) The expression r 2 - 6r is undefined when its denominator, r 2 4
4, is 0. Here
r2
4 = (r - 2)(r + 2) = 0
implies that the denominator is 0 when r = - 2 or r = 2. (d) In the expression x 2 4 1 the denominator, x 2 1, is never 0 because any real number squared plus 1 is always greater than or equal to 1. Thus this rational expression is defined for all real numbers x.
Now Try Exercises 25, 27, 31, 33
STUDY TIP
You may want to review your notes on fractions. Many of the mathematical concepts that apply to fractions also apply to rational expressions. Fractions were discussed in Chapter 1.
Simplifying Rational Expressions
In Chapter 1, we used the basic principle of fractions, a a#c = . b#c b
8 For example, this basic principle allows us to simplify the fraction 12 as
2 2#4 8 = # = . 12 3 4 3
EXAMPLE 3
Simplifying fractions
Simplify each fraction by applying the basic principle of fractions. 5 36 (a) (b) 10 48
Solution 5 1 5 1 (a) = = 10 2 5 2
# #
(b) -
36 3 # 12 3 = - # = 48 4 12 4
Now Try Exercises 39, 43
ISBN 1-256-49082-2
We can also apply this basic principle to rational expressions. For example, x(x 4(x provided that x 1. 1) x = , 1) 4
NOTE: The simplification at the bottom of the previous page is not valid when x = 1 because the expression is undefined for this x-value. When simplifying a rational expression, we assume that values of the variable that make the rational expression undefined are excluded, unless stated otherwise.
BASIC PRINCIPLE OF RATIONAL EXPRESSIONS
The following property can be used to simplify rational expressions, where P, Q, and R are polynomials. P P#R = Q#R Q
Q and R are nonzero.
P # P NOTE: Q # R = Q R
P P # R = Q # 1 = Q, provided that Q R
0 and R
0.
Like fractions, rational expressions can be written in lowest terms. For example, the 2 rational expression x 2 x 2 x 1 1 can be written in lowest terms by factoring the numerator + + and the denominator and then applying the basic principle of rational expressions.
READING CHECK
• How do you know when a rational expression is written in lowest terms?
(x - 1)(x x2 - 1 = (x + 1)(x x + 2x + 1 x - 1 = x + 1
2
1) 1)
Factor the numerator and the denominator. Apply
PR QR
=
P Q
with R = x + 1.
Because the basic principle of rational expressions cannot be applied further to x - 1, we x + 1 say that this expression is written in lowest terms.
EXAMPLE 4
Simplifying rational expressions
Simplify each expression. 8y 2x + 6 (b) (a) 2 3x + 9 4y (z + 1)(z - 5) (z - 5)(z + 3) x2 - 9 2 x 2 + 7x + 3
(c)
(d)
Solution (a) Factor out the greatest common factor, 4y, in the numerator and the denominator.
2 # 4y 8y 2 = # = 2 y y 4y 4y
Apply
PR QR
=
P Q
with R = 4y.
(b) Start by factoring the numerator and denominator. 2(x 2x + 6 = 3x + 9 3(x 3) 2 = 3) 3
Apply
PR QR
=
P Q
with R = x + 3.
P PR (c) The commutative property allows us to write QR as R Q. R 5) (z + 1)(z z + 1 PR Apply RQ = = (z 5)(z + 3) z + 3
P Q
with R = z - 5.
ISBN 1-256-49082-2
(d) Start by factoring the numerator and the denominator. (x - 3)(x x2 - 9 = 2 (2 x + 1)(x 2 x + 7x + 3
Now Try Exercises 51, 55, 61, 79
Expressions and equations are different concepts. An expression does not contain an equals sign, whereas an equation is a statement that two expressions are equal and always contains an equals sign. For example,
READING CHECK
• How do rational expressions and rational equations differ?
x x are two rational expressions, and x x 4 4
and
2 x
An equation must contain an equals sign.
=
T 2
x
is a rational equation. In this section we evaluate and simplify rational expressions. Later, we will solve rational equations by finding x-values that make the equations true.
A negative sign can be placed in a fraction in a number of ways. For example, 5 = 7 5 7 = 5 7
illustrates three fractions that are equal. This property can also be applied to rational expressions, as demonstrated in the next example.
EXAMPLE 5
Distributing a negative sign
Simplify each expression. -x - 6 10 - z (a) (b) 2 x + 12 z - 10 (c) 5 - x x - 5
Solution (a) Factor - 1 out of the numerator and 2 out of the denominator.
- 1(x -x - 6 = 2 x + 12 2(x (b) Factor - 1 out of the numerator.
6) 1 = 6) 2
- 1( - 10 + z) - 1(z 10 - z = = z - 10 z - 10 z
10) = -1 10
(c) Rewrite the expression with the negative sign in the numerator and then apply the distributive property. Be sure to include parentheses around the numerator. ISBN 1-256-49082-2
- (5 - x) 5 - x x -5 + x = = = x - 5 x - 5 x - 5 x
5 = 1 5
The same answer can be obtained by distributing the negative sign in the denominator. 5 - x 5 - x 5 - x 5 = = = x - 5 - (x - 5) -x + 5 5 x = 1 x
NOTE: The result for Example 5(b) becomes more obvious if we substitute a number for z. For example, if we let z = 6, then
10 - 6 4 10 - z = = = - 1. z - 10 6 - 10 -4
MAKING CONNECTIONS
Negative Signs and Rational Expressions
In general, (b - a) equals - 1(a - b). Thus if a
b, then b - a = - 1. a - b
Applications
N REAL-WORLD CONNECTION The next example is based on the discussion in A Look Into Math for this section and illustrates modeling traffic flow with a rational expression.
EXAMPLE 6
Modeling traffic flow
Suppose that 10 cars per minute can pass through a construction zone. If traffic arrives randomly at an average rate of x cars per minute, the average time T in minutes spent waiting in line and passing through the construction zone is given by T = 1 , 10 - x
where x 6 10. (Source: N. Garber and L. Hoel, Traffic and Highway Engineering.) (a) Complete Table 7.1 by finding T for each value of x.
TABLE 7.1 Waiting in Traffic
x (cars/minute) T (minutes) (b) Interpret the results.
5
7
9
9.5
9.9
9.99
Solution (a) When x = 5 cars per minute, then T = 10 1 5 = 1 minute. Other values are found 5 similarly and are shown in Table 7.2. TABLE 7.2 Waiting in Traffic
x (cars/minute) T (minutes)
5
1 5
7
1 3
9 1
9.5 2
9.9 10
9.99 100
(b) As the average traffic rate increases from 9 cars per minute to 9.9 cars per minute, the time needed to pass through the construction zone increases from 1 minute to 10 minutes. As x nears 10 cars per minute, a small increase in x increases the waiting time dramatically.
Now Try Exercise 99
ISBN 1-256-49082-2
This nonlinear effect for traffic congestion in Example 6 is shown in Figure 7.1, where points from Table 7.2 have been plotted and a curve passing through them has been sketched. A vertical dashed line was also sketched at x = 10. This dashed line is called a
vertical asymptote and indicates that the rational expression is undefined at this value of x. Near the left side of the vertical asymptote, the waiting time T increases dramatically for small increases in x. The graph of T does not intersect or cross this vertical asymptote.
Length of Wait in Traffic
T
10 9 8 7 6 5 4 3 2 1
Wait (minutes)
T=
1 10 – x x = 10
0 1 2 3 4 5 6 7 8 9 10
x
Traffic Rate (cars/minute)
Figure 7.1
TECHNOLOGY NOTE
Making Tables Table 7.2 can also be created with a graphing calculator by using the Ask feature, as illustrated in the following displays.
TABLE SETUP TblStart 5 Tbl 1 Indpnt: Auto Ask Depend: Auto Ask X
5 7 9 9.5 9.9 9.99
CALCULATOR HELP
To make a table of values, see Appendix A (pages AP-2 and AP-3).
Y1
.2 .33333 1 2 10 100
Y 1 1/ ( 10 X )
N REAL-WORLD CONNECTION When a new species of animal is introduced into an area that it did not previously inhabit, its population may grow quickly at first, and then level off over time. Rational expressions can be used to model such situations, as demonstrated in the next example.
EXAMPLE 7
Modeling a fish population
Suppose that a small fish species is introduced into a pond that had not previously held this type of fish, and that its population P in thousands is modeled by P = 3x + 1 , x + 4
where x Ú 0 represents time in months. (a) Complete Table 7.3 by finding P for each value of x. Round to 3 decimal places.
TABLE 7.3 Fish Population
x (months)
ISBN 1-256-49082-2
0
6
12
36
72
P (thousands) (b) How many fish were initially introduced into the pond? (c) Interpret the results shown in your completed table.
Solution 3(0) + 1 (a) When x = 0, the population is P = 0 + 4 = 1 = 0.25 thousand fish. The other 4 values are found similarly and are shown in Table 7.4. TABLE 7.4 Fish Population
x (months) P (thousands)
0 0.25
6 1.9
12 2.313
36 2.725
72 2.855
(b) Table 7.4 shows that initially (when x = 0) there were 0.25 thousand, or 250 fish. (c) The fish population increased quickly at first but then leveled off. This population growth is shown graphically in Figure 7.2, where the population appears to be leveling off at 3 thousand fish.
Fish Population
P
Population (thousands)
3 2 1
P=3
P = 3x + 1 x+4
0
12
24
36
48
60
72
x
Time (months)
Figure 7.2
Now Try Exercise 101
AN APPLICATION INVOLVING PROBABILITY (OPTIONAL) If 10 marbles, one blue and nine red, are placed in a jar, then the probability, or likelihood, of picking the blue mar1 ble at random is 1 chance in 10, or 10. The probability of drawing a red marble at random is 9 9 chances in 10, or 10. Probability is a real number from 0 to 1. A probability of 0, or 0%, indicates that an event is impossible, whereas a probability of 1, or 100%, indicates that an event is certain. Rational expressions are often used to describe probability.
EXAMPLE 8
Calculating probability
Suppose that n balls, numbered 1 to n, are placed in a container and only one ball has the winning number. (a) What is the probability of drawing the winning ball at random? (b) Calculate this probability for n = 100, 1000, and 10,000. (c) What happens to the probability of drawing the winning ball as the number of balls increases?
Solution (a) There is 1 chance in n of drawing the winning ball, so the probability is 1. n 1 1 1 (b) For n = 100, 1000, and 10,000, the probabilities are 100, 1000, and 10,000. (c) As the number of balls increases, the probability of picking the winning ball decreases.
Now Try Exercise 105
ISBN 1-256-49082-2
An expression of the form where P and Q are polynomials with Q 0 A rational expression is undefined for any value of the variable that makes the denominator equal to 0. Factor the numerator and the denominator completely. Then apply P P#R # R = Q. Q
x 2 + 3x - 5 1 x - 3 2x + 9 , , and , x 2x2 - 1 5x 1
1 x - 3 is undefined when x = 3. 5y is undefined when y = 1 y2 - 1
Exercises
11. 13. y + 1 ; y = -2 - 1 4 y2 7z ; z = -2 z - 4
2
CONCEPTS AND VOCABULARY
12.
1. A rational expression can be written as _____, where P P and Q are _____ with Q 0. Q; polynomials 2. Is 2 x 2 x+ 1 a rational expression? Why Yes; both x and 2 x 2 + 1 are polynomials. or why not?
3y - 1 ; y = - 1 -2 y2 + 1 5 ; z = -1 z - 3z + 2
2
14.
Undefined
3. A rational expression is undefined whenever the _____ is equal to 0. denominator
1 4. The rational expression x - a is undefined whenever x = a .
5 15. ; t = -2 3t + 6 4 - x 17. ; x = - 2 -1 x - 4 19. 6 - x ;x = 0 1 x - 6
Undefined
5 6
16. 18. 20.
Undefined
4t 5 ;t = 2t + 5 2 x - 7 ; x = 4 -1 7 - x
5. The basic principle of fractions states that a fraction # a can be simplified by using a # c = b . b c 6. The basic principle of rational expressions can be used to simplify _____ expressions. rational
EVALUATING RATIONAL EXPRESSIONS
8 - 2x ; x = - 5 -1 2x - 8
Exercises 21–24: Complete the table for the given expression. If a value is undefined, place a dash in the table. 21. x x x + 1
Exercises 7–20: If possible, evaluate the expression for the given value of the variable.
ISBN 1-256-49082-2
Exercises 51–88: Simplify the expression. 51. 53. 55. 5x 4 10x 6 8xy 3 6 x 2y 2
1 2x2
52. 54.
1 2
6y 2 9y
5
2y 3
24.
x
2x - 1 x2 - 1
-2
-5 3
-1
—
0
1
1
—
2
1
4y 3x
36 x 2y 5 6x y
6y 4 x3
Exercises 25–38: Find any values of the variable that make the expression undefined. 25. 27. 29. 31. 33. 35. 37. 8 0 x 26. 28. 30. 32. 34. 36. 38. 7 -1 x + 1 7 - z 7 z - 7 3 + y 3y - 7
2
7 3
Exercises 89 and 90: Thinking Generally Complete the statement involving a rational expression. 89. The expression x - a simplifies to -1 . a - x
? 90. The expression - x - a simplifies to x + a. a - x x + a
101. Frog Population Suppose that a frog species is introduced into a wetland area and its population in hundreds is modeled by P = 7x + 3 , x + 6
Exercises 91–98: Refer to Making Connections on page 423. (a) Decide whether you are given an expression or an equation. (b) If you are given an expression, simplify it. If you are given an equation, solve it. 91. x + 1 = 7
(a) Equation (b) 6
where x Ú 0 is time in months. (a) Complete the table by finding P for each given value of x. Round to 2 decimal places. x (months) P (hundreds) 0
0.5
12
4.83
36
6.07
72
6.5
92. x - 4 = 0
2
x 93. x(x + 1)
(a) Expression (b)
95. 97.
(a) Expression
x2 - 4 x + 2
1 x + 1
x - 2 94. (x - 2)(x - 8) 96. x - 4 8 - 2x
(a) Equation
(b) -2, 2
(b) What was the initial frog population? 50 (c) Interpret the results in your completed table.
The population increased quickly at first, but then leveled off.
(a) Expression (b)
1 x - 8
102. Insect Population Suppose that an insect population in thousands per acre is modeled by P = 5x + 2 , x + 1
(b) x - 2
(a) Expression (b) - 1 2
(a) Equation
x = 1 2(1 + 3)
98.
(b) 8
(a) Equation
x 8 + 2 = 2 2
(b) 4
APPLICATIONS of 3 vehicles/min, the average wait is 1 min. 2
99. (a) 1; when traffic arrives at an average rate 2
where x Ú 0 is time in months. (a) Complete the table by finding P for each given value of x. Round to 3 decimal places. x (months) P (thousands) 0
2
12
4.769
36
4.919
60
4.951
99. Modeling Traffic Flow (Refer to Example 6.) Five vehicles per minute can pass through a construction zone. If the traffic arrives randomly at an average rate of x vehicles per minute, the average time T in minutes spent waiting in line and passing through 1 the construction zone is given by T = 5 - x for x 6 5. (Source: N. Garber.) (a) Evaluate T for x = 3 and interpret the result. (b) Complete the table and interpret the results.
As x nears 5 vehicles/min, a small increase in x increases the wait dramatically.
(b) What was the initial insect population? 2000 (c) Interpret the results in your completed table.
The population increased quickly at first, but then leveled off.
103. Probability Suppose that a coin is flipped. What is the probability that a head appears? 1 2 104. Probability A die shows the numbers 1, 2, 3, 4, 5, and 6. If each number has an equal chance of appearing on any given roll, what is the probability that a 2 or 4 appears? 2, or 1 3 6
x T
2
1 3
4
1
4.5
2
4.9
10
4.99
100
100. Standing in Line A worker at a poolside store can serve 20 customers per hour. If children arrive randomly at an average rate of x per hour, then the average number of children N waiting in line is given by 2 N = 400 x 20x for x 6 20. (Source: N. Garber.) (a) Complete the table. x N
ISBN 1-256-49082-2
5
0.083
10
0.5
18
8.1
19
18.05
20
—
(b) Compare the number of children waiting in line if the average rate increases from 18 to 19 children per hour. It increases by almost 10 children.
105. Probability (Refer to Example 8.) Suppose that there are n balls in a container and that three balls have a winning number. If a ball is drawn randomly, do each of the following. (a) Write a rational expression that gives the probability of drawing a winning ball. 3 n (b) Write a rational expression that gives the probability of not drawing a winning ball. Evaluate your expression for n = 100 and interpret the result. n - 3 97 n ; 100 ; there is a 97% chance that a winning ball will not be drawn.
106. Surface Area of a Cylinder If a cylindrical container has a volume of p cubic feet, then its surface area S in square feet (excluding the top and bottom) is given by S = 2p, where r is the radius of the cylinder. r r (a) Give the equation of the vertical asymptote. (b) Explain how the graph relates to traffic flow. 109.
T
6 5
Wait (minutes)
T=
4 3 2 1 0 1 2 3
1 5–x
(a) x = 5 (b) As the average arrival rate nears 5 cars/min, a small increase in x increases the waiting time dramatically.
(a) Calculate S when r = 1 foot. 4p 12.6 ft 2 2 (b) What happens to this surface area when r becomes large? Sketch this situation.* It becomes very small. (c) What happens to the surface area when r becomes small (nearly 0)? Sketch this situation.*
It becomes very large.
4
5
6
x
Traffic Rate (cars/minute)
110.
T
12 10
Wait (minutes)
107. Distance and Time A car is traveling at 60 miles per hour. (a) 6 hr (a) How long does it take the car to travel 360 miles? (b) Write a rational expression that gives the time M that it takes the car to travel M miles. 60 108. Distance and Time A bicyclist rides uphill at 10 miles per hour for 5 miles and then rides downhill at 20 miles per hour for 5 miles. What is the bicyclist’s average speed? (Hint: Average speed equals distance divided by time.) 13.3 mph Exercises 109 and 110: Traffic Flow (Refer to Example 6.) The figure shows a graph of the waiting time T in minutes at a construction zone when cars are arriving randomly at an average rate of x cars per minute.
T=
1 8–x
8 6 4 2 0 2 4 6 8 10 12
(a) x = 8 (b) As the average arrival rate nears 8 cars/min, a small increase in x increases the waiting time dramatically.
x
Traffic Rate (cars/minute)
WRITING ABOUT MATHEMATICS
111. What is a rational expression? When is a rational expression undefined?
5x 5x 112. Does the rational expression 10x + 24 equal 10x + 2 ? + 4 Explain your answer.
Group Activity
Working with Real Data
Directions: Form a group of 2 to 4 people. Select someone to record the group’s responses for this activity. All members of the group should work cooperatively to answer the questions. If your instructor asks for your results, each member of the group should be prepared to respond.
(a) No; the data are not linear.
Students Per Computer In the early years of personal computers, school districts could not afford to buy a computer for every student. As the price of computers decreased, more and more school districts were able to move toward this goal. The following table lists numbers of students per computer during these early years. Year Students/Computer Year Students/Computer 1983 125 1991 18 1985 50 1993 14 1987 32 1995 10 1989 22 1997 6
(a) Make a scatterplot of the data. Would a straight line model the data accurately? Explain.* (b) Discuss how well the formula
S models the data quite well.
S =
125 , 1 + 0.7(x - 1983)
x Ú 1983
models these data, where S represents the students per computer and x represents the year. (c) In what year does the formula suggest that there were about 17 students per computer? 1992
7.2 MULTIPLICATION AND DIVISION OF RATIONAL EXPRESSIONS
431
7.2
Multiplication and Division of Rational Expressions
Review of Multiplication and Division of Fractions ● Multiplication of Rational Expressions ● Division of Rational Expressions
A LOOK INTO MATH N
Stopping distance for a car can vary depending on the road conditions. If the road is slippery, more distance is needed to stop. Also, cars that are traveling downhill require additional stopping distance. Rational expressions are frequently used by highway engineers to estimate the stopping distance of a car on slippery surfaces or on hills. (See Example 4.) In previous chapters we reviewed how to add, subtract, multiply, and divide real numbers and polynomials. In this section we show how to multiply and divide rational expressions; in the next section we discuss addition and subtraction of rational expressions.
Review of Multiplication and Division of Fractions
To multiply two fractions we use the property a b
#c
d
=
ac . bd
In the next example we review multiplication of fractions. (For a full review of fractions, see Section 1.2.)
Multiplication of Rational Expressions
Multiplying rational expressions is similar to multiplying fractions.
PRODUCTS OF RATIONAL EXPRESSIONS
To multiply two rational expressions, multiply the numerators and multiply the denominators. That is, A B where B and D are nonzero.
READING CHECK
• How do we multiply rational expressions?
#C
D
=
AC , BD
EXAMPLE 3
Multiplying rational expressions
Multiply and simplify to lowest terms. Leave your answers in factored form. x - 1 # x + 3 3 # 2x - 5 (b) (a) x x - 1 4x x - 1 (c) x2 - 4 x + 3
#x+3 x + 2
(d)
4 x + 3x + 2
2
#x
2
+ 2x + 1 8
Solution 3 2x (a) x x
#
3(2 x 5 = 1 x(x
5) 1) 1)(x + 3) 4 x(x 1)
Multiply the numerators and the denominators. Multiply the numerators and the denominators. Simplify.
(b)
x - 1 4x
#x+3 x - 1
= =
(x
x + 3 4x (x - 2)(x + 2) x + 3
(c)
STUDY TIP
Have you been to your instructor’s office lately? Visit with your instructor about your progress in the class to be sure that your work is complete and your grades are up-to-date.
x2 - 4 x + 3
#x+3 x + 2
= =
#x+3 x + 2
Factor. Multiply the numerators and the denominators. Simplify. Multiply the numerators and the denominators. Factor. Simplify.
ISBN 1-256-49082-2
7.2 MULTIPLICATION AND DIVISION OF RATIONAL EXPRESSIONS
433
EXAMPLE 4
Estimating stopping distance
If a car is traveling at 60 miles per hour on a slippery road, then its stopping distance D in feet can be calculated by D = 3600 30
# 1, x where x is the coefficient of friction between the tires and the road and 0 6 x … 1. The more slippery the road is, the smaller the value of x. (Source: L. Haefner, Introduction to Transportation Systems.) (a) Multiply and simplify the formula for D. (b) Compare the stopping distance on an icy road with x = 0.1 to the stopping distance on dry pavement with x = 0.4.
Solution (a) Because
3600 30
#1 x =
120 # 30 120 3600 = = , x 30x x # 30
it follows that D = 120. x (b) When x = 0.1, D = 120 = 1200 feet, and when x = 0.4, D = 120 = 300 feet. A car 0.1 0.4 traveling at 60 miles per hour on an icy road requires a stopping distance that is 4 times that of a car traveling at the same speed on dry pavement.
Now Try Exercise 73
Division of Rational Expressions
Dividing rational expressions is similar to dividing fractions.
QUOTIENTS OF RATIONAL EXPRESSIONS
READING CHECK
• How do we divide rational expressions?
To divide two rational expressions, multiply by the reciprocal of the divisor. That is, C A A , = B D B where B, C, and D are nonzero.
# D,
C
EXAMPLE 5
Dividing rational expressions
Divide and simplify to lowest terms. 10 x2 - 9 5 , (x - 3) , (b) 2 (a) 2x x - 4 x + 4
Solution
(c)
x2 - x x , 2 x - 2 x - x - 2
(a)
ISBN 1-256-49082-2
5 2x
10 x 4
= = =
5 2x
#x
4 10
Multiply by the reciprocal of the divisor. Multiply the numerators and the denominators. Simplify. Note that
5 20
Multiply by the reciprocal of the divisor. Multiply the numerators and the denominators. Factor the numerator and the denominator. Simplify.
(x 2 - x)(x - 2) (x 2 - x - 2)x 2) 2)
x(x - 1)(x x(x + 1)(x x - 1 = x + 1
Now Try Exercises 49, 57, 65
7.2
Putting It All Together
EXPLANATION EXAMPLES
CONCEPT
Multiplication of Rational Expressions
If A, B, C, and D represent rational expressions where B and D are nonzero, multiply the numerators and multiply the denominators: A B
4x x - 1
#x-1 x + 1
= =
4 x(x - 1) (x - 1)(x + 1) 4x x + 1
#C
D
=
AC . BD
Then simplify the result to lowest terms. Division of Rational Expressions If A, B, C, and D represent rational expressions where B, C, and D are nonzero, multiply the first expression by the reciprocal of the second expression: A C A , = B D B x + 1 x + 1 x + 1 , = x - 3 x - 5 x - 3 = =
7.2 MULTIPLICATION AND DIVISION OF RATIONAL EXPRESSIONS
435
7.2
Exercises
37.
(t + 1)(t + 2)
CONCEPTS AND VOCABULARY
1. To multiply rational expressions, multiply the _____ and multiply the _____. numerators; denominators 2. To divide two rational expressions, multiply the first expression by the _____ of the second expression. reciprocal (t + 1)2 t + 2 x2 x2 + 4 z2 - 1 z2 - 4
x - 2 2 - x , -2 3x 6x z + 2 z + 2 , z + 1 z - 1 z - 1 z + 1
8y( y + 7) 25. 12y( y + 7) 27.
6( y + 1) 26. 12( y + 1) x + 2 x - 2
x 2x - 1
x + 1 x + 1 , x 2x - 1 z + 7 z + 7 , 1 z - 4 z - 4
x(x + 2)(x + 3) x(x - 2)(x + 3)
28.
2(x + 1)(x - 1) 4(x + 1)(x - 1)
1 2
55.
3y + 4 3y + 4 , 2y + 1 y + 2
y + 2 2y + 1
y + 5 y , y - 2 y + 3
1 4t
Exercises 29–48: Multiply and simplify to lowest terms. Leave your answers in factored form. 29. 8 x
t + 1 t2 - 1 , 57. 2 4 t + 1 59.
4(t - 1) t2 + 1
4 8 58. 3 , 2 2t t 60.
(y + 5)( y + 3) y( y - 2)
#x+1 x 8(x + 1) x2
30.
7 2x
#
x x - 1
7 2(x - 1)
y + 3 y2 - 9 , 2 y + 5 y - 25
2
y - 3 y - 5
8 + x 31. x
ISBN 1-256-49082-2
#
x - 3 x + 8 z - 7
x - 3 x
5x 2 + x 32. 2x - 1 34. 2z + 1 3z x + 1 x + 3
# #
y + 1 y2 - 1 , 2 y - 4 y - 16y
1 x
5x + 1 2x - 1
61.
5 2x - 4x x - 2 x - 4 , , 2 x 62. 2 2x - 1 2x - 1 x + 1 x + x z - 3 2z 2 - 5z - 3 , 2 z - 4 z + z - 20 z + 3 z 2 + 12z + 27 , 2 z + 2 z - 5z - 14
2z + 1 z + 5
t2 - 1 , (t + 1) t 2 + 5t - 6 t 2 - 2t - 3 , (t - 3) t 2 - 5t - 6 a - b a - b , a + b 2a + 3b
1 t + 6
D =
2500 30
#
1 , x + 0.3
1 t - 6
2a + 3b a + b
where x equals the slope of the hill. (Source: L. Haefner.) (a) Multiply and simplify the formula for D. D = 3(x 2500.3) + (b) Compare the stopping distance for an uphill slope of x = 0.1 to a downhill slope of x = - 0.1.
About 208.3 ft; about 416.7 ft; downhill is twice as long.
x 3 - y3 x 2 + xy + y 2 , x - y x 2 - y2
2
x - y x + y
x - y 1 , x - y x + 2 xy + y 2 (x + y)2 a2 - b 2 a - b , 2 2 2a + 3b 4a - 9b a + b 2a - 3b
75. Probability Suppose that one jar holds n balls and that a second jar holds n + 1 balls. Each jar contains one winning ball. (a) The probability, or chance, of drawing the winning ball from the first jar and not drawing it 1 from the second jar is n + 1 1 n
#
n . n + 1
1 100
c 71. Thinking Generally Simplify a - b # b - b . 1 b - c - a
72. Thinking Generally Simplify
APPLICATIONS
a - b b - c
,
b - a a - b a - b. c - b
Simplify this expression. (b) Find this probability for n = 99.
73. Stopping on Slippery Roads (Refer to Example 4.) If a car is traveling at 30 miles per hour on a slippery road, then its stopping distance D in feet can be calculated by D = 900 30
# 1, x where x is the coefficient of friction between the tires and the road and 0 6 x … 1. (Source: L. Haefner.) (a) Multiply and simplify the formula for D. D = 30 x (b) Compare the stopping distance on an icy road with x = 0.1 and on dry pavement with x = 0.4.
300 ft; 75 ft; dry pavement is one-fourth as long.
76. U.S. AIDS Cases The cumulative number of AIDS cases C in the United States from 1982 to 1994 can be modeled by C = 3200x 2 + 1586, and the cumulative number of AIDS deaths D from 1982 to 1994 can be modeled by D = 1900x 2 + 619. In these equations x = 0 corresponds to 1982, x = 1 corresponds to 1983, and so on until x = 12 corresponds to 1994.* (Source: U.S. Department of Health.) (a) Write the rational expression D in terms of x. C (b) Evaluate your expression for x = 4, 7, and 10. Round your answers to the nearest thousandth. Interpret the results. (c) Explain what the rational expression D represents. C
WRITING ABOUT MATHEMATICS
74. Stopping on Hills If a car is traveling at 50 miles per hour on a hill with wet pavement, then its stopping distance D is given by
77. Explain how to multiply two rational expressions. 78. Explain how to divide two rational expressions.
SECTIONS 7.1 and 7.2
Checking Basic Concepts
5. Waiting in Line Customers are waiting in line at a department store. They arrive randomly at an average rate of x per minute. If the clerk can wait on 2 customers per minute, then the average time in min1 utes spent waiting in line is given by T = 2 - x for x 6 2. (Source: N. Garber, Traffic and Highway Engineering.) (a) Complete the table. x T (b) What happens to the waiting time as x increases but remains less than 2? 0.5 1.0 1.5 1.9
ISBN 1-256-49082-2
1. If possible, evaluate the expression x 2 3 1 for x = - 1 and x = 3. 2. Simplify to lowest terms. 6 x 3y 2 5x - 15 (b) (a) 2 3 x - 3 15x y x2 - x - 6 x 2 + x - 12
(c)
3. Multiply and simplify to lowest terms. 2x + 4 # x + 1 4 # 2x (b) 2 (a) 3x 6 x - 1 x + 2 4. Divide and simplify to lowest terms. 14 x2 + x x 7 , 3 (b) , (a) 2 x - 3 x - 3 3z 5z
7.3 ADDITION AND SUBTRACTION WITH LIKE DENOMINATORS
437
7.3
Addition and Subtraction with Like Denominators
Review of Addition and Subtraction of Fractions ● Rational Expressions with Like Denominators
A LOOK INTO MATH N
When companies manufacture a large number of items, quality control is important. For example, suppose that a company makes computer flash drives. Because it is not practical to check every flash drive to make sure that it works properly, inspectors often check a random sample. By using mathematics and rational expressions, this technique helps determine the likelihood that all the flash drives are good. (See Example 6.) In this section we discuss methods for adding and subtracting rational expressions with like denominators. These methods are similar to the ones used to add and subtract fractions with like denominators.
Review of Addition and Subtraction of Fractions
In Section 1.2 we demonstrated how the property b a + b a + = c c c can be used to add fractions with like denominators. For example, 2 3 + 2 5 3 + = = 7 7 7 7 b a - b a = c c c is used. For example, 4 2 - 4 2 2 = = 5 5 5 5
STUDY TIP
By this time in the semester, it is likely that you know some of your classmates. Have you started or joined a study group? Be sure not to miss the opportunity to study math with your classmates.
Add the numerators.
To subtract two fractions with like denominators, the property
TECHNOLOGY NOTE
Arithmetic of Fractions Many calculators have the capability to perform addition and subtraction of fractions, as illustrated in the following figures. Compare these results with those from Example 1.
( 3/8 ) ( 4/8 ) Frac 7/8 ( 5/9 ) ( 1/9 ) Frac 2/3 ( 12/5 ) ( 7/5 ) Fra c
CALCULATOR HELP
To express a result as a fraction on a calculator, see Appendix A (pages AP-1 and AP-2).
1 ( 23/20 ) ( 13/20 ) Frac 1/2
Rational Expressions with Like Denominators
Addition and subtraction of rational expressions with like denominators are similar to addition and subtraction of fractions. The following property can be used to add two rational expressions with like denominators.
SUMS OF RATIONAL EXPRESSIONS
To add two rational expressions with like denominators, add their numerators. Keep the same denominator. A B A + B + = C C C
C is nonzero.
When we add rational expressions with like denominators, we add the numerators. Then we combine like terms and simplify the resulting expression by applying the basic - x principle of rational expressions. For example, we can add x 2 x 1 and 1 + 1 as follows. + x 2x x 1 + 1 - x 2x + 1 - x = x 1 x 1 = = 2x - x + 1 x + 1 x x 1 1
Add the numerators. Commutative property Combine like terms. Simplify.
READING CHECK
• How do we add rational expressions with like denominators?
= 1
- x It is important to understand that the expressions x 2 x 1 + 1 + 1 and 1 are equivalent + x expressions. That is, they are equal for every value of x except - 1, for which the first expression is undefined. In the next example, we add rational expressions with like denominators and simplify the result to lowest terms.
EXAMPLE 2
Adding rational expressions with like denominators
Add and simplify to lowest terms. 3 2 z 2 (a) + (b) + b b z + 2 z + 2 (c) 1 x - 1 + 2 2 x + x x + x (d) t2 + t 1 - 3t + t - 1 t - 1
ISBN 1-256-49082-2
7.3 ADDITION AND SUBTRACTION WITH LIKE DENOMINATORS
439
Solution 3 2 3 + 2 5 (a) + = = b b b b z 2 z + 2 (b) + = z 2 z 2 z 2
Add the numerators. Add the numerators. Simplify. Add the numerators. Factor the denominator. Simplify.
= 1 (c) 1 x - 1 + 1 x - 1 + 2 = x2 x x x x2 x = = (d) x x(x + 1) 1 x + 1
t2 + t 1 - 3t t 2 + t + 1 - 3t + = t 1 t 1 t 1 = = t 2 - 2t + 1 t - 1 (t - 1)(t 1) t 1
Add the numerators. Combine like terms. Factor the numerator. Simplify.
= t - 1
Now Try Exercises 19, 25, 33, 35
EXAMPLE 3
Adding rational expressions with two variables
Add and simplify to lowest terms. b 5 a 4 + 2 + (b) 2 (a) 2 xy xy a - b a - b2
Solution 4 5 4 + 5 9 (a) + = = xy xy xy xy a b a + b (b) 2 + 2 = 2 2 2 a b a b a b2
(c)
1 -1 + x - y y - x
Add the numerators. Add the numerators. Factor the denominator. Simplify.
= =
a b (a - b)(a 1 a - b
b)
1 (c) First write x - y + y - 1 x with a common denominator. Note that if we multiply the second term by 1, written in the form - 1, it becomes -1
-1 y - x
#
( - 1)( - 1) 1 1 1 = = = . x - y 1 ( y - x)( - 1) -y + x
Thus the given sum can be simplified as follows.
ISBN 1-256-49082-2
1 -1 1 1 + = + x - y y - x x y x y =
Now Try Exercises 51, 53, 55
Next we consider subtraction of rational expressions with like denominators.
DIFFERENCES OF RATIONAL EXPRESSIONS
To subtract two rational expressions with like denominators, subtract their numerators. Keep the same denominator. B A - B A = C C C
C is nonzero.
READING CHECK
• How do we subtract rational expressions with like denominators?
Subtraction of rational expressions with like denominators is similar to addition except that instead of adding numerators, we subtract them. For example, the expressions x 3x 4 and x 2 x 4 have like denominators and can be subtracted as follows. 3x x 4 2x x 4 = = 3x - 2 x x 4 x x 4
Subtract the numerators. Combine like terms.
In the next example, we subtract rational expressions with like denominators and simplify the result to lowest terms.
EXAMPLE 4
Subtracting rational expressions with like denominators
Subtract and simplify to lowest terms. 3y 2y 1 a + 1 (b) (a) a a 3y - 1 3y - 1
Solution a + 1 1 a + 1 - 1 a (a) = = = 1 a a a a
(c)
-2 1 + x 2 2 x + 5x - 3 2 x + 5x - 3
2
2y - 3y y y = or 3y 1 3y 1 3y 1 3y - 1 3y - 1 1 + x - ( - 2) -2 1 + x = (c) 2 2 2x 5x 3 2x 5x 3 2x2 5x 3 x 3 = (2 x - 1)(x 3) (b) 2y 3y = =
Now Try Exercises 21, 27, 67
1 2x - 1
If the numerator of the second fraction in a difference has more than one term, it is important to put parentheses around the second numerator. x + 6 2x 1 3 2x (3 x + 6 x = 1 2x 1 = = = ( x + 6 - 3 2x + 1 x + 6 - 3 2x + 1 2x + 3 2x + 1 x x) x)
Subtract the numerators; insert parentheses. Distributive property
ISBN 1-256-49082-2
7.3 ADDITION AND SUBTRACTION WITH LIKE DENOMINATORS
441
NOTE: If parentheses were not inserted in the previous calculation, the numerator would be
x + 6 - 3 which would give an incorrect result.
x = 3,
EXAMPLE 5
Subtracting rational expressions with like denominators
Subtract and simplify to lowest terms. x + y x - y x - 1 2x (b) (a) x + 1 x + 1 3y 3y
Solution 2x (a) x 1
x x
(x 1) 2x 1 = 1 x 1 = = 2x - x + 1 x + 1 x x 1 1 (x 3y y)
Subtract the numerators. Distributive property Simplify the numerator. Simplify. Subtract the numerators. Distributive property Simplify the numerator. Simplify.
= 1 (b) x + y 3y x 3y y = = = =
Now Try Exercises 31, 59
x + y
x + y - x + y 3y 2y 3y 2 3
N REAL-WORLD CONNECTION A Look Into Math for this section discusses how applications involving quality control use rational expressions. The next example illustrates one way this can occur.
EXAMPLE 6
Analyzing quality control
A container holds a mixture of 8-GB and 16-GB computer flash drives. In this container, there is a total of n flash drives, including 2 defective 8-GB flash drives and 4 defective 16-GB flash drives. If a flash drive is picked at random by a quality control inspector, then the probability, or chance, of one of the defective flash drives being chosen is given by the expression 2 + 4. n n (a) Simplify this expression. (b) Interpret the result.
Solution (a) Because the denominators are the same, we simply add the numerators.
ISBN 1-256-49082-2
4 2 + 4 6 2 + = = n n n n
Add the numerators.
(b) There are 6 chances in n that a defective flash drive is chosen.
Now Try Exercise 73
For polynomials A, B, and C, where C is nonzero, A B A + B + = . C C C
x 1 - x x + 1 - x 1 + = = x + 1 x + 1 x + 1 x + 1 2x x 2x + x 3x + 2 = 2 = 2 x - 1 x - 1 x - 1 x - 1
2
Subtraction of Rational Expressions
For polynomials A, B, and C, where C is nonzero, A B A - B = . C C C If B consists of more than one term, put parentheses around B and apply the distributive property.
2 x - (x + 2) x + 2 2x - 2 = x - 4 x - 4 x2 - 4
2
= = =
2x - x - 2 x2 - 4 x - 2 (x + 2)(x - 2) 1 x + 2
7.3
Exercises
ADDITION AND SUBTRACTION OF RATIONAL EXPRESSIONS
CONCEPTS AND VOCABULARY
1. When adding two rational expressions with like denominators, add their _____. The _____ do not change. numerators; denominators
Exercises 19–70: Simplify to lowest terms. 19. 21. 23. 25. 2 1 + x x
3 x
2. When subtracting two rational expressions with like denominators, subtract their _____. The _____ do not change. numerators; denominators A B + = 3. C C
A + B C
20.
1 2
9 7 x x
2 x
A B = 4. C C
A - B C
7 7 + 2x 4x 4x
22. 24. 26.
4z 4z + 3
2x + 1 x - 1 + 5x 5x
3 5
ADDITION AND SUBTRACTION OF FRACTIONS
y + 3 2y - 12 + 3 y - 3 y - 3 x -3 + 1 x - 3 x - 3 5z z 4z + 3 4z + 3 t + 5 t + 7 + 2 t + 6 t + 6 5x 3x - 3 1 2x + 3 2x + 3 4 x - 4 + 2 x2 - x x - x
1 x - 1
5 - y 3y - 1 + 2 y + 2 y + 2 2x 1 + 1 2x + 1 2x + 1 z 1 - z 2z + 1 2z + 1 4t - 13 1 - t + 3 t - 4 t - 4 x 2x - 5 1 5 - x 5 - x 1 2x - 2 + 4x2 - 1 4x2 - 1
ISBN 1-256-49082-2
2z - 1 2z + 1
7.3 ADDITION AND SUBTRACTION WITH LIKE DENOMINATORS
443
3 - 3z 3x z2 - 1 x2 + 2 + + z - 1 36. z - 2 z - 2 x + 1 x + 1x + 2 x2 + 4x - 1 x2 - 4x - 5 37. 2 4x + 2 4x + 2 35. 2x2 - x + 5 x 2 - x + 14 38. 1 x2 - 9 x2 - 9 3y 2y - 5 39. + y - 1 5 5 41. x + y x - y + 4 4 x 2
67. 68. 69. 70.
2x + 6 3x + 7 2 3x - 2 x - 5 3x - 2 x - 5
2
1 3x - 5
3x - 1 5x - 4 2 2 x - 7x + 6 2 x - 7x + 6
2
1 x - 2
9y 2 4x2 2 x - 3y 2 x + 3y 2 x + 3y y3 x3 - 2 x - y x 2 + xy + y 2 x + xy + y 2
3y - 22 8y 40. + y - 2 11 11 42. x + y x - y 4 4 y 2
3 5 2 71. Thinking Generally If 3 + x + 3 + x equals 10 , what must be true about x? It equals 7. 8 4 72. Thinking Generally If 6 + x - 3 + y equals 0, what must be true about x and y? x = 2y
z2 + 4 4z z 2 + 2z 1 43. z - 2 44. + z - 2 z - 2 z + 1 z + 1 45. 2 x - 5x 3 2x + 1 2x + 1
2 2
z + 1
46.
x - 3
2x + 5
2x 9x + 10 + x + 2 x + 2
7n 2n2 - n + 5
APPLICATIONS
4n 3n + 47. 2 2 2n - n + 5 2n - n + 5 48. 49. 50. 51. 53. 1 n - 2 n + n + 1 n + n + 1
2
n - 1 n2 + n + 1
73. Quality Control (Refer to Example 6.) A container holds a total of n + 1 batteries. In this container, there are 6 defective AA batteries, 5 defective C batteries, and 3 defective D batteries. If a battery is chosen at random by a quality control inspector, the probability, or chance, of one of the defective batteries being chosen is 6 5 3 + + . n + 1 n + 1 n + 1 (a) Simplify this expression. n 14 1 + (b) Evaluate the simplified expression for n = 99 7 and interpret the result. 50; when there are 100 batter-
1 2 3 + + x + 3 x + 3 x + 3
6 x + 3
x 1 2x + 1 + 2x - 5 2x - 5 2x - 5 8 1 + ab ab
9 ab
3x 2x - 5
52. 54.
6 9 + xy xy
15 xy
ies, there are 7 chances in 50 that a defective battery is chosen.
y x + 2 (x + y) (x + y)2 5 -5 + x - y y - x
1 x + y
x - 2y y + 2 2 2 x - y x - y2 -4 4 + y - x x - y
1 x + y
55. 57. 59. 61. 63.
ISBN 1-256-49082-2
10 x - y
56. 58. 60. 62. 64.
8 x - y
8 8 + 0 a - b b - a a + b a - b 4a 4a b 2a
6 6 0 + x - y y - x x - y x + 9y 2y - x 5x 5x y x 1 x - y x - y a2 2ab + b 2 + a + b a + b
1 x + 2
74. Intensity of a Light Bulb The farther a person is from a light bulb, the less intense its light. The equa19 tion I = 4d 2 approximates the light intensity from a 60-watt light bulb at a distance of d meters, where I is measured in watts per square meter. (Source: R. Weidner.) (a) Find I for d = 2 meters and interpret the result. (b) The intensity of light from a 100-watt light bulb 32 is about I = 4d 2. Find an expression for the sum of the intensities of light from a 100-watt bulb and a 60-watt bulb. (a) 19 = 1.1875; the intensity is 16
1.1875 W/m2 at 2 m. (b)
51 4d 2
75. Explain how to add two rational expressions with like denominators. Give an example. 76. Explain how to subtract two rational expressions with like denominators. Give an example.
Addition and Subtraction with Unlike Denominators
Finding Least Common Multiples ● Review of Fractions with Unlike Denominators ● Rational Expressions with Unlike Denominators
A LOOK INTO MATH N
The sum and difference of rational expressions frequently occur in the design of electrical devices, such as smart phones and HD televisions. (See Example 7 and the discussion preceding it.) In Section 7.3, we added and subtracted rational expressions with like denominators. Although the denominators of rational expressions are often unlike, rational expressions can still be added and subtracted after a common denominator is found. One way to find the least common denominator for a sum or difference of rational expressions is to find the least common multiple of the denominators. In the following subsection we show how to find the least common multiple.
Finding Least Common Multiples
N REAL-WORLD CONNECTION Two friends work part-time at a store. The first person works every sixth day, and the second person works every eighth day. If they both work today, how many days will pass before they work on the same day again? We can answer this question by listing the days that each person works.
First person: Second person:
6, 8,
12, 18, 24, 16, 24, 32,
30, 36, 42, 40, 48, 56,
48, 64
54
After 24 days, the two friends work on the same day. The next time is after 48 days. The numbers 24 and 48 are common multiples of 6 and 8. (Find another.) However, 24 is the least common multiple (LCM) of 6 and 8 because it is the smallest common multiple. Another way to find the least common multiple of 6 and 8 is to factor each number into prime numbers. 6 = 2 # 3 and 8 = 2 # 2 # 2
STUDY TIP
The methods discussed in Chapter 1 for finding the LCM of two numbers can also be applied to find the LCM of two polynomials.
To find the least common multiple, first list each factor the greatest number of times that it occurs in either factorization. Then find the product of these numbers. For this example, the factor 2 occurs three times in the factorization of 8 and only once in the factorization of 6, so list 2 three times. The factor 3 appears only once in the factorization of 6, so list it once: The least common multiple is their product: 2 # 2 # 2 # 3 = 24. This same procedure can also be used to find the least common multiple of two or more polynomials. 2, 2, 2, 3.
FINDING THE LEAST COMMON MULTIPLE
READING CHECK
• How do we find the least common multiple of two polynomials?
The least common multiple (LCM) of two or more polynomials can be found as follows.
STEP 1: STEP 2: STEP 3:
Factor each polynomial completely. List each factor the greatest number of times that it occurs in any factorization. Find the product of this list of factors. The result is the LCM.
EXAMPLE 1
Finding least common multiples
Find the least common multiple of each pair of expressions. (a) 2 x, 5x 2 (b) x 2 - x, x - 1 (c) x + 2, x - 3 (d) x 2 + 2 x + 1, x 2 + 3x + 2
7.4 ADDITION AND SUBTRACTION WITH UNLIKE DENOMINATORS
445
Solution (a) STEP 1: Factor 2 x and 5x 2 completely.
2x = 2 # x
and 5x 2 = 5 # x # x
In either factorization, the factor 2 occurs at most once, the factor 5 occurs at most once, and the factor x occurs at most twice. The list of factors is 2, 5, x, x. STEP 3: The LCM equals the product
STEP 2:
2 # 5 # x # x = 10x 2. 1)
(b) STEP 1: Factor x 2 - x and x - 1 completely. Note that x - 1 cannot be factored. x 2 - x = x(x
STEP 2: STEP 3:
and x - 1 = x
1
(c) STEP 1:
STEP 2: STEP 3: STEP 1:
(d)
Both factors, x and x - 1, occur at most once in either factorization. The list of factors is x, (x 1). The LCM is the product x(x 1), or x 2 - x. Neither x + 2 nor x - 3 can be factored. The list of factors is (x + 2), (x - 3). The LCM is the product (x + 2)(x - 3), or x 2 - x - 6. Factor x 2 + 2 x + 1 and x 2 + 3x + 2 completely. 1)(x 1) and x 2 + 3x + 2 = (x + 1)(x 2)
x 2 + 2 x + 1 = (x
STEP 2:
In either factorization, the factor (x + 1) occurs at most twice and the factor (x + 2) occurs at most once. The list is (x 1), (x 1), (x 2). STEP 3: The LCM is the product (x 1)2(x 2).
Now Try Exercises 15, 19, 27, 29
USING A STEP DIAGRAM TO FIND THE LCM The least common multiple for two polynomials can be found using a step diagram similar to that used in Section 1.2 for finding the least common multiple of two numbers. The next example illustrates this process.
EXAMPLE 2
Using a step diagram to find the LCM
Use a step diagram to find the LCM of 2 x 3 - 8 x 2 and 2 x 3 - 2 x 2 - 24 x.
Solution Start by factoring 2 x 3 - 8 x 2 and 2 x 3 - 2 x 2 - 24 x completely.
2 x 3 - 8 x 2 = 2 x 2(x - 4)
and 2 x 3 - 2 x 2 - 24 x = 2 x(x + 3)(x - 4)
Write the factored form of each expression on the top step of the diagram. We find the expressions in each of the following steps of the diagram by factoring out any common factor from the expressions in the previous step. The process continues until no common factor can be found, as shown in Figure 7.3.
2 Common factors x x–4 2x2(x – 4) x2(x – 4) x(x – 4) x
ISBN 1-256-49082-2
Start: Factor out 2. Factor out x. Factor out x – 4. Stop: No common factors.
Figure 7.3 Factor Step Diagram for Finding the LCM
We find the LCM by multiplying the expressions along the side and at the bottom of the diagram. The LCM is 2 # x # (x 4) # x # (x 3) = 2 x 2(x - 4)(x + 3).
Now Try Exercise 31
Review of Fractions with Unlike Denominators
Before we can find the sum 1 + 1 by hand, we need to rewrite these fractions by using their 2 3 least common denominator. The least common denominator (LCD) of 1 and 1 corresponds 2 3 to the least common multiple of the denominators 2 and 3, which is 6. As a result, we can rewrite these fractions as 1 2
#3
3
=
3 6
and
1 3
#2
2
=
2 . 6
Their sum is 1 + 1 = 3 + 2 = 5 . 2 3 6 6 6
EXAMPLE 3
Adding and subtracting fractions with unlike denominators
Simplify each expression. (a) 4 3 + 10 15 (b) 7 1 8 6
Solution 3 4 (a) The LCD for 10 and 15 equals the LCM of 10 and 15, which is 30. We rewrite these fractions as
3 10 Their sum is
#3
3
=
9 30
and
4 15
#2
2
=
8 . 30
3 4 9 8 17 . + = + = 10 15 30 30 30 (b) The LCD for 7 and 1 is the LCM of 8 and 6, which is 24. We rewrite these fractions as 8 6 21 1 # 4 4 7 # 3 = and = . 8 3 24 6 4 24 Their difference is 7 1 21 4 17 . = = 8 6 24 24 24
Now Try Exercises 45, 47
Rational Expressions with Unlike Denominators
The first step in adding or subtracting rational expressions with unlike denominators is to rewrite each expression by using the least common denominator. Then the sum or difference can be found by using the techniques discussed in Section 7.3.
NOTE: The LCD of two or more rational expressions equals the LCM of their denominators.
To add or subtract rational expressions, we often rewrite a rational expression with a different denominator. This technique is demonstrated in the next example and is used in future examples.
EXAMPLE 4
Rewriting rational expressions
Rewrite each rational expression so it has the given denominator D. 1 3 , D = 8x2 (b) , D = x2 - 1 (a) 2x x + 1
7.4 ADDITION AND SUBTRACTION WITH UNLIKE DENOMINATORS
447
Solution (a) We need to write 23x so that it is equivalent to 8? 2 . x
* 4x
? 3 = 2x 8x2 Because 8x 2 = 2 x # 4x, we can multiply 23x by 1 in the form 4 x as follows. 4x 3 2x
* 4x
# 4x
4x
=
12 x 8x2
Multiply rational expressions.
1 (b) We must write x + 1 so that it is equivalent to x 2 ? 1 . -
* (x - 1)
? 1 = 2 x + 1 x - 1
* (x - 1)
Because x - 1 = (x + 1)(x follows.
2
1 1), we can multiply x + 1 by 1 in the form x - 1 as x - 1
1 x + 1
Now Try Exercises 37, 39
#x x x - 1 1 = 2 1 x - 1
Multiply rational expressions.
EXAMPLE 5
Adding rational expressions with unlike denominators
Find each sum and leave your answer in factored form. 1 1 7 1 x 5 + + 2 (b) + (c) 2 (a) 8y x - 1 x + 1 x + 1 4y x + 2x + 1
Solution (a) First find the LCM of 8y and 4y 2.
Thus the LCM is 2 # 2 # 2 # y # y = 8y 2. Now, because 8y 2 = 8y # y and y 8y = 2 # 2 # 2 # y
and 4y 2 = 2 # 2 # y # y 8y 2 = 4y 2 # 2,
we multiply the first expression by y and the second expression by 2 . 2 5 7 5 + 2 = 8y 8y 4y =
ISBN 1-256-49082-2
(b) The LCM for x - 1 and x + 1 is their product, (x 1 1 1 + = x - 1 x + 1 x - 1 = = =
1)(x 1 1
1).
#x x 1 1 + 1 x + 1
#x x Rewrite by using the LCD. Multiply rational expressions. Add the numerators. Simplify the numerator.
x + 1 x - 1 + (x - 1)(x + 1) (x + 1)(x - 1) x + 1 + x - 1 (x - 1)(x + 1) 2x (x - 1)(x + 1)
(c) First find the LCM for x 2 + 2 x + 1 and x + 1. Because x 2 + 2 x + 1 = (x + 1)(x + 1), their LCM is (x + 1)(x + 1) = (x + 1)2. 1 1 x x + = + 2 x + 1 x + 1 x + 2x + 1 (x + 1)
2
#x x 1 Rewrite by using the LCD. 1
Multiply rational expressions. Add the numerators.
= =
Now Try Exercises 53, 65, 71
x x + 1 + 2 (x + 1) (x + 1)2 2x + 1 (x + 1)2
Subtraction of rational expressions is performed in a manner similar to addition and is illustrated in the next example.
EXAMPLE 6
Subtracting rational expressions with unlike denominators
Simplify each expression. Write your answer in lowest terms and leave it in factored form. 5 z 5 1 (a) (b) - 2 z z - 1 x + 1 x - 1 1 3 5 3 x - 2 (d) + - 2 (c) 2 x x - 1 x - 2x x + 2x x - x
Solution (a) The LCD is z(z
(d) The given expression contains three rational expressions. Begin by finding the LCM of the three denominators: x - 1, x 2 - x, and x. Because x 2 - x = x(x 1), the LCM is x(x - 1). 5 3 3 3 + = - 2 x x - 1 x - 1 x - x = = = =
ISBN 1-256-49082-2
#x x -
5 3 + x x(x - 1)
#x x 1 1
Rewrite by using the LCD. Multiply rational expressions. Combine the expressions. Simplify the numerator. Factor the numerator. Simplify to lowest terms.
N REAL-WORLD CONNECTION Sums of rational expressions occur in applications involving electricity. The flow of electricity through a wire can be compared to the flow of water through a hose. Voltage is the force “pushing” the electricity and corresponds to water pressure in a hose. Resistance is the opposition to the flow of electricity and corresponds to the diameter of a hose. More resistance results in less flow of electricity. An ordinary light bulb is an example of a resistor in an electrical circuit. Resistance is often measured in units called ohms. For example, a standard 60-watt light bulb has a resistance of about 200 ohms. Suppose that two light bulbs are wired in parallel so that electricity can flow through either light bulb, as illustrated in Figure 7.4. If the individual resistances of the light bulbs are R and S, then the total resistance of the circuit is found by adding the expression
Figure 7.4
1 1 + , R S and then taking the reciprocal. (Source: R. Weidner and R. Sells, Elementary Classical Physics, vol. 2.)
EXAMPLE 7
Modeling electrical resistance
1 Add R + 1, and then find the reciprocal of the result. S
Solution 1 1 The LCD for R and S is RS.
1 1 1 + = R S R
.
#S
S
+
1 S
#R
R
Rewrite by using the LCD. Multiply rational expressions. Add the numerators.
= =
S R + RS RS S + R RS
CRITICAL THINKING
Find the reciprocal of the sum x + 1 . x
+ In general, the reciprocal of a is b, so the reciprocal of S RS R is a b
RS . S + R This final expression can be used to find the total resistance of the circuit.
Now Try Exercise 101
7.4
w
Putting It All Together
EXPLANATION EXAMPLES
CONCEPT
Least Common Multiple (LCM) of Polynomials
1. Factor each polynomial completely. 2. List each factor the greatest number of times that it occurs in any factorization. 3. The LCM is the product of this list. The LCD of two or more rational expressions equals the least common multiple (LCM) of their denominators.
For x - 2 x and x 2 - 4x + 4, 1. x 2 - 2 x = x(x - 2) x 2 - 4 x + 4 = (x - 2)(x - 2) 2. x, (x - 2), (x - 2) 3. LCM = x(x - 2)(x - 2).
2
Least Common Denominator (LCD)
3 2 The LCD of x 2 - 2 x and x 2 - 4 x + 4 is x(x - 2)(x - 2) because the LCM of x 2 - 2 x and x 2 - 4 x + 4 is x(x - 2)(x - 2), as shown above.
7.4 ADDITION AND SUBTRACTION WITH UNLIKE DENOMINATORS
451
CONCEPT
EXPLANATION
EXAMPLES
Addition and Subtraction of Rational Expressions with Unlike Denominators
First rewrite each expression by using the LCD. Then add or subtract the expressions. Finally, write your answer in lowest terms. It is often helpful to leave the result in factored form.
The LCM of x and x + 2 is x(x + 2). 2 5 2 = + x x x + 2 = =
Exercises
25. 4 x 2 - 1, 2 x + 1 (2 x - 1)(2 x + 1) 26. x 2 + 4 x + 3, x + 3 (x + 1)(x + 3) 27. x 2 - 1, x + 1 (x - 1)(x + 1) 28. x 2 - 4, x - 2 (x - 2)(x + 2) 29. 2 x 2 + 7x + 6, x 2 + 5x + 6 (2 x + 3)(x + 2)(x + 3) 30. x 2 - 3x + 2, x 2 + 2 x - 3 (x - 2)(x - 1)(x + 3) 31. 3y 2 + 6y, 3y 3 + 3y 2 - 6y 3y( y + 2)( y - 1) 32. y 2 + 3y + 2, y 4 - 4y 2 y 2( y + 2)( y - 2)( y + 1)
ADDITION AND SUBTRACTION OF RATIONAL EXPRESSIONS
CONCEPTS AND VOCABULARY
1. Give a common multiple of 6 and 9 that is not the least common multiple. 2. The LCM of x and y is
Examples include 36 and 54 (answers may vary). xy .
3. To rewrite 3 with denominator 12, multiply 3 by 1 4 4 written as the fraction _____. 3 3
4 4. To rewrite x - 1 with denominator x 2 - 1, multiply 4 x - 1 by 1 written as the rational expression _____. x + 1 x + 1
Exercises 33–44: Rewrite the rational expression so it has the given denominator D. For example, if the fraction 1 2 4 is written with denominator D = 8, it becomes 8 . 33. 1, D = 9 3 35. 5, D = 21 7 37. 39. 41.
3 9 15 21 2x2 8x3
34. 3, D = 24 4 36. 4, D = 30 5 38. 40. 42.
18 24 24 30 15x 9x 2
Exercises 13–32: Find the least common multiple. Leave your answer in factored form. 13. 4 x, 6 x 12 x 15. 5x, 10x 2 10x 2 17. x, x + 1 x(x + 1) 19. 2 x + 1, x + 3
(2 x + 1)(x + 3)
2 2
ISBN 1-256-49082-2
14. 6 x, 9x 18 x 16. 4 x 2, 12 x 12 x 2 18. 4 x, x - 1 4 x(x - 1) 20. 5x + 3, x + 9
(5x + 3)(x + 9)
2 2
7.4 ADDITION AND SUBTRACTION WITH UNLIKE DENOMINATORS
453
APPLICATIONS
101. Electricity In Example 7, we demonstrated that the 1 1 expressions 75; 75; yes 1 1 + R S and S + R RS
as illustrated in the figure. Write the difference 1 1 D - F F - D as one term. FD
D Lens
S Film
are equivalent. Evaluate both expressions by using R = 120 and S = 200. Are your answers the same?
32 102. Intensity of a Light Bulb The formula I = 4d 2 approximates the intensity of light from a 100-watt light bulb at a distance of d meters, where I is in watts per square meter. For light from a 40-watt 16 bulb the equation for intensity becomes I = 5d 2.
(Source: R. Weidner.)
104. Geometry Find the sum of the areas of the two rectangles shown in the figure. Write your answer 3x 2 3x - 2 in factored form. (x - -1)(x + 1)
(a) Find an expression for the sum of the intensities 56 of light from the two light bulbs. 5d 2 (b) Find the combined intensity of their light at 56 d = 5 meters. 125 = 0.448 W/m2 103. Photography A lens in a camera has a focal length, which is important for focusing the camera. If an object is at a distance D from a lens that has a focal length F, then to be in focus, the distance S between the lens and the film should satisfy the equation 1 1 1 = - , S F D
1 x–1 x–2
2 x+1 x
WRITING ABOUT MATHEMATICS
105. Explain how to find the least common multiple of two polynomials. 106. Explain how to subtract two rational expressions with unlike denominators.
SECTIONS 7.3 and 7.4
Checking Basic Concepts
3. Simplify each expression. 5 5 1 1 (b) + + (a) x x + 1 x - 3 3 - x x + 2 -4 (c) 4x + 2 2x + 1 a b 4. Simplify the expression . a - b a + b
1. Simplify each expression. 2 2 x x + (b) (a) x + 2 x + 2 3x 3x z z2 + z + (c) z + 2 z + 2 2. Find the least common multiple of each pair of expressions. (a) 3x, 5x (b) 4 x, x 2 + x (c) x + 1, x - 1
If two and a half pizzas are cut so that each piece equals one-eighth of a pizza, then there are 20 pieces of pizza. The quotient two and a half divided by one-eighth can be written as 2 1 2 1 8 2 + or 1 8 1 2
.
These expressions are examples of complex fractions. Typically, we want to simplify a coma plex fraction by rewriting it as a standard fraction in the form b. In this section we show how to simplify complex fractions.
NEW VOCABULARY n Complex fraction n Basic complex fraction
Basic Concepts
A complex fraction is a rational expression that contains fractions in its numerator, denominator, or both. Examples of complex fractions include 1 1 1 2 , 1 3 2x 3 x x 1 x 1 1
5z 7 z 4 2z
, and
.
Complex fractions
c The expression a , d can be written as a basic complex fraction, where both the numerab tor and denominator of the complex fraction are single fractions.
a b . c d Because
Basic complex fraction
c a a , = b d b
# d, c we can simplify basic complex fractions by multiplying the numerator and the reciprocal of the denominator. This strategy is summarized as follows.
READING CHECK
• How do we simplify a basic complex fraction?
SIMPLIFYING BASIC COMPLEX FRACTIONS
For any real numbers a, b, c, and d, a b a = c b d where b, c, and d are nonzero.
Simplifying Complex Fractions
There are two methods of simplifying a complex fraction. The first is to simplify both the numerator and the denominator and then divide the resulting two fractions. The second is to multiply the numerator and denominator by the least common denominator of all fractions within the complex fraction.
STUDY TIP
Two methods for simplifying complex fractions are discussed in this section. If your instructor does not require you to use a specific method, choose the one that works best for you.
SIMPLIFYING THE NUMERATOR AND DENOMINATOR The following steps outline Method I for simplifying complex fractions.
SIMPLIFYING COMPLEX FRACTIONS (METHOD I)
To simplify a complex fraction, perform the following steps.
STEP 1: STEP 2: STEP 3:
Write the numerator as a single fraction; write the denominator as a single fraction. Divide the denominator into the numerator by multiplying the numerator and the reciprocal of the denominator. Simplify the result to lowest terms.
READING CHECK
• What is the first step when using Method I to simplify complex fractions?
Sometimes a complex fraction already has both its numerator and denominator written as single fractions. As a result, we can start with Step 2. This situation is shown in Example 1.
(b) Start by writing 3 3 and 1 1 as improper fractions. 4 2 3 15 3 4 4 = Write as improper fractions (Step 1). 1 3 1 2 2 = = 15 4 5 2
#2
3
Multiply by the reciprocal of
3 2
(Step 2).
Multiply and simplify (Step 3).
2y 3 3 x x (c) We can skip Step 1. To simplify 4 , 2y, multiply 4 by the reciprocal of 2y, or 3 . x 4 x 2y 3 = # Multiply by the reciprocal of 2y (Step 2). 3 4 3 2y
= =
2 xy 12 xy 6
(x - 1)2 4
Multiply the fractions. Simplify (Step 3).
8 by the reciprocal of x - 1, or x - 1. 8
(d) We can skip Step 1. Multiply (x 4 x 8 = 1 1)2 = (x 4
1)2
#
8 x 1
Multiply by the reciprocal of
x - 1 8
(Step 2).
8(x - 1)(x - 1) 4(x - 1)
Multiply the fractions. Simplify (Step 3).
= 2(x - 1)
Now Try Exercises 13, 15, 19, 29
EXAMPLE 2
Simplifying complex fractions
Simplify. Write your answer in lowest terms. 4 1 2 1 1 x + + x x a b x - 1 (b) (a) (c) 1 4 2 1 1 x + x x a b x - 1 1 1 + x y (d) 1 1 - 2 2 x y
Solution (a) In Step 1, write the numerator as one fraction by using the LCD, ab.
1 a
a b a b 1 + = = b ab ab ab
ISBN 1-256-49082-2
Write the denominator as one fraction by using the LCD, ab. 1 a a b a b 1 = = b ab ab ab
(d) The LCD for the numerator is xy, and the LCD for the denominator is x 2y 2. y 1 1 x + + x y xy xy = 1 1 y2 x2 - 2 - 2 2 x2 y x 2y 2 x y y = xy y y xy
2
Write by using the LCD (Step 1).
x x2 x 2y 2 x
Combine terms.
= = = =
#
x 2y 2 y2 x2
Multiply by the reciprocal of y2 - x 2 x 2y 2
(Step 2).
x 2y 2( y + x) xy( y 2 - x 2) x 2y 2( y + x) xy( y - x)( y + x) xy y - x
Multiply.
Factor the denominator (Step 3). Simplify.
Now Try Exercises 31, 37, 41, 45
MULTIPLYING BY THE LCD In Method II for simplifying a complex fraction, we multiply the numerator and denominator of the complex fraction by the least common denominator of all the fractions within the complex fraction. The following steps outline Method II for simplifying complex fractions.
SIMPLIFYING COMPLEX FRACTIONS (METHOD II)
To simplify a complex fraction, perform the following steps.
STEP 1: STEP 2: STEP 3:
Find the LCD of all fractions within the complex fraction. Multiply the numerator and the denominator of the complex fraction by the LCD. Simplify the result to lowest terms.
READING CHECK
• When using Method II, by what do we multiply the numerator and denominator of the complex fraction?
ISBN 1-256-49082-2
We can use this method to simplify the complex fraction in Example 2(a). Because the LCD for the numerator and the denominator is ab, we multiply the complex fraction by 1, expressed in the form ab. This method is equivalent to multiplying the numerator and the ab denominator by ab.
1 1 1 1 a + b ab + a a b b = 1 1 1 1 a - b ab a a b b ab ab + a b = ab ab a b = b + a b - a
7.5 COMPLEX FRACTIONS
459
Multiply by
ab ab
= 1.
Distributive property
Simplify the fractions.
EXAMPLE 3
Simplifying complex fractions
Simplify. (a) 2z 4 3 + z z 1 x - 3 (b) 1 3 + x x - 3 1 1 a b (c) 1 1 2 2b 2a 2
Solution (a) The LCD for all fractions within the complex fraction is z, so multiply the numerator and denominator of the complex fraction by z (Step 1).
2z 4 3 + z z
=
(2z) z 4 3 a + bz z z 2z 2 4z 3z + z z 2z 2 4 + 3 2z 2 7
Multiply by
z z
= 1 (Step 2).
=
Distributive property
= =
Simplify the fractions (Step 3).
Add.
(b) The LCD for all fractions within the complex fraction is the product x(x - 3) (Step 1). 1 1 x - 3 x - 3 = 1 3 1 3 + a + b x x x - 3 x - 3
# x(x x(x 3) 3)
Multiply by
x(x - 3) x(x - 3)
= 1 (Step 2).
ISBN 1-256-49082-2
x(x 3) x - 3 = x(x 3) 3x(x 3) + x x - 3 x = (x - 3) + 3x = x 4x - 3
Distributive property
Simplify the fractions (Step 3). Simplify the denominator.
(c) The LCD for the numerator and the denominator is 2a 2b 2 (Step 1). 1 1 1 1 a a b b = 1 1 1 1 2b 2 2a 2 2b 2 2a 2
CRITICAL THINKING
Are the expressions a b + 1 1 and 1 + 1 equal? Explain. a b + 1 Are the expressions a and 1 + b a a b
# 2a 2b 2
2 2
2a b
Multiply by
2a 2b 2 2a 2b 2
= 1 (Step 2).
2a 2b 2 2a 2b 2 a b = 2 2 2a b 2a 2b 2 2b 2 2a 2 = = 2ab 2 - 2a 2b a2 - b 2 2ab(b - a) (a - b)(a + b) 2ab a + b
Distributive property
equal? Explain.
b
Simplify the fractions (Step 3).
Factor.
= Now Try Exercises 33, 35, 49
Simplify.
7.5
Putting It All Together
EXPLANATION EXAMPLES
CONCEPT
Complex Fraction
A rational expression that contains fractions in its numerator, denominator, or both
3 +
1 x + 1 1 3 x + 1 2 x 2 = x 4 x - 1
and
y x x y y x + x y x - 1 2x
Simplifying Basic Complex Fractions
When b, c, and d are nonzero, a a b = c b d
# d. c #x-1
4
=
Method I: Simplifying the Numerator and Denominator
Combine the terms in the numerator, combine the terms in the denominator, and then multiply the numerator by the reciprocal of the denominator. Multiply the numerator and the denominator by the LCD of all fractions within the expression.
1 3 4 + x x x 4 = = x 5 1 4 y y y
#y
1
=
4y x
Method II: Multiplying the Numerator and Denominator by the LCD
2 1 2 1 + a + b xy x x y y = 4 1 4 1 a - b xy x y x y = 2y + x 4x - y
ISBN 1-256-49082-2
3. A complex fraction is a rational expression that contains _____ in its numerator, denominator, or both. fractions x 1 4. Write the expression 2 , x - 1 as a complex fraction. c 5. Write the expression a , d as a complex fraction. b 1 6. What is the LCD for x + 2 and 1 ? x(x + 2) x a b c d
x 2 1 x - 1
6 m - 2 23. 3 2 m - 2 p + 1 p 25. p + 2 p 5 z2 - 1 27. z z2 - 1 y y - 9 29. 1 y + 3
2
3 n + 1 24. 6 n + 1
1 2
p + 1 p + 2
SIMPLIFYING COMPLEX FRACTIONS
Exercises 7–12: For the complex fraction, determine the LCD of all the fractions appearing within the expression. x 1 5 6 7. 30 2 - 3x 15 2 - x x + 1 9. 2 + x x - 1 1 1 x 2 8. 2x 1 1 + x 2 1 4 x 4x 10. 1 1 + 2x 3x
12x
2p 2p + 5 26. 4p 1 4p + 10 z z - 2 28. 1 z z - 2 2y 2y - 1 30. 2y(2y + 1) 1 4y 2 - 1 1 x 32. 1 4 + x 4 34. 5x 1 + 1 x
55. Resistance in Electricity Light bulbs are often wired so that electricity can flow through either bulb, as illustrated in the accompanying figure.
6b + 4a 6b - 15a
2 1 q q + 1 43. 1 q + 1 1 + x + 1 x 45. 1 x + 1 x
q + 2 q
4 - 5 9p - 25 - 25 5 - 5
1 1 1 + 2 x - 3 x + 3 2 x + 3 46. 1 1 1 - 2 + 2 x - 9 1 1 - 3 4x2 x 48. 1 1 + x - 1 x - 1 1 1 - 2 2 x y 50. 1 1 x y 52.
(x - 4)(x - 1) 8x3
In this way, if one bulb burns out, the other bulb still works. If two light bulbs have resistances T and S, their combined resistance R is R = 1 1 1 + T S .
1 1 2x - 1 2x + 1 47. x + 1 x
2x (x + 1)(2 x - 1)(2 x + 1)
6 x 2 - 10
Simplify this formula. R =
ST S + T
56. Resistance in Electricity (Refer to the preceding exercise.) Evaluate the formula R = 1 1 1 + T S
1 1 - 2 2 ab a b 49. 1 1 a b 51. a -1 1 + b -1
1 ab
x + y xy
when T = 100 and S = 200. R = 66.6
WRITING ABOUT MATHEMATICS
ab a + b
a2 - b 2 -a 2b 2 a -2 - b -2
APPLICATIONS
53. Annuity If P dollars are deposited every 2 weeks in an account paying an annual interest rate r expressed as a decimal, then the amount in the account after 2 r P(1 + 26)52 - P years can be approximated by r 52 r a Pa 1 + b - Pb , . 26 26 Write this expression as a complex fraction. 54. Annuity (Continuation of the preceding exercise) Use a calculator to evaluate the expression when r = 0.026 (2.6%) and P = $100. Interpret the result.
About 5334.84; if $100 is deposited every 2 wk in an account paying 2.6% annual interest, the account will have $5334.84 after 2 yr. r 26
57. A student simplifies a complex fraction as shown below. Explain the student's mistake and how you would simplify the complex fraction correctly. 1 + 1 x 1 x 1 x + 1 = 2 1 x
58. Explain one method for simplifying a complex fraction.
Rational Equations and Formulas
Solving Rational Equations ● Rational Expressions and Equations ● Graphical and Numerical Solutions ● Solving a Formula for a Variable ● Applications
A LOOK INTO MATH N
In Section 7.1 we demonstrated that if cars enter a construction zone at random at an average rate of x cars per minute and if 10 cars per minute can pass through the zone, then the average time T that a driver spends waiting in line and passing through the construction zone is T = 1 , 10 - x
where T is in minutes and x 6 10. If the highway department wants to limit the average wait for a car to 1 minute, then mathematics can be used to determine the corresponding 2 value of x. (Source: N. Garber and L. Hoel, Traffic and Highway Engineering.)
NEW VOCABULARY n Rational equation n Basic rational equation n Extraneous solutions
Solving Rational Equations
N REAL-WORLD CONNECTION If an equation contains one or more rational expressions, it is called a rational equation. Rational equations occur in mathematics whenever a rational expression is set equal to a constant. For example, the rational expression
1 10 x
Rational expression
from A Look Into Math can be used to estimate the average time that drivers wait to get through a construction site. If the wait is 1 minute, we determine x by solving the rational 2 equation 1 10 x = 1 . 2
Rational equation
To solve this equation, we multiply each side by the LCD: 2(10 2(10 10 x) 2(10 = x 2 2 = 10 - x x = 8
1 2
x).
x)
Multiply by the LCD. Simplify. Add x; subtract 2.
The average wait is minute when cars arrive randomly at an average rate of 8 cars per minute. In general, if a rational equation is in the form c a = , b d we can multiply each side of this equation by the common denominator bd to obtain a(bd) c(bd) = , b d
ISBN 1-256-49082-2
which simplifies to ad = cb. This technique can be used to solve some types of basic rational equations, which have a single rational expression on each side of the equals sign.
SOLVING BASIC RATIONAL EQUATIONS
The equations c a = b d and ad = bc
READING CHECK
• Why do we cross multiply?
are equivalent, provided that b and d are nonzero. Note that converting the first equation to the second equation is sometimes called cross multiplying.
EXAMPLE 1
Solving rational equations
Solve each equation. 4 3x x + 1 5 = = (b) (a) x 3 5 2
Solution (a)
(c) x =
4 3x - 4
(d)
1 2 3 + = x x 7
Given equation Cross multiply. Divide by 5.
5 4 = x 3 5x = 12 x = 12 5
The solution is 12. 5 (b) x 5 2(x 3x 2 1) = 15x = 2 = 13x 2 = x 13
2 The solution is 13.
1
Given equation Cross multiply. Distributive property Subtract 2x. Divide by 13.
3x - 4 x = 4 3x - 4 x - 4 = 0 (3x + 2)(x - 2) = 0 3x + 2 = 0 x = 2 3 or or x - 2 = 0 x = 2 1 2 3 + = x x 7 3 3 = x 7 21 = 3x 7 = x The solution is 7.
Now Try Exercises 9, 15, 31, 41
The solutions are - 2 and 2. 3 (d)
Given equation Add the rational expressions.
ISBN 1-256-49082-2
Another technique for solving rational equations is to multiply each side by the least common denominator. Unlike cross multiplying, this technique can always be used.
EXAMPLE 2
Multiplying by the LCD
Solve each equation. Check your answer. 1 1 1 (a) = x x - 1 9x 1 1 12 (b) + = 2 x - 1 x + 1 x - 1
Solution (a) Start by multiplying each term by the LCD, 9x(x
1).
Given equation Multiply each term by the LCD. Simplify each rational expression. Distributive property; simplify. Add 1. Substitute 10 for x. The LCD is 90.
1 1 1 = x x - 1 9x 9x(x 1) 9x(x x x - 1 1) = 9x(x 1) 9x
In Example 2 we checked our answers. Although the answer checked in both cases, it may not always check. When multiplying a rational equation by the LCD, it is possible to obtain extraneous solutions that do not satisfy the given equation. This situation is demonstrated in Example 3. Before solving Example 3, we present a step-by-step strategy for solving rational equations.
STEPS FOR SOLVING A RATIONAL EQUATION
STEP 1:
Find the LCD of the terms in the equation. Multiply each side of the equation by the LCD. Simplify each term. Solve the resulting equation. Check each answer in the given equation. Any value that makes a denominator equal 0 should be rejected because it is an extraneous solution.
READING CHECK
• What is the first step in solving a rational equation?
STEP 2: STEP 3: STEP 4: STEP 5:
c NOTE: If the rational equation is in the form a = d , or “fraction equals fraction,” cross b multiplying to obtain ad = bc may be helpful. However, be sure to check your answers.
EXAMPLE 3
Solving an equation having an extraneous solution
If possible, solve
Solution The LCD is x 2
1 1 4 . + = 2 x - 2 x + 2 x - 4 4 = (x 2)(x 2) (Step 1).
Given equation Multiply by the LCD (Step 2). Simplify each term (Step 3). Combine like terms (Step 4). Divide by 2.
2
Note that terms on both sides of the equation are undefined because it is not possible to divide by 0. Therefore 2 is not a solution; rather, it is an extraneous solution. That is, there are no solutions to the given equation.
Now Try Exercise 53
Rational Expressions and Equations
Rational expressions and rational equations are not the same concepts. For example, 3 x and x x2 - x
are both rational expressions. Neither one contains an equals sign. We often simplify or evaluate rational expressions. By factoring the denominator and applying the basic principle of rational expressions, we can simplify the second expression. x x = x(x - 1) x - x
2
Factor the denominator. Simplify.
=
1 x - 1
1 Thus the expressions x 2 x x and x - 1 are equal for every value of x (except 0 and 1). Expressions can also be evaluated. For example, if we replace x with 3, the expression 3 x equals 3 , or 1. 3 When two expressions are set equal to each other, an equation is formed that is typically 1 true for only a limited number of x-values. For example, 3 = x - 1 is a rational equation x that is true for only one x-value.
3(x
1) = x # 1 2x = 3 x = 3 2
1 3 = x x 1
Given equation Cross multiply. Distributive property Add 3; subtract x. Divide each side by 2.
3x - 3 = x
The only solution is 3 . (Check this solution.) If we replace x with any value other than 3 , the 2 2 1 equation 3 = x - 1 is a false statement. x
EXAMPLE 4
Identifying expressions and equations
Determine whether you are given an expression or an equation. If it is an expression, simplify it and then evaluate it for x = 5. If it is an equation, solve it. x x2 - 2x 1 x - 1 = (b) + (a) x + 1 x + 3 x - 1 x - 1
Solution (a) There is an equals sign, so it is an equation that can be solved.
x x (x
2
1 x = 1 x 3 3) = x(x
2
Given equation
1)(x
1)
Cross multiply. Multiply. Subtract x 2. Subtract x; add 3. Replace x with 3 in given equation. The answer checks.
x + 2x - 3 = x + x 2x - 3 = x x = 3 Check:
ISBN 1-256-49082-2
STUDY TIP
Be sure to look for an equals sign before you work on any problem that involves rational expressions. If the problem has an equals sign, it is an equation. If it does not, it is an expression. Refer to Making Connections on this page.
(b) There is no equals sign, so it is an expression that can be simplified. The common denominator is x - 1, so add the numerators. 1 x2 - 2x + 1 x2 - 2x + = x - 1 x - 1 x - 1 (x - 1)(x - 1) x - 1 = x - 1 =
Add the numerators. Factor the numerator. Simplify.
The given expression simplifies to x - 1. When x equals 5, the expression evaluates to 5 - 1, or 4.
Now Try Exercises 63, 67
MAKING CONNECTIONS
Expressions versus Equations
1. If a problem does not contain an equals sign, you are probably adding, subtracting, multiplying, dividing, or otherwise simplifying an expression. Your answer will be an expression, not a value for x. 2. If the problem has an equals sign, it is an equation to be solved. Your answer will be a value (or values) for x that makes the equation a true statement. Be sure to check all your answers.
Graphical and Numerical Solutions
Like other types of equations, rational equations can also be solved graphically and numerically. Graphs of rational expressions are not lines. They are typically curves that can be graphed by plotting several points and then sketching a graph or by using a graphing calculator.
EXAMPLE 5
Solving a rational equation graphically and numerically
Solve 2 = x + 1 graphically and numerically. x
Solution
Graphical Solution To solve x = x + 1, graph y1 = x and y2 = x + 1. The graph of y2 is a
2 2
line with slope 1 and y-intercept 1. To graph y1 make a table of values, as shown in Table 7.5. Then plot the points and connect them with a smooth curve. Note that y1 is undefined for x = 0. Generally, it is a good idea to plot at least three points on each side of an asymptote. These points were plotted and the curves sketched in Figure 7.5. Note that the graphs of y1 and y2 intersect at ( 2, - 1) and (1, 2). The solutions to the equation are 2 and 1. Check these solutions. TABLE 7.5 x -3 -2 x = 0 is an asymptote h
2 x
Solving a Formula for a Variable
N REAL-WORLD CONNECTION Formulas in applications often involve both the use of rational expressions and the need to solve an equation for a variable. This idea is illustrated in the next two examples.
EXAMPLE 6
Finding time in a distance problem
If a person travels at a speed, or rate, r for time t, then the distance d traveled is d = rt. (a) How far does a person travel in 2 hours when traveling at 60 miles per hour? (b) Solve the formula d = rt for t. (c) How long does it take a person to go 250 miles when traveling at 40 miles per hour?
Solution (a) A person traveling at 60 miles per hour for 2 hours travels
d = rt = 60 # 2 = 120 miles.
Given equation Divide each side by r.
(b) To solve d = rt for t, divide each side by r. d = rt d = t r
(c) At 40 miles per hour, a person can travel 250 miles in
ISBN 1-256-49082-2
t =
25 250 d = = 6.25 hours. = r 40 4
Note that 6.25 hours equals 6 hours and 15 minutes.
Now Try Exercise 87
Solving a formula for a variable
Solve each equation for the specified variable. (b) P = nRT for V (c) S = 2prh + pr 2 for h (a) A = bh for b V 2
Solution (a) First, multiply each side by 2.
bh 2 2A = bh A = 2A = b h (b) Begin by multiplying each side by V. nRT V PV = nRT nRT V = P P = (c) Begin by subtracting pr 2 from each side. S = 2prh + pr 2 S - pr = 2prh S - pr 2 = h 2pr
2
Given equation Multiply each side by 2. Divide each side by h.
Given equation Multiply each side by V. Divide each side by P.
Given equation Subtract pr 2 from each side. Divide each side by 2pr.
Now Try Exercises 89, 91, 97
Applications
N REAL-WORLD CONNECTION Rational equations sometimes occur in time and rate problems, as demonstrated in the next two examples.
EXAMPLE 8
Mowing a lawn
Two people are mowing a large lawn. One person has a riding mower, and the other person has a push mower. The person with the riding mower can cut the lawn alone in 4 hours, and the person with the push mower can cut the lawn alone in 9 hours. How long does it take for them, working together, to cut the lawn?
Solution The first person can cut the entire lawn in 4 hours, so this person can cut 1 of the lawn in 4 t 1 hour, 2 of the lawn in 2 hours, and in general, 4 of the lawn in t hours. The second person 4 t can cut the lawn in 9 hours, so (using similar reasoning) this person can cut 9 of the lawn in t hours. Together they can cut t t + 4 9
ISBN 1-256-49082-2
of the lawn in t hours. The job is complete when the fraction of the lawn cut reaches 1. To find out how long this task takes, solve the equation t t + = 1. 4 9
Begin by multiplying each side by the LCD, or 36.
CRITICAL THINKING
If one person can mow a lawn in x hours and another person can mow it in y hours, how long does it take them, working together, to mow the lawn?
t t + = 1 4 9 36t 36t + = 36(1) 4 9 9t + 4t = 36 13t = 36 t = 36 13
Equation to be solved Multiply each term by 36. Simplify. Combine like terms. Divide each side by 13.
Working together they can cut the lawn in 36 13
Now Try Exercise 103
2.8 hours.
EXAMPLE 9
Solving a distance problem
Suppose that the winner of a 600-mile car race finishes 12 minutes ahead of the secondplace finisher. If the winner averages 5 miles per hour faster than the second racer, find the average speed of each racer.
Solution Here we apply the four-step method for solving an application problem.
STEP 1:
Identify any variables. x: x + 5: speed of slower car in miles per hour speed of faster car in miles per hour
STEP 2:
Write an equation. To determine the time required for each car to finish the race, we use the equation t = d. Because the race is 600 miles, the time for the slower r car is 600 and the time for the faster car is x 600 5. The difference between these x times is 12 minutes, or 12 = 1 hour. Thus to determine x, we solve 60 5 1 600 600 = . x x 5 5
NOTE: Speeds are in miles per hour, so time must be in hours, not minutes.
STEP 3:
Solve the equation. Start by multiplying by the LCD, 5x(x 1 600 600 = x x + 5 5 5) =
5).
Equation to be solved
600 # 5x(x x
5)
-
600 # 5x(x x + 5
1 # 5x(x 5
2 2
5)
Multiply each term by the LCD. Simplify. Distributive property Combine like terms. Rewrite the equation. Factor. Zero-product property Solve.
The slower car travels at 120 miles per hour, and the faster car travels 5 miles per hour faster, or 125 miles per hour. (The solution - 125 has no meaning in this problem.)
Check your answer. The slower car travels 600 miles at 120 miles per hour, which takes 600 = 5 hours. The faster car travels 600 miles at 125 miles per hour, which 120 requires 600 = 4.8 hours. The difference between their times is 5 - 4.8 = 0.2 hour, 125 or 0.2 * 60 = 12 minutes, so the answer checks.
Now Try Exercise 105
CALCULATOR HELP
To make a table, see Appendix A (pages AP-2 and AP-3).
TECHNOLOGY NOTE
Solving Rational Equations Numerically Tables can be used to find solutions to rational equations. The following displays show the positive solution of 120 from Example 9. Note the use of parentheses for entering the formula for Y1.
\Y 1 600/ X ( X 5) \Y 2 1 / 5 \Y 3 \Y 4 \Y 5 \Y 6
Plot1 Plot2 Plot3
To solve an equation of the form c a = , b d b 0, d 0
3 5 = is equivalent to 6 x = 30. 6 2x x 8 = is equivalent to x 2 = 16. x 2 (Cross multiplication works only for equations with one rational expression on each side: “fraction equals fraction.”)
cross multiply to obtain ad = bc.
Rational Equations
To solve a rational equation, use the following steps.
STEP 1: STEP 2:
Solve
1 2 5 = . x 3x 6
Find the LCD.
Multiply each side of the equation by the LCD. STEP 3: Simplify each term.
STEP 4: STEP 5:
The LCD is 6 x. 1(6 x) 2(6 x) 5(6 x) STEP 2: = x 3x 6
STEP 1:
Solve the resulting equation. Check each possible solution.
Exercises
29. 31. 33. 35. 37. 39. 3 1 = 1 - x 1 + x 1 = - z -1 z + 2 x 3 -1 = - , -1 2x + 5 3 2 x x + = 3 4 2 4 x 3x = 1 4 4 2 1 4 + = -1 t + 1 t + 1 -6 1 2 1 + = 6 x x 2 2 4 + 1 = 2 x - 1 x - 1
-3 (extraneous: 1)
1 2
CONCEPTS AND VOCABULARY
30. 32. 34. 36. 38. 40. 42. 44.
1. If an equation contains one or more rational expressions, it is called a(n) _____ equation. rational 2. Give an example of a rational expression and an + + example of a rational equation. 2 x 3x 5; 2 x 3x 5 = 9 c 3. The equation a = d is equivalent to _____, provided b that b and d are nonzero. ad = bc 2 4. Are the equations x - 1 = 5 and 5(x - 1) = 2 equivalent provided that x 1? Yes 5 5. One way to solve the equation 3x + 43x = 1 is to multiply each side by the LCD, which is 12x . R 6. To solve the equation T = SV for V, multiply each side by the variable V and then divide each side by the variable T .
1 2x = 1 - 2x 2
1 6
z 1 = -1, 3 z - 2 3 1 x = -2, 1 3 2 3x + 5 x x = 1 -12 4 3 x 2x + = 6 6 3 3 5 2 = 3 t - 5 t - 5 4 1 2 = -3 x 2x
1 2
(answers may vary)
41. 43.
SOLVING RATIONAL EQUATIONS
Exercises 7–62: Solve and check your answer. 7. 9. 11. 13. 15. 17. 19. 21. 23.
ISBN 1-256-49082-2
x 3 = 2 4 6 3 = z 5
3 2
8. 10. 12.
1 4
2x 2 = 3 5 2 1 = z 7
7 2
1 1 + 2 = 2 x x + x 3
- 2 (extraneous: 0)
3 5
1 4 45. = - 1 x + 2 4 - x2
1 (extraneous: -2)
46. 48. 50. 52.
1 6 + 1 = 2 x - 3 x - 9
-4 (extraneous: 3)
5 2
47. 49. 51.
7y 3y = 0 4 2 2 1 = 3 2x + 1 5 8 = 2x x + 2
5y y = 0 6 3 1 3 7 = x + 4 5 3 1 5 = x - 1 3x
5 2
5 2 = 1 4z 3z
7 12
3 1 = 2 z + 1 z + 1 0 6 6 + = 5 - 3, 2 5 y y + 1 2 6 = -1 x 1 2x - 1
- 2, 2
4 1 6 1 + ,5 = y y - 1 5 6 1 1 = 0 3 2x x + 3
14. 16. 18. 20. 22. 24. 26. 28.
10 11
1 2 = 3 z - 1 z + 1 3 2 = 25 n + 5 n - 5 5 m = 5 m - 1 4 5x 1 = 5 - x 3
5 16
4 2 = 7 z + 3 z - 2 4 1 = 6 3n + 2 n - 1 3 3 5m = 2m - 1 2 4 x + 2 4 = 3x 3 x + 1 = 6 x
2 3
1 1 2 53. + = 2 x - 1 x + 1 x - 1
No solutions (extraneous: 1)
1 1 2 + = 2 2x + 1 2x - 1 4x - 1 No solutions 1extraneous: 1 2 2 1 1 6 55. + = 2 3 x - 2 x + 2 x - 4 54. 56. 57. 2 1 1 = 2 10 x + 3 x - 3 x - 9 1 1 1 No solutions + = 2 p + 1 p + 2 p + 3p + 2 (extraneous: -1) 1 1 1 No solutions = 2 p - 1 p + 3 p + 2p - 3
Exercises 63–70: Expressions and Equations (Refer to Example 4.) Determine whether you are given an expression or an equation. If it is an expression, simplify (if possible) and evaluate it for x = 2. If it is an equation, solve it. 1 1 - x 63. x x
Expression; 1; 1
1 64. - x = 0 x
Equation; −1, 1
Exercises 83–86: Solve the rational equation graphically to the nearest thousandth. 83. 1 22 + = 1.3 px - 2 2.1x 84. p 3 6 + = 2x 2 x - 4
2
65. 67.
1 1 1 = 2x 4x 8
Equation; 2
66. 68.
1 1 2x 4x
Expression;
0.300, 1.100
1 1 4 x; 8
Equation; 4
x + 1 2x - 3 = x - 1 2x - 5
2
Equation; 0
2x - 1 x + 1 = 4x + 1 2x - 1
p 1 85. 3 = 0 4 x
1.084
-2.619, 0.506, 3.160
86.
1 x = 1 2 2 2px
-1.913, -0.454, 0.367
4x + 4 x 69. + x + 2 x + 2
Expression; x + 2; 4
x2 - 2 x 70. x - 2 x - 2
SOLVING AN EQUATION FOR A VARIABLE
Expression; x + 1; undefined
Exercises 87–98: (Refer to Examples 6 and 7.) Solve the equation for the specified variable. 87. m = a =
F m
GRAPHICAL AND NUMERICAL SOLUTIONS
Exercises 71–74: Use the graph to solve the given equation. Check your answers. 71. 1 = 4 x - 1, 1 2 2 x y F for a a
88. m =
K =
72.
x = x 0, 2 x - 1 y V 89. I = for r R + r r =
90.
2A 91. h = for b b 3 z 93. = for z k z + 5 b =
2A h
V I
- R
r 1 = for R T R - r
R = r(T + 1)
mv 2
2K for K v2 2
92. h = b1 =
3 2 1 1
y = 4x x
y= x 2 x–1
–3 –2 –1 –1 –2 –3 2 3
3
2
3
x
z =
15 k - 3
5 t + r 94. for t = r t 2 t = r 5 - r
2A h
2A for b1 b1 + b2
- b2
y= 1 x
95. T = b =
y=x
73.
x - 1 = - 2 x -1, 1 2 x y 74.
3 = 1 -2, 2 x - 1
2
97.
3 1 2 for x = x y k x = ky 3y + 2k
aT a - T
ab for b a + b
96. A = b =
98.
1 1 1 + for R1 = R R1 R2
R1 =
RR2 R2 - R
aA A + 2
2b for b a - b
y
4 3 2 2 3 4
APPLICATIONS
2
–1 y = xx
2 3
y=1 x
–4 –3 –2
99. Waiting in Traffic (Refer to A Look Into Math for this section.) Solve the equation 9 cars/min
ISBN 1-256-49082-2
–3 –2 –1 –1 –2 –3
x
y = –2x
–1 –2
y=
3 x2 – 1
1 = 1 10 - x to determine the traffic rate x in cars per minute corresponding to an average waiting time of 1 minute.
100. Waiting in Line At a post office, customers arrive at random at an average rate of x people per minute. The clerk can wait on 4 customers per minute. The average time T in minutes spent waiting in line is given by T = 1 , 4 - x
106. Freeway Travel Two drivers travel 150 miles on a freeway and then stop at a wayside rest area. The first driver travels 5 miles per hour faster and arrives 1 7 hour ahead of the second. Find the average speed of each car. 75 mph; 70 mph 107. Braking Distance If a car is traveling downhill at 30 miles per hour on wet pavement, then the braking distance B in feet for this car is given by B = 30 , 0.3 + m
(a) 120; the braking distance is 120 feet when the slope of the road is -0.05.
where x 6 4. (a) 1, 10, and 100; it increases dramatically. (a) Evaluate T for x = 3, 3.9, and 3.99. What happens to the waiting time as the arrival rate nears 4 people per minute? (b) Find x when the waiting time is 5 minutes. 3.8 101. Shoveling a Sidewalk It takes an older employee 4 hours to shovel the snow from a sidewalk, but a younger employee can shovel the same sidewalk in 3 hours. How long will it take them to clear the walk if they work together? 12 1.7 hr 7 102. Pumping Water One pump can empty a pool in 5 days, whereas a second pump can empty the pool in 7 days. How long will it take the two pumps, working together, to empty the pool? 35 2.9 days 12 103. Painting a House One painter can paint a house in 8 days, yet a more experienced painter can paint the house in 4 days. How long will it take the two painters, working together, to paint the house? 8 2.7 days 3 104. Highway Curves To make a highway curve safe, highway engineers often bank it, as shown in the figure. If a curve is designed for a speed of 50 miles per hour and is banked with positive slope m, then a minimum radius R in feet for the curve is given by R = 2500 . 15m + 2
Introduction to Transportation Systems.)
where m 6 0 is the slope of the hill. (Source: L. Haefner,
(a) Find the braking distance for m = - 0.05 and interpret the result. (b) Find m if the braking distance is 150 feet. -0.1 108. Slippery Roads If a car is traveling at 30 miles per hour on a level road, then its braking distance in feet is 30, where x is the coefficient of friction between x the road and the tires. The variable x is positive and satisfies x … 1. The closer the value of x is to 0, the more slippery the road is. (Source: L. Haefner.) (a) Evaluate the expression for x = 1, 0.5, and 0.1. Interpret the results. (b) Find x for a braking distance of 150 feet. 0.2
(a) 30 ft, 60 ft, 300 ft; braking distance increases as x approaches 0.
109. River Current A boat can travel 36 miles upstream in the same time that it can travel 54 miles downstream. If the speed of the current is 3 miles per hour, find the speed of the boat without a current. 15 mph 110. Airplane Speed An airplane can travel 380 miles into the wind in the same time that it can travel 420 miles with the wind. If the wind speed is 10 miles per hour, find the speed of the airplane without any wind. 200 mph 111. Airplane Speed An airplane can travel 450 miles into the wind in the same time that it can travel 750 miles with the wind. If the wind speed is 50 miles per hour, find the speed of the airplane without any wind. 200 mph 112. River Current A boat can travel 114 miles upstream in the same time that it can travel 186 miles downstream. If the speed of the current is 6 miles per hour, find the speed of the boat without a current. 25 mph 113. Running and Jogging An athlete runs 10 miles and then jogs home. The trip home takes 1 hour longer than it took to run that distance. If the athlete runs 5 miles per hour faster than she jogs, what are her average running and jogging speeds?
10 mph running; 5 mph jogging
(Source: N. Garber and L. Hoel, Traffic and Highway Engineering.)
(a) Find R for m = 0.1. Interpret the result. (b) If R = 500, find m. Interpret the result.
(a) R 714.3; when the slope of the bank is 0.1, the minimum curve radius is about 714.3 ft.
m = 0.2; when the minimum curve radius is 500 ft, the slope of the bank should be 0.2.
ISBN 1-256-49082-2
105. Bicycle Race The winner of a 6-mile bicycle race finishes 2 minutes ahead of a teammate and travels, on average, 2 miles per hour faster than the teammate. Find the average speed of each racer. 20 mph; 18 mph
114. Speed Limit A person drives 390 miles on a stretch of road. Half the distance is driven traveling 5 miles per hour below the speed limit, and half the distance is driven traveling 5 miles per hour above the speed limit. If the time spent traveling at the slower speed exceeds the time spent traveling at the faster speed by 24 minutes, find the speed limit. 70 mph
WRITING ABOUT MATHEMATICS
115. Do all rational equations have solutions? Explain. 116. Why is it important to check your answer when solving rational equations? Explain.
SECTIONS 7.5 and 7.6
Checking Basic Concepts
4. Braking Distance If a car is traveling uphill at 60 miles per hour on wet pavement, then the braking distance D in feet for this car is given by D = 120 , 0.3 + m
1. Simplify each complex fraction. x 2 1 3 2x 3x (a) (b) 2x 6x 5 1 1 1 1 - 2 2 a b r t (d) (c) 1 2 1 2 + a r b t 2. Solve each equation. Check your answer. 1 3 x 4 (a) = (b) = 2x x + 1 2x + 3 5 1 3 3 2 (c) + = 1 (d) = -2 x 2x 2x x + 1 1 1 2 (e) = 2 x - 1 2 x - 1 3. Solve each equation for the specified variable. ax (a) - 3y = b for x 2 1 k (b) = for m m 2m - 1
where m 7 0 is the slope of the hill. (Source: L. Haefner,
Introduction to Transportation Systems.)
(a) Find D for m = 0.1 and interpret the result. (b) Find the slope of the road if D is 200 feet. Interpret the result.
7.7
Proportions and Variation
Proportions ● Direct Variation ● Inverse Variation ● Analyzing Data ● Joint Variation
A LOOK INTO MATH N
Proportions are frequently used to solve applications. The following are a few examples. • If someone earns $120 per day, then that person can earn $600 in 5 days. • If a car goes 280 miles on 10 gallons of gas, then it can go 560 miles on 20 gallons of gas. • If a person walks 1 mile in 20 minutes, then that person can walk 1 mile in 10 minutes. 2 In this section we discuss several applications of proportions.
NEW VOCABULARY n Ratio N n Proportion n Directly proportional (varies directly) n Constant of proportionality/ variation n Inversely proportional (varies inversely) n Varies jointly
Proportions
REAL-WORLD CONNECTION A ratio is a comparison of two quantities. For example, a math class might have 7 boys for every 8 girls. Thus the boy–girl ratio in this class is 7 to 8, or 7. 8 In mathematics, ratios are typically expressed as fractions. Ratios and proportions are sometimes used to find how much space remains for music on a compact disc (CD). A 700-megabyte CD can store about 80 minutes of music. Suppose that some music has been recorded on the CD and that 256 megabytes are still available. Using ratios and proportions, we can find how many more minutes of music could be recorded. A proportion is a statement that two ratios are equal. Let x represent the number of minutes available on a CD. Then 80 minutes are to 80 x 700 megabytes as x minutes are to 256 megabytes. By setting the ratios 700 and 256 equal to each other, we obtain the proportion
80 x = . 700 256 Solving this equation for x gives 700x = 80(256) x = 80 # 256 700
About 29 minutes are available to record on the CD.
STUDY TIP
Look back at your progress so far this semester. Are there parts of your study process that need some adjustment? Are your notes and assignments organized? Are you spending enough time on homework and practice problems?
MAKING CONNECTIONS
Proportions and Fractional Parts
We could have solved the preceding problem by noting that the fraction of the CD still available for recording music is 256. So 256 of 80 minutes is 700 700 256 700
# 80
29.3 minutes.
EXAMPLE 1
Calculating the water content in snow
Six inches of light, fluffy snow are equivalent to about half an inch of rain in terms of water content. If 15 inches of this type of snow fall, estimate the water content.
Solution Let x be the equivalent amount of rain. Then 6 inches of snow are to 1 inch of rain as 15 inches 2 of snow are to x inches of rain, which can be written as the proportion
6
1 2
=
15 . x
Snow Snow = Rain Rain
Solving this equation gives
ISBN 1-256-49082-2
6x =
15 2
or
x =
15 = 1.25. 12
Thus 15 inches of light, fluffy snow are equivalent to about 1.25 inches of rain.
Now Try Exercise 65
Proportions frequently occur in geometry when we work with similar figures. Two triangles are similar if the measures of their corresponding angles are equal. Corresponding sides of similar triangles are proportional. Figure 7.7 shows two right triangles that are similar because each has angles of 30 , 60 , and 90 . We can find the length of side x by using proportions. Side x is to 16 as 5.5 is to 8, which can be written as the proportion x 5.5 = . 16 8 Solving yields the equation 8 x = 5.5(16) x = 11.
Cross multiply. Divide each side by 8. Hypotenuse Shorter leg = Hypotenuse Shorter leg
Figure 7.7
NOTE: Proportions can be set up in different ways and still produce the correct result. For example, we could say that x is to 5.5 in the smaller triangle as 16 is to 8 in the larger triangle.
16 x = 5.5 8
Hypotenuse Hypotenuse = Shorter leg Shorter leg
Solving, we obtain 8 x = 5.5(16), or x = 11, which is the same answer.
EXAMPLE 2
Calculating the height of a tree
A 6-foot-tall person casts a 4-foot-long shadow. If a nearby tree casts a 36-foot-long shadow, estimate the height of the tree. See Figure 7.8.
6 feet 4 feet 36 feet NOT TO SCALE
Figure 7.8
h 6 feet
Solution The triangles shown in Figure 7.9 represent the situation shown in Figure 7.8. These triangles are similar because the measures of the corresponding angles are equal. Therefore their sides are proportional. Let h be the height of the tree.
4 feet
36 feet
h 36 = 6 4 4h = 6(36) h = 6(36) 4
Height Shadow length = Height Shadow length Cross multiply. Divide each side by 4.
ISBN 1-256-49082-2
Figure 7.9
h = 54 The tree is 54 feet tall.
Now Try Exercise 56
Direct Variation
N REAL-WORLD CONNECTION If your wage is $12 per hour, then your total pay P is proportional to the number of hours H that you work, which can be represented by the equation
P 12 = , H 1 or, equivalently,
Pay Pay = Hours Hours
P = 12H. We say that your pay P is directly proportional to the number of hours H that you work. The constant of proportionality is 12.
DIRECT VARIATION
Let x and y denote two quantities. Then y is directly proportional to x, or y varies directly with x, if there is a nonzero number k such that y = kx. The number k is called the constant of proportionality, or the constant of variation.
The following 4-step process is helpful when solving variation applications. This process is used in Examples 3 and 4 and can also be used for other types of variation problems.
SOLVING A VARIATION APPLICATION
When solving a variation problem, the following steps can be used.
STEP 1: STEP 2: STEP 3: STEP 4:
Write the general equation for the type of variation problem that you are solving. Substitute given values in this equation so the constant of variation k is the only unknown value in the equation. Solve for k. Substitute the value of k in the general equation in Step 1. Use this equation to find the requested quantity.
EXAMPLE 3
Solving a direct variation problem
Let y be directly proportional to x, or vary directly with x. Suppose y = 7 when x = 5. Find y when x = 11.
Solution STEP 1: The general equation for direct variation is y = kx. STEP 2: Substitute 7 for y and 5 for x in y = k x. Solve for k.
7 = k(5) 7 = k 5
ISBN 1-256-49082-2
Let y = 7 and x = 5. Divide each side by 5.
STEP 3: STEP 4:
Replace k with 7 in the equation y = kx to obtain y = 7 x. 5 5 To find y when x = 11, let x = 11 in y = 7 x. 5 y = 7 77 (11) = = 15.4 5 5
Solving a direct variation application
The amount of weight that a beam of wood can support varies directly with its width. A beam that is 2.5 inches wide can support 800 pounds. How much weight can a similar beam support if its width is 3.2 inches?
Solution STEP 1: The general equation for direct variation is y = kx, where y is the weight and x is the width. STEP 2: Substitute 800 for y and 2.5 for x in y = k x. Solve for k.
800 = k(2.5) 800 = k 2.5 k = 320
STEP 3: STEP 4:
Let y = 800 and x = 2.5. Divide each side by 2.5. Simplify; rewrite the equation.
Replace k with 320 in y = kx to obtain y = 320x. To find the weight y when x = 3.2, substitute 3.2 for x in y = 320x. y = 320(3.2) = 1024 pounds
Now Try Exercise 67
The graph of y = kx is a line passing through the origin, as illustrated in Figure 7.10. Sometimes data in a scatterplot indicate that two quantities are directly proportional. The constant of proportionality k corresponds to the slope (which may be negative) of the line passing through the points in the scatterplot.
Direct Variation y READING CHECK
• How does the constant of proportionality for data that represent direct variation compare to the slope of a line that passes through the data?
y = kx, k > 0
x
Figure 7.10
EXAMPLE 5
Modeling college tuition
Table 7.7 lists the tuition for taking various numbers of credits. (a) A scatterplot of the data is shown in Figure 7.11. Could the data be modeled using a line?
Table 7.7 y Tuition
(b) Explain why tuition is directly proportional to the number of credits taken. (c) Find the constant of proportionality. Interpret your result. (d) Predict the cost of taking 16 credits.
Solution (a) The data are linear and suggest a line passing through the origin. (b) Because the data can be modeled by a line passing through the origin, tuition is directly proportional to the number of credits taken. Hence doubling the credits will double the tuition and tripling the credits will triple the tuition. (c) The slope of the line equals the constant of proportionality k. If we use the first and last data points (3, 189) and (17, 1071), the slope is
k =
1071 - 189 = 63. 17 - 3
That is, tuition is $63 per credit. If we graph the line y = 63x, it models the data, as shown in Figure 7.12. This graph can also be created with a graphing calculator.
Tuition
y
1250
Tuition (dollars)
1000 750 500 250 0 4 8 12 16 20
x
Credits
Figure 7.12
(d) If y represents tuition and x represents the credits taken, 16 credits would cost y = 63(16) = $1008.
Now Try Exercise 73
MAKING CONNECTIONS
Ratios and the Constant of Proportionality
The constant of proportionality in Example 5 can also be found by calculating the ratios y x, where y is the tuition and x is the credits taken. Note that each ratio in the table is 63 y because the equation y = 63x is equivalent to the equation x = 63. x y y x
3 189 63
5 315 63
8 504 63
11 693 63
17 1071 63
ISBN 1-256-49082-2
Inverse Variation
N REAL-WORLD CONNECTION When two quantities vary inversely, an increase in one quantity results in a decrease in the second quantity. For example, at 30 miles per hour a car travels 120 miles in 4 hours, whereas at 60 miles per hour the car travels 120 miles in 2 hours.
Doubling the speed (or rate) decreases the travel time by half. Distance equals rate times time, so d = rt. Thus 120 = rt, or equivalently, t = 120 . r
We say that the time t to travel 120 miles is inversely proportional to the speed or rate r. The constant of proportionality or constant of variation is 120.
READING CHECK
• When two quantities are inversely proportional, how does an increase in one quantity affect the other quantity?
INVERSE VARIATION
Let x and y denote two quantities. Then y is inversely proportional to x, or y varies inversely with x, if there is a nonzero number k such that k y = . x
The data shown in Figure 7.13 represent inverse variation and are modeled by y = k. x Note that, as x increases, y decreases. We assume k is positive.
Inverse Variation y y =k , k > 0 x
x
Figure 7.13
EXAMPLE 6
Solving an inverse variation problem
Let y be inversely proportional to x, or vary inversely with x. Suppose y = 5 when x = 6. Find y when x = 21.
Solution k STEP 1: The general equation for inverse variation is y = x . k STEP 2: Because y = 5 when x = 6, substitute 5 for y and 6 for x in y = x . Solve for k.
5 =
k 6
Let y = 5 and x = 6. Multiply each side by 6.
ISBN 1-256-49082-2
30 = k
STEP 3:
Replace k with 30 in the equation y = k to obtain y = 30 . x x 30 10 STEP 4: To find y, let x = 21. Then y = 21 = 7 .
Now Try Exercise 39
N REAL-WORLD CONNECTION A wrench is commonly used to loosen a nut on a bolt. See Figure 7.14. If the nut is difficult to loosen, a wrench with a longer handle is often helpful.
L
F
Figure 7.14
EXAMPLE 7
Illustrating inverse variation with a wrench
Table 7.8 lists the force F necessary to loosen a particular nut with wrenches of different lengths L.
Table 7.8
L (inches) F (pounds)
6 12
8 9
12 6
18 4
24 3
(a) Make a scatterplot of the data and discuss the graph. Are the data linear? (b) Explain why the force F is inversely proportional to the handle length L. Find k so that k F = L models the data. (c) Predict the force needed to loosen the nut with a 15-inch wrench.
Solution (a) The scatterplot shown in Figure 7.15 reveals that the data are nonlinear. As the length L of the wrench increases, the force F necessary to loosen the nut decreases.
Wrench Force
F
12
Figure 7.15 k (b) If F is inversely proportional to L, then F = L, or FL = k. That is, the product of F and L equals the constant of proportionality k. In Table 7.8, the product of F and L always equals 72 for each data point. Thus F is inversely proportional to L with constant of proportionality k = 72, so F = 72 . L (c) If L = 15, then F = 72 = 4.8. A wrench with a 15-inch handle requires a force of 15 4.8 pounds to loosen the nut.
TECHNOLOGY NOTE
Scatterplots and Graphs A graphing calculator can be used to create scatterplots and graphs. A scatterplot of the data in Table 7.8 on the previous page is shown in the first figure. In the second figure, the data and the equation y = 72 are graphed. Note that each tick mark represents 5 units. x
CALCULATOR HELP
To make a scatterplot, see Appendix A (pages AP-3 and AP-4).
Analyzing Data
So far in this section, we have discussed direct and inverse variation. Table 7.9 gives a summary of these two types of variation.
TABLE 7.9 Direct and Inverse Variation
Type of Variation y varies directly with x
READING CHECK
• How can the constant of variation be used to determine whether data are directly or inversely proportional?
Equation y = kx y = k x
Constant of Variation k = x k = xy y y varies inversely with x
The last column in Table 7.9 shows that a set of data represents direct variation when y the quotients x equal a constant, and it represents inverse variation when the products xy equal a constant. In the next example, we determine if tables of data represent direct variation, inverse variation, or neither.
EXAMPLE 8
Analyzing data
Determine whether the data in each table represent direct variation, inverse variation, or neither. (a) x 4 (b) x 2 5 10 20 5 9 11 y (c) 40 32 16 8 y 18 45 81 99
x y
2 8
4 20
6 30
8 56
Solution (a) As x increases, y decreases. Because xy = 160 for each data point in the table, the equation y = 160 models the data. The data represent inverse variation. x y (b) Because x = 9 for each data point in the table, the equation y = 9x models the data in the table. These data represent direct variation. y (c) Neither the products xy nor the ratios x are constant for the data in the table. Therefore these data represent neither direct variation nor inverse variation.
Now Try Exercises 51(a), 53(a), 55(a)
Joint Variation
N REAL-WORLD CONNECTION In many applications a quantity depends on more than one variable. In joint variation a quantity varies with the product of more than one variable. For example, the formula for the area A of a rectangle is given by
A = WL, where W and L are the width and length, respectively. Thus the area of a rectangle varies jointly with the width and length.
JOINT VARIATION r Let x, y, and z denote three quantities. Then z varies jointly with x and y if there is a nonzero number k such that z = kxy. h Figure 7.16
Sometimes joint variation can involve a power of a variable. For example, the volume V of a cylinder is given by V = pr 2h, where r is its radius and h is its height, as illustrated in Figure 7.16. In this case we say that the volume varies jointly with the height and the square of the radius. The constant of variation is k = p.
EXAMPLE 9
Finding the strength of a rectangular beam
The strength S of a rectangular beam varies jointly with its width w and the square of its thickness t. See Figure 7.17. If a beam 3 inches wide and 5 inches thick supports 750 pounds, how much can a similar beam 2 inches wide and 6 inches thick support?
t w
Figure 7.17
Solution STEP 1: The strength of a beam is modeled by S = kwt 2, where k is a constant of variation. STEP 2: We can find k by substituting S = 750, w = 3, and t = 5 in the formula.
750 = k # 3 # 52 750 k = 3 # 52 = 10
2
Substitute in S = kwt 2. Solve for k; rewrite. Simplify.
STEP 3: STEP 4:
The equation S = 10wt models the strength of this type of beam. When w = 2 and t = 6, the beam can support S = 10 # 2 # 62 = 720 pounds.
Now Try Exercise 83
ISBN 1-256-49082-2
CRITICAL THINKING
Compare the increased strength of a beam if the width doubles and if the thickness doubles. What happens to the strength of a beam if both the width and thickness triple?
A statement that two ratios are equal Two quantities x and y vary according to the equation y = kx, where k is a nonzero constant. The constant of proportionality (or variation) is k.
8 49 = x 17 y = 4x x y 1 4
and
x 3 = 6 14 y = 4 x 4 16
Direct Variation
or 2 8
Note that if x doubles, then y also doubles. Inverse Variation Two quantities x and y vary according to the equation y = k, where k is x a nonzero constant. The constant of proportionality (or variation) is k. y = x y 3 x 1 3 or 3 1 xy = 3 6
1 2
Note that if x doubles from 3 to 6, then y decreases by half. Joint Variation Three quantities x, y, and z vary according to the equation z = kxy, where k is a nonzero constant. The area A of a triangle varies jointly with b and h according to the equation A = 1 bh, where b is its base and h is 2 its height. The constant of variation is k = 1. 2
7.7
Exercises
7. Would the food bill B generally vary directly or inversely with the number of people N being fed? Explain your reasoning.
Directly; doubling the number being fed doubles the bill.
CONCEPTS AND VOCABULARY
1. What is a proportion? A statement that two ratios are equal 2. If 5 is to 6 as x is to 7, write a proportion that allows x you to find x. 5 = 7 6 3. Suppose that y is directly proportional to x. If x doubles, what happens to y? It doubles. 4. Suppose that y is inversely proportional to x. If x doubles, what happens to y? It is halved. 5. If y varies directly with x, then x equals a(n) constant constant y 8. Would the time T needed to paint a building vary directly or inversely with the number of painters N working on the job? Explain your reasoning.
Inversely; doubling the number of painters will halve the time.
9. If xy equals a constant for every data point (x, y) in a table, then the data represent _____ variation. inverse 10. If x equals a constant for every data point (x, y) in a table, then the data represent _____ variation. direct y ISBN 1-256-49082-2
. .
6. If y varies inversely with x, then xy equals a(n)
33. y = 4 when x = 2 35. y = 3 when x = 2
(a)
15 7 3 2
34. y = 5 when x = 10 36. y = 11 when x = 55
(a)
1 5
Exercises 11–22: Solve the proportion. x 5 11. = 15 24 8 14 2 13. 21 = x 3 15. 17. 3 h = 48 16 256 3 2x = 4 7
21 8
(a) 2
(b) 12 (b) 9
(a)
1 2
(b) 3 (b)
6 5
x 3 12. = 5 7 4 9 14. = x 9 16. 18.
37. y = - 60 when x = 8 (a) - 15 (b) -45 2 38. y = - 17 when x = 68 (a) - 1 (b) - 3 4 2
81 4
20 15 = a 4 7 5 = 3z 4
16 3
28 15
Exercises 39–44: Inverse Variation Suppose that y is inversely proportional to x. (a) Use the given information to find the constant of proportionality k. (b) Then use y = k to find y for x = 8. x 39. y = 6 when x = 4 41. y = 80 when x = 1 2 43. y = 20 when x = 20
(a) 400 (b) 50 (a) 40 (b) 5 (a) 24 (b) 3
40. y = 2 when x = 24 42. y = 1 when x = 32 4 44. y = 8 when x = 12 3
(a) 32 (b) 4 (a) 8 (b) 1 (a) 48 (b) 6
x 8 19. = -4, 4 6 3x 21. x 7 = - 7, 7 7 4x 2 2
4 4x 20. -3, 3 = x 9 22. 2 27x 4 4 = - 9, 9 3x 8 ad c
c 23. Thinking Generally Solve a = d for b. b = b + 24. Thinking Generally Solve a c 2 b = 1 for b. 2 c 2 - 2a
Exercises 45–50: Joint Variation Let z vary jointly with x and y. (a) Find the constant of variation k. (b) Use z = kxy to find z when x = 5 and y = 7. 45. z = 6 when x = 3 and y = 8 46. z = 135 when x = 2.5 and y = 9 47. z = 5775 when x = 25 and y = 21 48. z = 1530 when x = 22.5 and y = 4
(a) k = 11 (a) k = 17 (a) k = 10 (a) k = 4 (b) z = 385 (b) z = 595 (a) k = 6 (b) z = 210 (a) k = 0.25 (b) z = 8.75
b =
2
Exercises 25–32: Do the following. (a) Write a proportion that models the situation. (b) Solve the proportion for x. 25. 5 is to 8 as 9 is to x. (a) 5 = 8
9 x 7 4
(b)
72 5 77 4
x 26. x is to 11 as 7 is to 4. (a) 11 =
(b)
49. z = 25 when x = 1 and y = 5 2 50. z = 12 when x = 1 and y = 12 4
(b) y = 140 (b) y = 350
27. A triangle has sides of 4, 7, and 10. In a similar triangle the shortest side is 8 and the longest side is x.
(a)
28. A rectangle has sides of 5 and 12. In a similar rectangle the longer side is 10 and the shorter side is x.
(a)
5 x
4 8
=
10 x
(b) 20
ANALYZING DATA
=
29. If you earn $98 in 7 hours, then you can earn x dollars in 11 hours.
(a)
98 7
12 10
(b)
25 6
=
x 11
(b) $154
30. If 14 gallons of gasoline contain 1.4 gallons of ethanol, then 22 gallons of gasoline contain x gallons of ethanol.
(a)
14 1.4
Exercises 51–56: (Refer to Example 8.) (a) Determine whether the data represent direct variation, inverse variation, or neither. (b) If the data represent either direct or inverse variation, find an equation that models the data. (c) Graph the equation and the data when possible.* 51. x y 52. x y 53. 2 3 10 12 3 12 3 4.5 20 6 6 6 4 6 30 5 9 4 5 7.5 40 4 12 3
(a) Direct (b) y = 3 x 2
=
22 x
(b) 2.2 gal
31. If 3 MP3 players can hold 750 songs, then 7 similar MP3 players can hold x songs.
(a)
32. If a gas pump fills a 25-gallon tank in 6 minutes, it can fill a 14-gallon tank in x minutes.
(a)
25 6
3 750
=
7 x
(b) 1750 (a) Neither (b) NA (c) NA
=
14 x
(b)
84 25
= 3.36 min
VARIATION
ISBN 1-256-49082-2
Exercises 33–38: Direct Variation Suppose that y is directly proportional to x. (a) Use the given information to find the constant of proportionality k. (b) Then use y = kx to find y for x = 6.
Exercises 57–62: Use the graph to determine whether the data represent direct variation, inverse variation, or neither. Find the constant of variation whenever possible. 57.
10 8 6 4 2 0 2 4 6 8 10
y
Direct; 2
58.
7 6 5 4 3 2 1
y
Inverse; 6
65. Water Content in Snow (Refer to Example 1.) Eight inches of heavy, wet snow are equivalent to an inch of rain. Estimate the water content in 13 inches of heavy, wet snow. 1.625 in. 66. Wages If a person working for an hourly wage earns $143 in 13 hours, how much will that person earn in 15 hours? $165
x
0 1 2 3 4 5 6 7
x
59.
10 8 6 4 2
y
Neither
60.
7 6 5 4 3 2 1
y
Neither
67. Strength of a Beam (Refer to Example 4.) The strength of a metal beam varies directly with its width. A beam that is 6.2 inches wide can support 2800 pounds. How much weight can a similar beam support if it is 4.7 inches wide? About 2123 lb 68. Strength of a Beam The strength of a wood beam varies inversely with its length. A beam that is 33 feet long can support 1200 pounds. How much weight can a similar beam support if it is 23 feet long? About 1722 lb
0
2
4
6
8 10
x
0 1 2 3 4 5 6 7
x
61.
10 8 6 4 2
y
Inverse; 8
62.
6 5 4 3 2 1
y
Direct; 1
69. Making Fudge If 2 2 cups of sugar can make 14 3 pieces of fudge, how much sugar is needed to make 49 pieces of fudge? 9 1 c 3 70. Making Coffee If 6 tablespoons of coffee grounds make 10 cups of coffee, how many tablespoons of coffee grounds are needed to make 35 cups of coffee?
21 tbsp
0
2
4
6
8
10
x
0
1 2
3
4 5
6
x
APPLICATIONS
63. Recording Music A 750-megabyte CD can record 85 minutes of music. How many minutes can be recorded on 420 megabytes? 47.6 min 64. Height of a Tree (Refer to Example 2.) A 5-foot-tall person casts an 8-foot-long shadow, and a nearby tree casts a 30-foot-long shadow. See the figure at the top of the next column. Estimate the height of the tree.
18.75 ft
71. Rolling Resistance of Cars If you were to try to push a car, you would experience rolling resistance. This resistance equals the force necessary to keep the car
moving slowly in neutral gear. The following table shows the rolling resistance R for passenger cars of different gross weights W. (Source: N. Garber and L. Hoel,
Traffic and Highway Engineering.)
L (inches) F (pounds)
Source: Tires Plus.
8 150
10 120
16 75
W (pounds) R (pounds)
2000 24
2500 30
3000 36
3500 42
k (a) Model the data, using the equation F = L. F = 1200 L (b) How much force should be used with a wrench 15 inches long? 80 lb
(a) Do the data represent direct or inverse variation? R Explain. Direct; the ratios W always equal 0.012. (b) Find an equation that models the data. Graph the equation with the data.* R = 0.012W (c) Estimate the rolling resistance of a 3200-pound car. 38.4 lb 72. Transportation Costs The use of a particular toll bridge varies inversely according to the toll. If the toll is $0.50, then 8000 vehicles use the bridge. Estimate the number of users if the toll is $0.80. (Source: N. Garber.)
5000 users
73. Flow of Water The gallons of water G flowing in 1 minute through a hose with a cross-sectional area A are shown in the table. A (square inch) G (gallons) 0.2 5.4 0.3 8.1 0.4 10.8 0.5 13.5
76. Ozone and UV Radiation Ozone in the atmosphere filters out approximately 90% of the harmful ultraviolet (UV) rays from the sun. Depletion of the ozone layer increases the amount of UV radiation reaching Earth’s surface. An increase in UV radiation is associated with skin cancer. The following graph shows the percentage increase y in UV radiation for a decrease in the ozone layer of x percent. (Source: R. Turner, D. Pearce, and I. Bateman, Environmental Economics.)
(a) Do the data represent direct or inverse variation? Explain. Direct; the ratios G always equal 27. A (b) Find an equation that models the data. Graph the equation with the data.* G = 27A (c) Interpret the constant of variation k. For each square-inch increase in the cross-sectional area, the flow increases by 27 gal/min.
Ozone Depletion y 16
Increase in UV Radiation (%)
74. Hooke’s Law The table shows the distance D that a spring stretches when a weight W is hung on it. W (pounds) D (inches) 2 1.6 6 4.8 9 7.2 15 12
14 12 10 8 6 4 2 0 2 4 6 8 10 12 14 16
(a) Do the data represent direct or inverse variation? D Explain. Direct; the ratios W always equal 0.8. (b) Find an equation that models the data. D = 0.8W (c) How far will the spring stretch if an 11-pound weight is hung on it? 8.8 in. 75. Tightening Lug Nuts (Refer to Example 7.) When a tire is mounted on a car, the lug nuts should not be over-tightened. The table at the top of the next column shows the maximum force used with wrenches of different lengths.
x
Decrease in Ozone Layer (%)
ISBN 1-256-49082-2
(a) Does this graph represent direct or inverse variation? Direct (b) Find an equation for the line in the graph. y = 1.5x (c) Estimate the percentage increase in UV radiation if the ozone layer decreases by 5%. 7.5%
77. Air Temperature and Altitude In the first 6 miles of Earth’s atmosphere, air cools as the altitude increases. The following graph shows the temperature change y in degrees Fahrenheit at an altitude of x miles. (Source:
A. Miller and R. Anthes, Meteorology.)
swept out an area 12 feet in diameter and the wind speed was 15 miles per hour? About 18,263 watts
y
200
Temperature Change (°F)
150 100 50 0 -50 -100 -150 2 3 4 5 6
x
(5, −95)
Altitude (miles)
83. Strength of a Beam (Refer to Example 9.) If a wood beam 5 inches wide and 3 inches thick supports 300 pounds, how much can a similar beam 5 inches wide and 2 inches thick support? About 133 lb 84. Carpeting The cost of carpet for a rectangular room varies jointly with its width and length. If a room 11 feet wide and 14 feet long costs $539 to carpet, find the cost to carpet a room 17 feet by 19 feet. Interpret the constant of variation k.
$1130.50; k represents a $3.50/ft 2 cost.
(c) Negative; for each 1-mile increase in altitude the temperature decreases by 19 F.
(a) Does this graph represent direct variation or inverse variation? Direct (b) y = -19x (b) Find an equation that models the data in the graph. (c) Is the constant of proportionality k positive or negative? Interpret k. (d) Find the change in air temperature 2.5 miles high.
47.5 F decrease
78. Cost of Tuition (Refer to Example 5.) The cost of tuition is directly proportional to the number of credits taken. If 6 credits cost $483, find the cost of 13 credits. What does the constant of proportionality represent? $1046.50; k represents the cost per credit. 79. Electrical Resistance The electrical resistance of a wire is directly proportional to its length. If a 30-footlong wire has a resistance of 3 ohms, find the resistance of an 18-foot-long wire. 1.8 ohms 80. Resistance and Current The current that flows through an electrical circuit is inversely proportional to the resistance. When the resistance R is 180 ohms, the current I is 0.6 amp. Find the current when the resistance is 54 ohms. 2 amps 81. Joint Variation The variable z varies jointly with the second power of x and the third power of y. Write a formula for z if z = 31.9 when x = 2 and y = 2.5. z = 0.5104 x 2y 3
85. Weight on the Moon The weight of a person on the moon is directly proportional to the weight of the person on Earth. If a 175-pound person weighs 28 pounds on the moon, how much will a 220-pound person weigh on the moon? About 35.2 lb 86. Weight Near Earth The weight W of a person near Earth is inversely proportional to the square of the person’s distance d from the center of Earth. If a person weighs 200 pounds when d = 4000 miles, how much does the same person weigh when d = 7000 miles? (Note: The radius of Earth is about 4000 miles.)
About 65.3 lb
87. Ohm's Law The voltage V in an electrical circuit varies jointly with the amperage I and resistance R. If V = 220 when I = 10 and R = 22, find V when I = 15 and R = 50. 750 88. Revenue The revenue R from selling x items at price p varies jointly with x and p. If R = $24,000 when x = 3000 and p = $8, find the number of items x sold when R = $30,000 and p = $6. 5000
WRITING ABOUT MATHEMATICS
82. Wind Power The electrical power generated by a windmill varies jointly with the square of the diameter of the area swept out by the blades and the cube of the wind velocity. If a windmill with a 10-foot diameter and a 16-mile-per-hour wind generates 15,392 watts, how much power would be generated if the blades
89. Explain what it means for a quantity y to be directly proportional to a quantity x. 90. Explain what it means for a quantity y to be inversely proportional to a quantity x.
ISBN 1-256-49082-2
Checking Basic Concepts
4. Decide whether the data in the table represent direct or inverse variation. Explain your reasoning. Find the constant of variation. (a) x 2 4 6 8 y (b) x y 3 2 12 6 4 6 9 6 4 12 8 3
1. Solve each proportion. x 2 4 5 (a) = (b) = 9 5 3 b 2. Write a proportion that models each situation. Then solve it. (a) 4 is to 6 as 8 is to x. (b) If 2 compact discs can record 148 minutes of music, then 5 compact discs can record x minutes of music. 3. Suppose that y is inversely proportional to x. If y = 4 when x = 15, find the constant of proportionality k. Find y when x = 10.
5. Wages If a person working for an hourly wage earns $272 in 17 hours, how much will the person earn in 10 hours?
CHAPTER 7
SECTION
Summary
7.1
.
INTRODUCTION TO RATIONAL EXPRESSIONS
Rational Expression P A rational expression can be written in the form Q, where P and Q are polynomials, and is defined whenever Q 0.
Example: x - 5 is a rational expression that is defined for all real numbers except x = 5. x2 Undefined Rational Expression A rational expression is undefined for any value of the variable that makes the denominator equal to 0.
Examples: x - 5 is undefined when x = 5.
3y y2 - 4 1
is undefined when y = - 2 or when y = 2.
Simplifying a Rational Expression To simplify a rational expression, factor the numerator and the denominator. Then apply the basic principle of rational expressions,
PR P = . QR Q
Examples:
Q and R nonzero
(x + 2)(x - 2) x2 - 4 x + 2 = = , 2 (x - 1)(x - 2) x - 1 x - 3x + 2
.
x - 3 x - 3 = = -1 3 - x x - 3
SECTION
7.2
MULTIPLICATION AND DIVISION OF RATIONAL EXPRESSIONS
Multiplying Rational Expressions To multiply two rational expressions, multiply the numerators and multiply the denominators.
ISBN 1-256-49082-2
Dividing Rational Expressions To divide two rational expressions, multiply by the reciprocal of the divisor.
C A A , = B D B
Example:
#D
C
=
AD BC
B, C, and D nonzero
x + 2 x + 2 x + 2 , = 2 x x(x + 3) x + 3x
7.3
.
#
x(x + 2) x 1 = = x + 2 x(x + 3)(x + 2) x + 3
SECTION
ADDITION AND SUBTRACTION WITH LIKE DENOMINATORS
Addition of Rational Expressions Having Like Denominators To add two rational expressions having like denominators, add their numerators. Keep the same denominator.
A B A + B + = C C C
Example:
C nonzero
5x 1 5x + 1 + = x + 4 x + 4 x + 4
Subtraction of Rational Expressions Having Like Denominators To subtract two rational expressions having like denominators, subtract their numerators. Keep the same denominator.
A B A - B = C C C
Example:
C nonzero
6 2 4 = 2x + 1 2x + 1 2x + 1
7.4
.
SECTION
ADDITION AND SUBTRACTION WITH UNLIKE DENOMINATORS
Finding Least Common Multiples The least common multiple (LCM) of two or more polynomials can be found as follows.
STEP 1: STEP 2: STEP 3:
Factor each polynomial completely. List each factor the greatest number of times that it occurs in any factorization. Find the product of this list of factors. It is the LCM. 2 x(x
2
Example: 4 x 2(x + 1) = 2
# 2 # x # x # (x + 1) - 1) = 2 # x # (x + 1) # (x - 1)
Listing each factor the greatest number of times and multiplying gives the LCM. 2 # 2 # x # x # (x + 1)(x - 1) = 4 x 2(x 2 - 1)
Finding the Least Common Denominator The least common denominator (LCD) is the least common multiple (LCM) of the denominators.
Example: From the preceding example, the LCD for
1 4 x 2(x + 1)
and 2 x(x 21 - 1) is 4 x 2(x 2 - 1).
Addition and Subtraction of Rational Expressions Having Unlike Denominators First write each rational expression by using the LCD. Then add or subtract the resulting rational expressions. Finally, write your answer in lowest terms.
Example:
x - (x - 1) 1 1 x x - 1 1 = = = x x - 1 x(x - 1) x(x - 1) x(x - 1) x(x - 1) Note that the LCD is x(x - 1).
Complex Fractions A complex fraction is a rational expression that contains fractions in its numerator, denominator, or both. The following equation can be used to simplify basic complex fractions.
a b a = c b d x 3 x Example: = x 3 x - 1
#d c b, c, and d nonzero
#x-1 x =
x(x - 1) x - 1 = 3x 3
Simplifying Complex Fractions Method I Combine terms in the numerator, combine terms in the denominator, and then multiply the numerator by the reciprocal of the denominator.
Method II Multiply the numerator and denominator by the LCD of all fractions within the expression and simplify the resulting expression. b - a 1 1 a b ab b - a = = 1 b + a ab 1 + a b ab
Example: Method I
#
ab b - a = b + a b + a
Method II
The least common denominator is ab. ab 1 1 1 ab 1 a - b ab a a a b b b b - a = = = 1 ab b + a 1 ab 1 1 + + a + b ab a a a b b b
SECTION
7.6
.
RATIONAL EQUATIONS AND FORMULAS
c Solving Rational Equations One way to solve the equation a = d is to cross multiply to obtain b ad = bc. (Check each answer.) A general way to solve rational equations is to follow these steps.
STEP 1: STEP 2: STEP 3: STEP 4: STEP 5:
Find the LCD of the terms in the equation. Multiply each side of the equation by the LCD. Simplify each term. Solve the resulting equation. Check each answer in the given equation. Reject any value that makes a denominator equal 0.
1 2 1 To solve 5 - 3x = 7 multiply each term by the LCD, 3x: x 3
Examples: 2 x = x + 3 implies that x + 3 = 4 x, or x = 1. This answer checks.
ISBN 1-256-49082-2
5(3x) 7(3x) 3x = , x 3x 3 which simplifies to 15 - 1 = 7x, or x = 2. This answer checks.
Solving for a Variable Many formulas contain more than one variable. To solve for a particular variable, use the rules of algebra to isolate the variable.
Example: To solve S = r for r, multiply each side by r to obtain Sr = 2p and then divide each
2p
side by S to obtain r = 2p. S
.
SECTION
7.7
5
PROPORTIONS AND VARIATION
Proportions A proportion is a statement that two ratios are equal.
Example: x = 7
4
Similar Triangles Two triangles are similar if the measures of their corresponding angles are equal. Corresponding sides of similar triangles are proportional.
Example: The following triangles are similar.
6 3 4 x
3 4 = x 6
Shorter leg Longer leg = Shorter leg Longer leg
By solving this proportion, we see that x = 8.
Direct Variation A quantity y is directly proportional to a quantity x, or y varies directly with x, if there is a nonzero constant k such that y = kx. The number k is called the constant of proportionality or the constant of variation.
Example: If y varies directly with x, then the ratios x always equal k. The following data satisfy y x y
= 4, so the constant of variation is 4. Thus y = 4 x. x y 1 4 2 8 3 12 4 16
Inverse Variation A quantity y is inversely proportional to a quantity x, or y varies inversely with x, if there is a nonzero constant k such that y = k. x
Example: If y varies inversely with x, then the products xy always equal k. The following data
satisfy xy = 12, so the constant of variation is 12. Thus y = 12. x x y 1 12 2 6 4 3 6 2
Joint Variation The quantity z varies jointly with x and y if z = kxy, k
0.
ISBN 1-256-49082-2
Example: The area A of a rectangle varies jointly with the width W and length L because A = LW.
Review Exercises
Exercises 19–24: Divide and write in lowest terms. 19. 21. 22. 23. x + 1 3x + 3 , 2x 5x x - 5 2 x - 10 , x + 2 x + 2
5 6
Exercises 1–4: If possible, evaluate the expression for the given value of x. 1. 3 x - 3 -x 7 - x x = -2 -3 5 x = 7 2. 4x 5 - x2
2
20.
1 2
4 x + 1 , 3 x 2x2
8 x(x + 1)
x = 3 -3 x = 2
3.
4.
Undefined
Undefined
4x x - 3x + 2
x - 1 x2 - 6x + 5 , 1 2 x + 5 x - 25 x 2 - y2 x - y x + y , x + y x + y a - b a3 - b 3 , 2(a 2 + ab + b 2) a + b 2a + 2b
5. Complete the table for the rational expression. If a value is undefined, place a dash in the table. x
3x x - 1
-2
2
-1
3 2
0
0
1
—
2 24.
6
6. Find the x-values that make x 2 8 4 undefined. -2, 2 Exercises 7–12: Simplify to lowest terms. 7. 9. 25x 3y 4 15x 5y
5y 3x 2
3
SECTION 7.3
Exercises 25–32: Add or subtract and write in lowest terms. 25. 2 8 + x + 10 x + 10 26. 28. 30. 9 8 x - 1 x - 1 x 3 + x + 3 x + 3
2
x 2 - 36 8. x - 6 x + 6 10. x + 3 x + 1
x - 9 -1 9 - x 2 x + 5x - 3 2x2 + x - 1
2
x - 5x 5x
2 2
x - 5 5
x - 2y x + 2y + 27. 2x 2x
1
x + 4 x + 1
10 x + 10
1 x - 1
1
11.
12.
3x + 10x - 8 3x 2 + x - 2
1 x - 2 29. 2 x - 1 x - 1 3 1 31. xy xy
2 xy 1 x + 1
2 x - 5
10 2x + 2 x - 25 x - 25 x + y x - y + 2y 2y
32.
Exercises 13 and 14: Do the following. (a) Decide whether you are given an expression or an equation. (b) If you are given an expression, simplify it. If you are given an equation, solve it. 13. x + 1 (x - 3)(x + 1)
(a) Expression (b)
x y
SECTION 7.4
Exercises 33–38: Find the least common multiple for the expressions. Leave your answer in factored form. 33. 3x, 5x 35. x, x - 5 37. x - 1, (x + 1)
2
14.
1 x - 3
(a) Equation
x = 2 3(4 - 1)
34. 5x 2, 10x 36. 10x 2, x 2 - x
2
15x
10x 2
(b) 18
SECTION 7.2
x(x - 5)
Exercises 15–18: Multiply and write in lowest terms. x - 3 15. x + 1
(x - 1)(x + 1)2
38. x 2 - 4 x, x 2 - 16 x(x - 4)(x + 4)
10x 2(x - 1)
# 2x + 2 x - 3
2
Exercises 39–44: Rewrite the rational expression by using the given denominator D. 39.
1 x + 5
Exercises 73–86: If possible, solve. Check your answer.
8 - 9x 6x2
1 1 49. + x + 1 x - 1
2x (x - 1)(x + 1)
4 3 50. 2 2x 3x 52.
2
73. 75.
1 3 1 + = 4 5x 5x 5 1 1 2 = 5 + x 3x 3
74. 76.
2x 1 + = 1 x - 1 x - 1 -2 2x 3 1 + = x + 3 x + 3 2
51.
1 + x 3 3x 2x
2x - 7 6x
1 x x - 1 x - 1
- (x
1 - 1)(x + 1) 13 6x
7
2 3 53. x - y x + y 3 1 55. + 2y 2x
SECTION 7.5
5y - x (x - y)(x + y) 3x + y 2 xy
2 2 1 + 54. x 2x 3x
5 1 3 77. = -2, 5 x x + 1 2 4 5 79. = 0 8 p p + 2 80.
1 1 1 78. = x - 1 x + 1 4
-3, 3
y x 56. -1 + y - x x - y
Exercises 57–66: Simplify the complex fraction. 3 4 57. 7 11 m n 59. 2m n2 x 5 58. 2x 7 81.
7 10
1 1 1 = 2 No solutions x - 3 x + 3 x - 9 1 -x = No solutions (extraneous: -1) x + 1 x + 1 2 2 - 4 - 3 (extraneous: 0) = 2 2 x x + x 1 1 2 + 2 = 2 -3 x - 2x x - 4 x + 2x
2
33 28
82. 83.
3(p + 1) p - 1
n 2
3 p - 1 60. 1 p + 1 2 2n + 1 62. 8 2n - 1 2 1 xy y 64. 2 1 + xy y
84. 85. 86.
2 - x 2 + x
1 1 3 - 2 = 2 -12 x 2 - 3x x - 9 x + 3x 1 1 5 = 2 - 2 2 x x + 4x x + 4x
2
4 5
3 m - 1 61. 2m - 2 m + 1 1 1 2x 3x 63. 2 1 3x 6x
3(m + 1) 2(m - 1)2
2n - 1 4(2n + 1)
1 3 5 - 2 = 2 -3 x - 1 x + 2x + 1 x - 1
1 3
1 1 x x + 1 65. x x + 1
SECTION 7.6
1 x2
2 1 x - 1 x + 1 66. 1 x + 3 2 x - 1
Exercises 87 and 88: Do the following. (a) Decide whether you are given an expression or an equation. (b) If you are given an expression, simplify it. If you are given an equation, solve it. 87. 4 - x = 0 x
(a) Equation
88.
(b) -2, 2
(a) Expression (b) x + 3
x2 9 x - 3 x - 3
Exercises 67–72: Solve and check your answer. 67. x 4 = 5 7
20 7
Exercises 89 and 90: Solve for the specified variable. x 1 3 2 90. y = for x 89. = for b + x - 1 c a b b =
2ac 3a - c
Exercises 91 and 92: Solve the proportion. 91. x 1 = 2 6 3 92. 5 7 = x 3
15 7
Exercises 103 and 104: Use the graph to determine whether the data represent direct or inverse variation. Find the constant of variation. 103.
8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8
y
Direct;
1 2
104.
12 10 8 6 4 2
y
Inverse; 12
Exercises 93 and 94: Proportions Do the following. (a) Write a proportion that models the situation. (b) Solve the proportion for x. 93. A rectangle has sides of 6 and 13. In a similar rectangle the longer side is 20 and the shorter side is x.
(a)
6 x
x
0
2
4
6
8 10 12
x
=
13 20
(b)
120 13
94. If you earn $341 in 11 hours, then you can earn x x dollars in 8 hours. (a) 341 = 8 (b) $248 11 Exercises 95 and 96: Direct Variation Suppose that y is directly proportional to x. (a) Use the given information to find the constant of proportionality k. (b) Then use y = kx to find y for x = 5. 95. y = 8 when x = 2
(a) 4 (b) 20
APPLICATIONS
105. Modeling Traffic Flow Fifteen vehicles per minute can pass through an intersection. If vehicles arrive randomly at an average rate of x per minute, the average waiting time T in minutes is given by T = 15 1 x for x 6 15. (Source: N. Garber, Traffic and Highway Engineering.)
96. y = 21 when x = 7
(a) 3 (b) 15
(a) Find T when x = 10 and interpret the result. (b) Complete the table. (a) 1 = 0.2; when the average rate 5 of arrival is 10 cars/min, the average wait is 0.2 min, or 12 sec.
x T
5
1 10
10
1 5
13
1 2
14
1
14.9
10
Exercises 97 and 98: Inverse Variation Suppose that y is inversely proportional to x. (a) Use the given information to find the constant of proportionality k. (b) Then use y = k to find y for x = 5. x 97. y = 2.5 when x = 4
(a) 10 (b) 2
(c) What happens to the waiting time as the traffic rate x approaches 15 vehicles per minute?
It increases dramatically.
98. y = 7 when x = 3
(a) 21 (b)
21 5
99. Joint Variation Suppose that z varies jointly with x and y. If z = 483 when x = 23 and y = 7, find the constant of variation k. k = 3 100. Joint Variation Suppose that z varies jointly with x and the square of y. If z = 891 when x = 22 and y = 3, find z when x = 10 and y = 4. z = 720 Exercises 101 and 102: Do the following. (a) Determine whether the data represent direct or inverse variation. (b) Find an equation that models the data. (c) Graph the data and your equation.* 101. x y 102. x
ISBN 1-256-49082-2
106. Distance and Time A car traveled at 50 miles per hour for 150 miles and then traveled at 75 miles per hour for 150 miles. What was the car’s average speed? 60 mph 107. Emptying a Swimming Pool A large pump can empty a swimming pool in 100 hours, whereas a small pump can empty the pool in 160 hours. How long will it take to empty the pool if both pumps are used? 800 61.5 hr 13 108. Running Two athletes run 10 miles. One of the athletes runs 2 miles per hour faster and finishes 10 minutes ahead of the other athlete. Find the average speed of each athlete. 10 mph and 12 mph 109. River Current A boat can travel 16 miles upstream in the same time that it can travel 48 miles downstream. If the speed of the current is 4 miles per hour, find the boat’s speed. 8 mph 110. Height of a Tree A 5-foot-tall person has a 6-footlong shadow, and a nearby tree has a 40-foot-long shadow. Estimate the tree’s height. About 33.3 ft
111. Transportation Costs Use of a toll road varies inversely with the toll. If the toll is $0.25, then 400 vehicles use the road. Estimate the number of users for a toll of $0.50. 200 vehicles
the water. How long could a person safely remain in air at the same temperature?
750 to 2250 sec, or 12.5 to 37.5 min
112. Cost of Carpet The cost of carpet is directly proportional to the amount of carpet purchased. If 17 square yards cost $612, find the cost of 13 square yards. $468 113. Tightening a Bolt The torque exerted on a nut by a wrench is inversely proportional to the length of the wrench’s handle. Suppose that a 12-inch wrench can be used to tighten a nut by using 30 pounds of force. How much force is necessary to tighten the same nut by using a 10-inch wrench? 36 lb 114. Water Content in Snow Twenty inches of extremely dry, powdery snow are equivalent to an inch of rain. Estimate the water content in 32 inches of this type of snow. 1.6 in. 115. Polar Plunge When a person swims in extremely cold water, the water removes body heat 25 times faster than air at the same temperature. To be safe, a person has between 30 and 90 seconds to get out of
116. Strength of a Beam The strength of a wood beam varies inversely with its length. A beam that is 18 feet long can support 900 pounds. How much weight can a similar beam support if its length is 21 feet? 771 lb 117. Wind Power The electric power generated by a windmill varies jointly with the square of the diameter of the area swept out by the blades and the cube of the wind velocity. If a windmill with 6-footdiameter blades and a 20-mile-per-hour wind generates 10,823 watts, how much power would be generated if the blades were 10 feet in diameter and the wind speed were 12 miles per hour? 6493.8 W 118. Strength of a Beam The strength of a beam varies jointly with its width w and the square of its thickness t. If a beam 8 inches wide and 5 inches thick supports 650 pounds, how much can a similar beam 6 inches wide and 6 inches thick support? 702 lb
CHAPTER 7
Test
9 5
Step-by-step test solutions are found on the Chapter Test Prep Videos available via the Video Resources , and on (search “RockswoldBeginAlg” and click on “Channels”). on DVD, in
1. Evaluate the expression 2 x 3x 1 for x = 3. -
9.
2. Find any x-value that makes x - 1 undefined. -2 x + 2 Exercises 3 and 4: Simplify the expression. 3. x 2 - 25 x + 5 x - 5 4. 3x 2 - 15x x - 5 3x
3x + 1 x + x + 4 x + 4
4x + 1 x + 4
10.
3t - 6 4t + 1 2t - 3 2t - 3
t + 7 2t - 3
11. Find the least common multiple for 6 x 2 and 3x 2 - 3x. 6 x 2(x - 1)
4 12. Rewrite the rational expression 7x by using the 2 4x - 4 denominator 7x - 7x. 7x 2 - 7x
Exercises 5–12: Simplify the expression. Write your answer in lowest terms. 5. x - 2 x + 4
Exercises 13 and 14: Simplify the expression. Write your answer in lowest terms. 13. y - 1 1 - 2 y + y y - y
2
Exercises 15 and 16: Simplify the complex fraction. a 3b 15. 5a b2 1 p - 1 16. 1 1 p - 1 1 +
b 15
p p - 2
27. Suppose that y is directly proportional to x. (a) If y = 14 when x = 4, find k so that y = kx. (b) Use y = kx to find y for x = 6. 21
7 2
28. Use the table to determine whether y varies directly or inversely with x. Find the constant of variation.
Inversely; 32
Exercises 17–24: Solve the equation and check your answer. 17. 19. 2 5 = x 7
35 2
x y
2 16
4 8
8 4
16 2
18. 20.
2 7
x + 3 = 1 3 2x 1 2 3 + = x - 1 x + 2 2
1 2 9 1 + = 2x 5x 10
4 3 1 = 21. 2 x + 1 x - 1 x - 1 22.
2
-1, 2
29. Emptying a Swimming Pool It takes a large pump 40 hours to empty a swimming pool, whereas a small pump can empty the pool in 60 hours. How long will it take to empty the pool if both pumps are used? 24 hr 30. Height of a Building A 5-foot-tall post has a 4-footlong shadow, and a nearby building has a 54-foot-long shadow. Estimate the height of the building. 67.5 ft 31. Standing in Line A department store clerk can wait on 30 customers per hour. If people arrive randomly at an average rate of x per hour, then the average number of customers N waiting in line is given by N = x2 , 900 - 30x
2 2 1 + 2 = 2 -12 x - 4x x - 16 x + 4x
x 1 - x 23. = No solutions (extraneous: 1) 2 2x - 1 2x - 1 24. x x 10x 0 (extraneous: 5) + = 2 x - 5 x + 5 x - 25
Exercises 25 and 26: Solve the equation for the specified variable. 25. y = 2 3x - 5 for x 26. a + b = 1 for b ab b = a a - 1
for x 6 30. Evaluate the expression for x = 24 and interpret your result.
= 3.2; when the arrival rate is 24 people/hr, there are about 3 people in line, on average.
16 5
x =
2 + 5y 3y
CHAPTER 7
Extended and Discovery Exercises
(b) For what value of x is N undefined? 15 (c) Plot the points from the table. Then graph x = 15 as a vertical, dashed line.* (d) Sketch a graph of N that passes through these points. Do not allow your graph to cross the vertical, dashed line, called an asymptote.* (e) Use the graph to explain why a small increase in x can sometimes lead to a long wait. (f) Explain what happens over a long period of time if the arrival rate x exceeds 15 cars per hour. (Hint: The formula is not valid for x Ú 15.)
The line of cars will continue to grow. (e) As x approaches 15, the wait increases dramatically.
1. Graph of a Rational Function A car wash can clean 15 cars per hour. If cars arrive randomly at an average rate of x per hour, the average number N of cars waiting in line is given by N = where x 6 15. way Engineering.) ISBN 1-256-49082-2
Exercises 2–6: Graphing Rational Functions Complete the following. (a) Use the given equation to complete the table of values for y.* x y (b) Determine any x-value that will make the expression undefined. (c) Sketch a dashed, vertical line (asymptote) in the xy-plane at any undefined values of x.* -4 -3 -2 -1 0 1 2 3 4
(d) Plot the points from the table.* (e) Sketch a graph of the equation. Do not let your graph cross the vertical, dashed line.* 2. y = 4. y = 6. y = 1 (b) 1 x - 1 4 (b) None x2 + 1 x (b) 1 x - 1 3. y = 5. y = 1 (b) -1 x + 1 x (b) -1 x + 1
CHAPTERS 1–7
Cumulative Review Exercises
75.4
1. Evaluate pr 2h when r = 2 and h = 6. 24p
2. Translate the phrase “two less than twice a number” into an algebraic expression using the variable x.
2x - 2
15. Determine which ordered pair is a solution to the system of equations: (2, - 6) or (1, - 2). (1, -2) 4x + y = 2 x - 4y = 9 16. Solve the system of equations. (2, -8) - 3r - t = 2 2r + t = - 4 Exercises 17 and 18: Determine if the system of linear equations has no solutions, one solution, or infinitely many solutions. 17. 2x - y = 5 - 2x + y = - 5
Infinitely many solutions
Exercises 3 and 4: Evaluate and simplify to lowest terms. 3. 1 , 5 2 4
2 5
4. 5 + 1 8 8
3 4
Exercises 5 and 6: Simplify the expression. 5. - 2 + 7x + 4 - 5x 2 x + 2 6. - 4(4 - y) + (5 - 3y) y - 11 7. Solve - 2 x + 11 = 13 and check the solution. -1 8. Solve - 3x + 1 Ú x. x …
1 4
18.
4 x - 6y = 12 - 6 x + 9y = 18
No solutions
Exercises 9 and 10: Graph the equation. Determine any intercepts.* 9. 2 x - 3y = 6 x-int: 3; y-int: -2
20. (ab)3 22. (x 2 - y 2)2 a 3b 3 x 4 - 2x 2y 2 + y 4
10. x = 1
11. Sketch a line with slope m = 3 passing through ( - 2, - 1). Write its slope–intercept form.* y = 3x + 5 12. The table lists points located on a line. Write the slope–intercept form of the line. y = 2 x - 1 x y -2 -5 -1 -3 0 -1 1 1
x-int: 1; y-int: none
21. (2y - 3)(5y + 2)
10y 2 - 11y - 6
Exercises 23 and 24: Simplify and write the expression using positive exponents. 23. (3x 2)-3
1 27x 6
24.
4x2 2x4
2 x2
25. Write 0.00123 in scientific notation. 1.23 * 10-3
2 26. Divide and check: 2 x x--x 1+ 3. 2 x + 1 +
4 x - 1
Exercises 13 and 14: Find the slope–intercept form for the line satisfying the given conditions. 13. Parallel to y = - 2x + 1 passing through (2, - 1) 3 14. Passing through ( - 1, 2) and (2, 4) y = 2 x + 3 y = -2x + 3
1 3 8 3
Exercises 31 and 32: Solve the equation. 31. y 2 + 5y - 14 = 0
-7, 2
APPLICATIONS
32. x 3 = 4 x
-2, 0, 2
Exercises 33 and 34: Simplify. 33. 3x 4y
#
y 9x 2
1 12 x
34.
2 x 2 x
2x x , x - 2 x - 4
2
35. Simplify
1 + 1 -
1 2x + 4
.
x + 2 x - 2 z + 2y 3
41. Shoveling the Driveway Two people are shoveling snow from a driveway. The first person shovels 12 square feet per minute, while the second person shovels 9 square feet per minute. (a) 21x (a) Write a simplified expression that gives the total square feet that the two people shovel in x minutes. (b) How long would it take them to clear a driveway with 1890 square feet? 90 min 42. Burning Calories An athlete can burn 10 calories per minute while running and 4 calories per minute while walking. If the athlete burns 450 calories in 60 minutes, how long is spent on each activity?
35 min running, 25 min walking
36. Solve z = 3x - 2y for x. x = Exercises 37 and 38: Solve. 4 3 37. = 1 3x 4x
7 12
2 3 1 + = 38. x - 1 x + 2 2
-1, 2
39. Suppose that y is directly proportional to x and that y = 7 when x = 14. Find y when x = 11. 5.5 40. Determine whether the data in the table represent direct or inverse variation. Find an equation that models the data. Inverse; y = 20 x x y 1 20 2 10 4 5 10 2
43. Triangle In an isosceles triangle, the measures of the two smaller angles are equal and their sum is 32° more than the largest angle. (a) Let x be the measure of one of the two smaller angles and y be the measure of the largest angle. Write a system of linear equations whose solution gives the measures of these angles. (b) Solve your system. (53, 74) or 53°, 53°, 74°
(a) 2 x + y = 180, 2 x - y = 32
44. Strength of a Beam The strength of a wood beam varies inversely with its length. A beam that is 10 feet long can support 1100 pounds. How much weight can a similar beam support if it is 22 feet long? About 500 lb
