Post on 17-Aug-2020
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Rational Expressions and Functions In the previous two chapters we discussed algebraic expressions, equations, and functions related to polynomials. In this chapter, we will examine a broader category of algebraic expressions, rational expressions, also referred to as algebraic fractions. Similarly as in arithmetic, where a rational number is a quotient of two integers with a denominator that is different than zero, a rational expression is a quotient of two polynomials, also with a denominator that is different than zero.
We start by introducing the related topic of integral exponents, including scientific notation. Then, we discuss operations on algebraic fractions, solving rational equations, and properties and graphs of rational functions with an emphasis on such features as domain, range, and asymptotes. At the end of this chapter, we show examples of applied problems, including work problems, that require solving rational equations.
RT.1 Integral Exponents and Scientific Notation
Integral Exponents
In section P.2, we discussed the following power rules, using whole numbers for the exponents.
product rule ππππ β ππππ = ππππ+ππ (ππππ)ππ = ππππππππ
quotient rule ππππ
ππππ= ππππβππ οΏ½
πππποΏ½ππ
=ππππ
ππππ
power rule (ππππ)ππ = ππππππ ππππ = ππ for ππ β ππ ππππ is undefined
Observe that these rules gives us the following result.
ππβππ = ππππβ(ππ+1) = ππππ
ππππ+1= ππππ
ππππβππ= ππ
ππ
Consequantly, ππβππ = (ππππ)β1 = ππππππ
.
Since ππβππ = ππππππ
, then the expression ππππ is meaningful for any integral exponent ππ and a
nonzero real base ππ. So, the above rules of exponents can be extended to include integral
exponents.
In practice, to work out the negative sign of an exponent, take the reciprocal of the base, or equivalently, βchange the levelβ of the power. For example,
3β2 = οΏ½13οΏ½2
= 12
32= 1
9 and 2
β3
3β1= 31
23= 3
8.
quotient rule product rule
power rule
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Attention! Exponent refers to the immediate number, letter, or expression in a bracket. For example,
ππβππ = ππππππ
, (βππ)βππ = 1(βπ₯π₯)2 = ππ
ππππ, ππππππ βππβππ = β ππ
ππππ.
Evaluating Expressions with Integral Exponents
Evaluate each expression.
a. 3β1 + 2β1 b. 5β2
2β5
c. β22
2β7 d. β2β2
3β2β3
a. 3β1 + 2β1 = 13
+ 12
= 26
+ 36
= ππππ
Caution! 3β1 + 2β1 β (3 + 2)β1, because the value of 3β1 + 2β1 is 56, as shown in the
example, while the value of (3 + 2)β1 is 15.
b. 5β2
2β5= 25
52= ππππ
ππππ
Note: To work out the negative exponent, move the power from the numerator to the denominator or vice versa.
c. β22
2β7= β22 β 27 = βππππ
Attention! The role of a negative sign in front of a base number or in front of an exponent is different. To work out the negative in 2β7, we either take the reciprocal of the base, or we change the position of the power to a different level in the fraction. So, 2β7 =
οΏ½12οΏ½7
or 2β7 = 127
. However, the negative sign in β22 just means that the number is negative.
So, β22 = β4. Caution! β22 β 14
d. β2β2
3β2β3= β23
3β22= βππ
ππ
Note: Exponential expressions can be simplified in many ways. For example, to simplify 2β2
2β3, we can work out the negative exponents first by moving the powers to a different level,
23
22 , and then reduce the common factors as shown in the example; or we can employ the
quotient rule of powers to obtain 2β2
2β3= 2β2β(β3) = 2β2+3 = 21 = 2.
Solution
1
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Simplifying Exponential Expressions Involving Negative Exponents
Simplify the given expression. Leave the answer with only positive exponents.
a. 4π₯π₯β5 b. (π₯π₯ + π¦π¦)β1
c. π₯π₯β1 + π¦π¦β1 d. (β23π₯π₯β2)β2
e. π₯π₯β4π¦π¦2
π₯π₯2π¦π¦β5 f. οΏ½β4ππ
5ππ3
24ππππβ6οΏ½β2
a. 4π₯π₯β5 = 4π₯π₯5
b. (π₯π₯ + π¦π¦)β1 = 1
π₯π₯+π¦π¦
c. π₯π₯β1 + π¦π¦β1 = 1
π₯π₯+ 1
π¦π¦
d. (β23π₯π₯β2)β2 = οΏ½β23
π₯π₯2οΏ½β2
= οΏ½ π₯π₯2
β23οΏ½2
= οΏ½π₯π₯2οΏ½2
(β1)2(23)2 = π₯π₯4
26
e. π₯π₯β4π¦π¦2
π₯π₯2π¦π¦β5= π¦π¦2π¦π¦5
π₯π₯2π₯π₯4= π¦π¦7
π₯π₯6
f. οΏ½β4ππ5ππ3
24ππππβ6οΏ½β2
= οΏ½βππ4ππ3ππ6
6οΏ½β2
= οΏ½(β1)ππ4ππ9
6οΏ½β2
= οΏ½ 6(β1)ππ4ππ9
οΏ½2
= 36ππ8ππ18
Scientific Notation
Integral exponents allow us to record numbers with a very large or very small absolute value in a shorter, more convenient form.
For example, the average distance from the Sun to the Saturn is 1,430,000,000 km, which can be recorded as 1.43 β 10,000,000 or more concisely as 1.43 β 109.
Similarly, the mass of an electron is 0.0000000000000000000000000009 grams, which can be recorded as 9 β 0.0000000000000000000000000001, or more concisely as 9 β10β28.
This more concise representation of numbers is called scientific notation and it is frequently used in sciences and engineering.
Definition 1.1 A real number ππ is written in scientific notation iff ππ = ππ β ππππππ , where the coefficient ππ is such that |ππ| β [ππ,ππππ), and the exponent ππ is an integer.
Solution exponent β5 refers to π₯π₯ only!
these expressions are NOT equivalent!
work out the negative exponents inside the
bracket
work out the negative exponents outside the
bracket
a βββ sign can be treated as a factor
of β1
power rule β multiply exponents
product rule β add exponents
6
4
(β1)2 = 1
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Converting Numbers to Scientific Notation Convert each number to scientific notation.
a. 520,000 b. β0.000102 c. 12.5 β 103
a. To represent 520,000 in scientific notation, we place a decimal point after the first
nonzero digit, 5 . 2 0 0 0 0
and then count the number of decimal places needed for the decimal point to move to its original position, which by default was after the last digit. In our example the number of places we need to move the decimal place is 5. This means that 5.2 needs to be multiplied by 105 in order to represent the value of 520,000. So, ππππππ,ππππππ =ππ.ππ β ππππππ.
Note: To comply with the scientific notation format, we always place the decimal point after the first nonzero digit of the given number. This will guarantee that the coefficient ππ satisfies the condition ππ β€ |ππ| < ππππ.
b. As in the previous example, to represent β0.000102 in scientific notation, we place a decimal point after the first nonzero digit,
β 0 . 0 0 0 1 . 0 2
and then count the number of decimal places needed for the decimal point to move to its original position. In this example, we move the decimal 4 places to the left. So the number 1.02 needs to be divided by 104, or equivalently, multiplied by 10β4 in order to represent the value of β0.000102. So, βππ.ππππππππππππ = βππ.ππππ β ππππβππ.
Observation: Notice that moving the decimal to the right corresponds to using a positive exponent, as in Example 3a, while moving the decimal to the left corresponds to using a negative exponent, as in Example 3b.
c. Notice that 12.5 β 103 is not in scientific notation as the coefficient 12.5 is not smaller than 10. To convert 12.5 β 103 to scientific notation, first, convert 12.5 to scientific notation and then multiply the powers of 10. So,
12.5 β 103 = 1.25 β 10 β 103 = ππ.ππππ β ππππππ
Solution
an integer has its decimal dot after
the last digit
multiply powers by adding exponents
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Converting from Scientific to Decimal Notation
Convert each number to decimal notation.
a. β6.57 β 106 b. 4.6 β 10β7 a. The exponent 6 indicates that the decimal point needs to be moved 6 places to the right.
So, β6.57 β 106 = β6 . 5 7 _ _ _ _ . = βππ,ππππππ,ππππππ
b. The exponent β7 indicates that the decimal point needs to be moved 7 places to the left. So,
4.6 β 10β7 = 0. _ _ _ _ _ _ 4 . 6 = ππ.ππππππππππππππππ
Using Scientific Notation in Computations
Evaluate. Leave the answer in scientific notation.
a. 6.5 β 107 β 3 β 105 b. 3.6 β 103
9 β 1014
a. Since the product of the coefficients 6.5 β 3 = 19.5 is larger than 10, we convert it to
scientific notation and then multiply the remaining powers of 10. So,
6.5 β 107 β 3 β 105 = 19.5 β 107 β 105 = 1.95 β 10 β 1012 = ππ.ππππ β ππππππππ
b. Similarly as in the previous example, since the quotient 3.69
= 0.4 is smaller than 1, we convert it to scientific notation and then work out the remaining powers of 10. So,
3.6 β 103
9 β 1014= 0.4 β 10β11 = 4 β 10β1 β 10β11 = ππ β ππππβππππ
Using Scientific Notation to Solve Problems
The average distance from Earth to the sun is 1.5 β 108 km. How long would it take a rocket, traveling at 4.7 β 103 km/h, to reach the sun?
Solution
fill the empty places by zeros
Solution
fill the empty places by zeros
divide powers by subtracting exponents
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To find time ππ needed for the rocket traveling at the rate π π = 4.7 β 103 km/h to reach the sun that is at the distance π·π· = 1.5 β 108 km from Earth, first, we solve the motion formula π π β ππ = π·π· for ππ. Since ππ = π·π·
π π , we calculate,
ππ =1.5 β 108
4.7 β 103β 0.32 β 105 = 3.2 β 104
So, it will take approximately ππ.ππ β ππππππ hours for the rocket to reach the sun.
RT.1 Exercises
Vocabulary Check Complete each blank with the most appropriate term or phrase from the given list:
numerator, opposite, reciprocal, scientific.
1. To raise a base to a negative exponent, raise the ______________ of the base to the opposite exponent.
2. A power in a numerator of a fraction can be written equivalently as a power with the ______________ exponent in the denominator of this fraction. Similarly, a power in a denominator of a fraction can be written equivalently as a power with the opposite exponent in the ______________ of this fraction.
3. A number with a very large or very small absolute value is often written in the ______________ notation. Concept Check True or false.
4. οΏ½34οΏ½β2
= οΏ½43οΏ½2 5. 10β4 = 0.00001 6. (0.25)β1 = 4
7. β45 = 145
8. (β2)β10 = 4β5 9. 2 β 2 β 2β1 = 18
10. 3π₯π₯β2 = 13π₯π₯2
11. β2β2 = β14 12. 510
5β12= 5β2
13. The number 0.68 β 10β5 is written in scientific notation. 14. 98.6 β 107 = 9.86 β 106 Concept Check
15. Match the expression in Row I with its equivalent expression in Row II. Choices may be used once, more than once, or not at all.
a. 5β2 b. β5β2 c. (β5)β2 d. β(β5)β2 e. β5 β 5β2
A. 25 B. 125
C. β25 D. β15 E. β 1
25
Concept Check Evaluate each expression.
16. 4β6 β 43 17. β93 β 9β5 18. 2β3
26 19. 2β7
2β5
Solution
258
20. β3β4
5β3 21. βοΏ½3
2οΏ½β2
22. 2β2 + 2β3 23. (2β1 β 3β1)β1
Concept Check Simplify each expression, if possible. Leave the answer with only positive exponents. Assume
that all variables represent nonzero real numbers. Keep large numerical coefficients as powers of prime numbers, if possible.
24. (β2π₯π₯β3)(7π₯π₯β8) 25. (5π₯π₯β2π¦π¦3)(β4π₯π₯β7π¦π¦β2) 26. (9π₯π₯β4ππ)(β4π₯π₯β8ππ)
27. (β3π¦π¦β4ππ)(β5π¦π¦β3ππ) 28. β4π₯π₯β3 29. π₯π₯β4ππ
π₯π₯6ππ
30. 3ππ5
ππππβ2 31. 14ππβ4ππβ3
β8ππ8ππβ5 32. β18π₯π₯β3π¦π¦3
β12π₯π₯β5π¦π¦5
33. (2β1ππβ7ππ)β4 34. (β3ππ2ππβ5)β3 35. οΏ½5π₯π₯β2
π¦π¦3οΏ½β3
36. οΏ½2π₯π₯3π¦π¦β2
3π¦π¦β3οΏ½β3
37. οΏ½ β4π₯π₯β3
5π₯π₯β1π¦π¦4οΏ½β4
38. οΏ½125π₯π₯2π¦π¦β3
5π₯π₯4π¦π¦β2οΏ½β5
39. οΏ½β200π₯π₯3π¦π¦β5
8π₯π₯5π¦π¦β7οΏ½β4
40. [(β2π₯π₯β4π¦π¦β2)β3]β2 41. 12ππβ2οΏ½ππβ3οΏ½
β2
6ππ7
42. (β2ππ)2ππβ5
(ππππ)β3 43. οΏ½2ππππ2οΏ½3οΏ½3ππ
4
ππβ4οΏ½β1
44. οΏ½ β3π₯π₯4π¦π¦6
15π₯π₯β6π¦π¦7οΏ½β3
45. οΏ½ β4ππ3ππ2
12ππ6ππβ5οΏ½β3
46. οΏ½β9β2π₯π₯β4π¦π¦
3β3π₯π₯β3π¦π¦2οΏ½8 47. (4βπ₯π₯)2π¦π¦
48. (5ππ)βππ 49. π₯π₯πππ₯π₯βππ 50. 9ππ2βπ₯π₯
3ππ2β2π₯π₯
51. 12π₯π₯ππ+1
β4π₯π₯2βππ 52. οΏ½π₯π₯ππβ1οΏ½3οΏ½π₯π₯ππβ4οΏ½β2 53. 25π₯π₯ππ+πππ¦π¦ππβππ
β5π₯π₯ππβπππ¦π¦ππβππ
Convert each number to scientific notation.
54. 26,000,000,000 55. β0.000132 56. 0.0000000105 57. 705.6 Convert each number to decimal notation.
58. 6.7 β 108 59. 5.072 β 10β5 60. 2 β 1012 61. 9.05 β 10β9 62. One gigabyte of computer memory equals 230 bytes. Write the number of bytes in 1 gigabyte, using decimal
notation. Then, using scientific notation, approximate this number by rounding the scientific notation coefficient to two decimals places.
Evaluate. State your answer in scientific notation.
63. (6.5 β 103)(5.2 β 10β8) 64. (2.34 β 10β5)(5.7 β 10β6)
65. (3.26 β 10β6)(5.2 β 10β8) 66. 4 β 10β7
8 β 10β3
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67. 7.5 β 109
2.5 β 104 68. 4 β 10β7
8 β 10β3
69. 0.05 β 160000.0004
70. 0.003 β 40,0000.00012 β600
Analytic Skills Solve each problem. State your answer in scientific notation.
71. A light-year is the distance that light travels in one year. Find the number of kilometers in a light-year if light travels approximately 3 β 105 kilometers per second.
72. In 2017, the national debt in Canada was about 6.4 β 1011 dollars. If Canadian population in 2017 was approximately 3.56 β 107, what was the share of this debt per person?
73. The brightest star in the night sky, Sirius, is about 4.704 β 1013 miles from Earth. If one light-year is approximately 5.88 β 1012 miles, how many light-years is it from Earth to Sirius?
74. The average discharge at the mouth of the Amazon River is 4,200,000 cubic feet per second. How much water is discharged from the Amazon River in one hour? in one year?
75. If current trends continue, world population ππ in billions may be modeled by the equation ππ = 6(1.014)π₯π₯, where π₯π₯ is in years and π₯π₯ = 0 corresponds to the year 2000. Estimate the world population in 2020 and 2025.
76. The mass of the sun is 1.989 β 1030 kg and the mass of the earth is 5.976 β 1024 kg. How many times larger in mass is the sun than the earth?
Discussion Point
77. Many calculators can only handle scientific notation for powers of 10 between β99 and 99. How can we compute (4 β 10220)2 ?
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RT.2 Rational Expressions and Functions; Multiplication and Division of Rational Expressions
In arithmetic, a rational number is a quotient of two integers with denominator different than zero. In algebra, a rational expression, offten called an algebraic fraction, is a quotient of two polynomials, also with denominator different than zero. In this section, we will examine rational expressions and functions, paying attention to their domains. Then, we will simplify, multiply, and divide rational expressions, employing the factoring skills developed in Chapter P.
Rational Expressions and Functions
Here are some examples of rational expressions:
β π₯π₯2
2π₯π₯π¦π¦, π₯π₯β1, π₯π₯2β4
π₯π₯β2, 8π₯π₯2+6π₯π₯β5
4π₯π₯2+5π₯π₯, π₯π₯β3
3βπ₯π₯, π₯π₯2 β 25, 3π₯π₯(π₯π₯ β 1)β2
Definition 2.1 A rational expression (algebraic fraction) is a quotient π·π·(ππ)πΈπΈ(ππ) of two polynomials ππ(π₯π₯) and
ππ(π₯π₯), where ππ(π₯π₯) β 0. Since division by zero is not permitted, a rational expression is defined only for the π₯π₯-values that make the denominator of the expression different than zero. The set of such π₯π₯-values is referred to as the domain of the expression.
Note 1: Negative exponents indicate hidden fractions and therefore represent rational expressions. For instance, π₯π₯β1 = 1
π₯π₯.
Note 2: A single polynomial can also be seen as a rational expression because it can be considered as a fraction with a denominator of 1.
For instance, π₯π₯2 β 25 = π₯π₯2β251
.
Definition 2.2 A rational function is a function defined by a rational expression,
ππ(ππ) =π·π·(ππ)πΈπΈ(ππ).
The domain of such function consists of all real numbers except for the π₯π₯-values that make
the denominator ππ(π₯π₯) equal to 0. So, the domain π«π« = β β {ππ|πΈπΈ(ππ) = ππ}
For example, the domain of the rational function ππ(π₯π₯) = 1
π₯π₯β3 is the set of all real
numbers except for 3 because 3 would make the denominator equal to 0. So, we write π·π· = β β {3}. Sometimes, to make it clear that we refer to function ππ, we might denote the domain of ππ by π·π·ππ , rather than just π·π·.
Figure 1 shows a graph of the function ππ(π₯π₯) = 1π₯π₯β3
. Notice that the graph does not cross
the dashed vertical line whose equation is π₯π₯ = 3. This is because ππ(3) is not defined. A closer look at the graphs of rational functions will be given in Section RT.5.
Figure 1
ππ(π₯π₯)
π₯π₯
1
3
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Evaluating Rational Expressions or Functions
Evaluate the given expression or function for π₯π₯ = β1, 0, 1. If the value cannot be calculated, write undefined.
a. 3π₯π₯(π₯π₯ β 1)β2 b. ππ(π₯π₯) = π₯π₯π₯π₯2+π₯π₯
a. If π₯π₯ = β1, then 3π₯π₯(π₯π₯ β 1)β2 = 3(β1)(β1 β 1)β2 = β3(β2)β2 = β3
(β2)2 = βππππ.
If π₯π₯ = 0, then 3π₯π₯(π₯π₯ β 1)β2 = 3(0)(0 β 1)β2 = ππ.
If π₯π₯ = 1, then 3π₯π₯(π₯π₯ β 1)β2 = 3(1)(1 β 1)β2 = 3 β 0β2 = ππππππππππππππππππ, as division by zero is not permitted.
Note: Since the expression 3π₯π₯(π₯π₯ β 1)β2 cannot be evaluated at π₯π₯ = 1, the number 1 does not belong to its domain.
b. ππ(β1) = β1(β1)2+(β1) = β1
1β1= ππππππππππππππππππ.
ππ(0) = 0(0)2+(0) = 0
0= ππππππππππππππππππ.
ππ(1) = 1(1)2+(1) = ππ
ππ.
Observation: Function ππ(π₯π₯) = π₯π₯π₯π₯2+π₯π₯
is undefined at π₯π₯ = 0 and π₯π₯ = β1. This is because
the denominator π₯π₯2 + π₯π₯ = π₯π₯(π₯π₯ + 1) becomes zero when the π₯π₯-value is 0 or β1.
Finding Domains of Rational Expressions or Functions
Find the domain of each expression or function.
a. 42π₯π₯+5
b. π₯π₯β2π₯π₯2β2π₯π₯
c. ππ(π₯π₯) = π₯π₯2β4π₯π₯2+4
d. ππ(π₯π₯) = 2π₯π₯β1π₯π₯2β4π₯π₯β5
a. The domain of 42π₯π₯+5
consists of all real numbers except for those that would make the denominator 2π₯π₯ + 5 equal to zero. To find these numbers, we solve the equation
2π₯π₯ + 5 = 0 2π₯π₯ = β5 π₯π₯ = βππ
ππ
Solution
Solution
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So, the domain of 42π₯π₯+5
is the set of all real numbers except for β52. This can be
recorded in set notation as β β οΏ½β πππποΏ½, or in set-builder notation as οΏ½πποΏ½ππ β βππ
πποΏ½, or in
interval notation as οΏ½ββ,βπππποΏ½ βͺ οΏ½β ππ
ππ,βοΏ½.
b. To find the domain of π₯π₯β2
π₯π₯2β2π₯π₯, we want to exclude from the set of real numbers all the
π₯π₯-values that would make the denominator π₯π₯2 β 2π₯π₯ equal to zero. After solving the equation
π₯π₯2 β 2π₯π₯ = 0 via factoring
π₯π₯(π₯π₯ β 2) = 0 and zero-product property
π₯π₯ = ππ or π₯π₯ = ππ,
we conclude that the domain is the set of all real numbers except for 0 and 2, which can be recorded as β β {ππ,ππ}. This is because the π₯π₯-values of 0 or 2 make the denominator of the expression π₯π₯β2
π₯π₯2β2π₯π₯ equal to zero.
c. To find the domain of the function ππ(π₯π₯) = π₯π₯2β4π₯π₯2+4
, we first look for all the π₯π₯-values that make the denominator π₯π₯2 + 4 equal to zero. However, π₯π₯2 + 4, as a sum of squares, is never equal to 0. So, the domain of function ππ is the set of all real numbers β.
d. To find the domain of the function ππ(π₯π₯) = 2π₯π₯β1
π₯π₯2β4π₯π₯β5, we first solve the equation
π₯π₯2 β 4π₯π₯ β 5 = 0 to find which π₯π₯-values make the denominator equal to zero. After factoring, we obtain
(π₯π₯ β 5)(π₯π₯ + 1) = 0
which results in π₯π₯ = 5 and π₯π₯ = β1
Thus, the domain of ππ equals to π«π«ππ = β β {βππ,ππ}.
Equivalent Expressions
Definition 2.3 Two expressions are equivalent in the common domain iff (if and only if) they produce the same values for every input from the domain.
Consider the expression π₯π₯β2π₯π₯2β2π₯π₯
from Example 2b. Notice that this expression can be
simplified to π₯π₯β2π₯π₯(π₯π₯β2) = 1
π₯π₯ by reducing common factors in the numerator and the
denominator. However, the domain of the simplified fraction, 1π₯π₯, is the set β β {0}, which
is different than the domain of the original fraction, β β {0,2}. Notice that for π₯π₯ = 2, the
expression π₯π₯β2π₯π₯2β2π₯π₯
is undefined while the value of the expression 1π₯π₯ is 1
2. So, the two
expressions are not equivalent in the set of real numbers. However, if the domain of 1π₯π₯ is
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resticted to the set β β {0,2}, then the two expressions produce the same values and as such, they are equivalent. We say that the two expressions are equivalent in the common domain.
The above situation can be illustrated by graphing the related functions, ππ(π₯π₯) = π₯π₯β2
π₯π₯2β2π₯π₯ and ππ(π₯π₯) = 1π₯π₯, as in Figure 2. The
graphs of both functions are exactly the same except for the hole in the graph of ππ at the point οΏ½2, 1
2οΏ½.
So, from now on, when writing statements like π₯π₯β2π₯π₯2β2π₯π₯
= 1π₯π₯, we keep in mind that they apply
only to real numbers which make both denominators different than zero. Thus, by saying in short that two expressions are equivalent, we really mean that they are equivalent in the common domain.
Note: The domain of ππ(π₯π₯) = π₯π₯β2π₯π₯2β2π₯π₯ = π₯π₯β2
π₯π₯(π₯π₯β2) = 1π₯π₯ is still β β {ππ,ππ}, even though the
(π₯π₯ β 2) term was simplified.
The process of simplifying expressions involve creating equivalent expressions. In the case of rational expressions, equivalent expressions can be obtained by multiplying or dividing the numerator and denominator of the expression by the same nonzero polynomial. For example,
βππ β ππβππππ
=(βπ₯π₯ β 3) β (β1)
(β5π₯π₯) β (β1) =ππ + ππππππ
ππ β ππππ β ππ
=(π₯π₯ β 3)
β1(π₯π₯ β 3) =1β1
= βππ
To simplify a rational expression:
Factor the numerator and denominator completely. Eliminate all common factors by following the property of multiplicative identity.
Do not eliminate common terms - they must be factors!
Simplifying Rational Expressions Simplify each expression.
a. 7ππ2ππ2
21ππ3ππβ14ππ3ππ2 b. π₯π₯2β9
π₯π₯2β6π₯π₯+9 c. 20π₯π₯β15π₯π₯2
15π₯π₯3β5π₯π₯2β20π₯π₯
a. First, we factor the denominator and then reduce the common factors. So,
7ππ2ππ2
21ππ3ππ β 14ππ3ππ2=
7ππ2ππ2
7ππ3ππ(3ππ β 2ππππ) =ππ
ππ(ππ β ππππ)
Solution
ππ(π₯π₯) =1π₯π₯
π₯π₯
1
2
ππ(π₯π₯) =π₯π₯ β 2π₯π₯2 β 2π₯π₯
π₯π₯
1
2
Figure 2
1
1
264
b. As before, we factor and then reduce. So,
π₯π₯2 β 9π₯π₯2 β 6π₯π₯ + 9
=(π₯π₯ β 3)(π₯π₯ + 3)
(π₯π₯ β 3)2 =ππ + ππππ β ππ
c. Factoring and reducing the numerator and denominator gives us
20π₯π₯ β 15π₯π₯2
15π₯π₯3 β 5π₯π₯2 β 20π₯π₯=
5π₯π₯(4 β 3π₯π₯)5π₯π₯(3π₯π₯2 β π₯π₯ β 4)
=4 β 3π₯π₯
(3π₯π₯ β 4)(π₯π₯ + 1)
Since 4β3π₯π₯3π₯π₯β4
= β(3π₯π₯β4)3π₯π₯β4
= β1, the above expression can be reduced further to
4 β 3π₯π₯(3π₯π₯ β 4)(π₯π₯ + 1) =
βππππ + ππ
Notice: An opposite expression in the numerator and denominator can be reduced to β1. For example, since ππ β ππ is opposite to ππ β ππ, then
ππβππππβππ
= βππ, as long as ππ β ππ.
Caution: Note that ππ β ππ is NOT opposite to ππ + ππ !
Multiplication and Division of Rational Expressions
Recall that to multiply common fractions, we multiply their numerators and denominators, and then simplify the resulting fraction. Multiplication of algebraic fractions is performed in a similar way.
To multiply rational expressions:
factor each numerator and denominator completely, reduce all common factors in any of the numerators and denominators, multiply the remaining expressions by writing the product of their numerators over
the product of their denominators. For instance,
Multiplying Algebraic Fractions
Multiply and simplify. Assume non-zero denominators.
a. 2π₯π₯2π¦π¦3
3π₯π₯π¦π¦2β οΏ½2π₯π₯
3π¦π¦οΏ½2
2(π₯π₯π¦π¦)3 b. π₯π₯3βπ¦π¦3
π₯π₯+π¦π¦β 3π₯π₯+3π¦π¦π₯π₯2βπ¦π¦2
Neither π₯π₯ nor 3 can be reduced, as they are
NOT factors ! 1
β1
2
3π₯π₯π₯π₯2 + 5π₯π₯
β3π₯π₯ + 15
6π₯π₯=
3π₯π₯π₯π₯(π₯π₯ + 5) β
3(π₯π₯ + 5)6π₯π₯
=3
2π₯π₯
265
a. To multiply the two algebraic fractions, we use appropriate rules of powers to simplify each fraction, and then reduce all the remaining common factors. So,
2π₯π₯2π¦π¦3
3π₯π₯π¦π¦2β
(2π₯π₯3π¦π¦)2
2(π₯π₯π¦π¦)3 =2π₯π₯π¦π¦
3β
4π₯π₯6π¦π¦2
2π₯π₯3π¦π¦3=
2π₯π₯π¦π¦ β 2π₯π₯3
3 β π¦π¦=ππππππ
ππ=ππππππππ
b. After factoring and simplifying, we have
π₯π₯3 β π¦π¦3
π₯π₯ + π¦π¦β
3π₯π₯ + 3π¦π¦π₯π₯2 β π¦π¦2
=(π₯π₯ β π¦π¦)(π₯π₯2 + π₯π₯π¦π¦ + π¦π¦2)
π₯π₯ + π¦π¦β
3(π₯π₯ + π¦π¦)(π₯π₯ β π¦π¦)(π₯π₯ + π¦π¦) =
πποΏ½ππππ + ππππ + πππποΏ½ππ + ππ
To divide rational expressions, multiply the first, the dividend, by the reciprocal of the second, the divisor.
For instance,
5π₯π₯ β 103π₯π₯
Γ·3π₯π₯ β 6
2π₯π₯2=
5π₯π₯ β 103π₯π₯
β 2π₯π₯2
3π₯π₯ β 6=
5(π₯π₯ β 2)3π₯π₯
β2π₯π₯2
3(π₯π₯ β 2) =10π₯π₯
9
Dividing Algebraic Fractions
Perform operations and simplify. Assume non-zero denominators.
a. 2π₯π₯2+2π₯π₯π₯π₯β1
Γ· (π₯π₯ + 1) b. π₯π₯2β25π₯π₯2+5π₯π₯+4
Γ· π₯π₯2β10π₯π₯+252π₯π₯2+8π₯π₯
β π₯π₯2+π₯π₯4π₯π₯2
a. To divide by (π₯π₯ + 1) we multiply by the reciprocal 1(π₯π₯+1). So,
2π₯π₯2 + 2π₯π₯π₯π₯ β 1
Γ· (π₯π₯ + 1) =2π₯π₯(π₯π₯ + 1)π₯π₯ β 1
β1
(π₯π₯ + 1) =2π₯π₯π₯π₯ β 1
b. The order of operations indicates to perform the division first. To do this, we convert
the division into multiplication by the reciprocal of the middle expression. Therefore,
π₯π₯2 β 25π₯π₯2 + 5π₯π₯ + 4
Γ·π₯π₯2 β 10π₯π₯ + 25
2π₯π₯2 + 8π₯π₯βπ₯π₯2 + π₯π₯
4π₯π₯2
=(π₯π₯ β 5)(π₯π₯ + 5)(π₯π₯ + 4)(π₯π₯ + 1) β
2π₯π₯2 + 8π₯π₯π₯π₯2 β 10π₯π₯ + 25
βπ₯π₯(π₯π₯ + 1)
4π₯π₯2
=(π₯π₯ β 5)(π₯π₯ + 5)
(π₯π₯ + 4) β2π₯π₯(π₯π₯ + 4)(π₯π₯ β 5)2 β
14π₯π₯
=(ππ + ππ)ππ(ππ β ππ)
Solution
follow multiplication rules
Solution
multiply by the reciprocal
equivalent answers
1
1 1 3
1
2
Recall: ππππ β ππππ = (ππ β ππ)οΏ½ππππ + ππππ + πππποΏ½
ππππ β ππππ = (ππ + ππ)(ππ β ππ)
1
1
2
Reduction of common factors can be done gradually, especially if there is many common factors to divide out.
266
RT.2 Exercises
Vocabulary Check Complete each blank with the most appropriate term or phrase from the given list: rational,
zero, common, factor, reciprocal.
1. If ππ(π₯π₯) and ππ(π₯π₯) are polynomials, then ππ(π₯π₯) = ππ(π₯π₯)ππ(π₯π₯)
is a ___________ function provided that ππ(π₯π₯) is not the
zero polynomial.
2. The domain of a rational function consists of all real numbers except for the π₯π₯-values that will make the denominator equal to________.
3. Two rational expressions are equivalent if they produce the same values for any inputs from their ______________ domain.
4. To simplify, multiply, or divide rational expressions, we first ______________ each numerator and denominator.
5. To divide by a rational expression, we multiply by its _______________.
Concept Check True or false.
6. ππ(π₯π₯) = 4βπ₯π₯β4
is a rational function. 7. The domain of ππ(π₯π₯) = π₯π₯β24
is the set of all real numbers.
8. π₯π₯β34βπ₯π₯
is equivalent to βπ₯π₯β3π₯π₯β4
. 9. ππ2+1ππ2β1
is equivalent to ππ+1ππβ1
.
Concept Check Given the rational function f, find ππ(β1), ππ(0), and ππ(2).
10. ππ(π₯π₯) = π₯π₯π₯π₯β2
11. ππ(π₯π₯) = 5π₯π₯3π₯π₯βπ₯π₯2
12. ππ(π₯π₯) = π₯π₯β2π₯π₯2+π₯π₯β6
Concept Check For each rational function, find all numbers that are not in the domain. Then give the domain,
using both set notation and interval notation.
13. ππ(π₯π₯) = π₯π₯π₯π₯+2
14. ππ(π₯π₯) = π₯π₯π₯π₯β6
15. β(π₯π₯) = 2π₯π₯β13π₯π₯+7
16. ππ(π₯π₯) = 3π₯π₯+25π₯π₯β4
17. ππ(π₯π₯) = π₯π₯+2π₯π₯2β4
18. β(π₯π₯) = π₯π₯β2π₯π₯2+4
19. ππ(π₯π₯) = 53π₯π₯βπ₯π₯2
20. ππ(π₯π₯) = π₯π₯2+π₯π₯β6π₯π₯2+12π₯π₯+35
21. β(π₯π₯) = 7|4π₯π₯β3|
Discussion Point
22. Is there a rational function ππ such that 2 is not in its domain and (0) = 5 ? If yes, how many such functions are there? If not, explain why not.
267
Concept Check
23. Which rational expressions are equivalent and what is the simplest expression that they are equivalent to?
a. 2π₯π₯+32π₯π₯β3
b. 2π₯π₯β33β2π₯π₯
c. 2π₯π₯+33+2π₯π₯
d. 2π₯π₯+3β2π₯π₯β3
e. 3β2π₯π₯2π₯π₯β3
Concept Check
24. Which rational expressions can be simplified?
a. π₯π₯2+2π₯π₯2
b. π₯π₯2+22
c. π₯π₯2βπ₯π₯π₯π₯2
d. π₯π₯2βπ¦π¦2
π¦π¦2 e. π₯π₯
π₯π₯2βπ₯π₯
Simplify each expression, if possible.
25. 24ππ3ππ3ππππ3
26. β18π₯π₯2π¦π¦3
8π₯π₯3π¦π¦ 27. 7βπ₯π₯
π₯π₯β7 28. π₯π₯+2
π₯π₯β2
29. ππβ5β5+ππ
30. (3βπ¦π¦)(π₯π₯+1)(π¦π¦β3)(π₯π₯β1) 31. 12π₯π₯β15
21 32. 18ππβ2
22
33. 4π¦π¦β124π¦π¦+12
34. 7π₯π₯+147π₯π₯β14
35. 6ππ+187ππ+21
36. 3π§π§2+π§π§18π§π§+6
37. ππ2β2520β4ππ
38. 9ππ2β34β12ππ2
39. π‘π‘2β25π‘π‘2β10π‘π‘+25
40. ππ2β36ππ2+12π‘π‘+36
41. π₯π₯2β9π₯π₯+8π₯π₯2+3π₯π₯β4
42. ππ2+8ππβ9ππ2β5ππ+4
43. π₯π₯3βπ¦π¦3
π₯π₯2βπ¦π¦2 44. ππ2βππ2
ππ3βππ3
Perform operations and simplify. Assume non-zero denominators.
45. 18ππ4
5ππ2β 25ππ
4
9ππ3 46. 28
π₯π₯π¦π¦Γ· 63π₯π₯3
2π¦π¦2 47. 12π₯π₯
49(π₯π₯π¦π¦2)3 β(7π₯π₯π¦π¦)2
8
48. π₯π₯+12π₯π₯β3
β 2π₯π₯β32π₯π₯
49. 10ππ6ππβ12
β 20ππβ4030ππ3
50. ππ2β14ππ
β 21βππ
51. π¦π¦2β254π¦π¦
β 25βπ¦π¦
52. (8π₯π₯ β 16) Γ· 3π₯π₯β610 53. (π¦π¦2 β 4) Γ· 2βπ¦π¦
8π¦π¦
54. 3ππβ9ππ2β9
β (ππ3 + 27) 55. π₯π₯2β16π₯π₯2
β π₯π₯2β4π₯π₯π₯π₯2βπ₯π₯β12
56. π¦π¦2+10π¦π¦+25π¦π¦2β9
β π¦π¦2β3π¦π¦π¦π¦+5
57. ππβ3ππ2β4ππ+3
Γ· ππ2βππππβ1
58. π₯π₯2β6π₯π₯+9π₯π₯2+3π₯π₯
Γ· π₯π₯2β9π₯π₯
59. π₯π₯2β2π₯π₯3π₯π₯2β5π₯π₯β2
β 9π₯π₯2β49π₯π₯2β12π₯π₯+4
60. π‘π‘2β49π‘π‘2+4π‘π‘β21
β π‘π‘2+8π‘π‘+15π‘π‘2β2π‘π‘β35
61. ππ3βππ3
ππ2βππ2Γ· 2ππβ2ππ
2ππ+2ππ 62. 64π₯π₯3+1
4π₯π₯2β100β 4π₯π₯+2064π₯π₯2β16π₯π₯+4
268
63. π₯π₯3π¦π¦β64π¦π¦π₯π₯3π¦π¦+64π¦π¦
Γ· π₯π₯2π¦π¦2β16π¦π¦2
π₯π₯2π¦π¦2β4π₯π₯π¦π¦2+16π¦π¦2 64. ππ3β27ππ3
ππ2+ππππβ12ππ2β ππ
2β2ππππβ24ππ2
ππ2β5ππππβ6ππ2
65. 4π₯π₯2β9π¦π¦2
8π₯π₯3β27π¦π¦3β 4π₯π₯
2+6π₯π₯π¦π¦+9π¦π¦2
4π₯π₯2+12π₯π₯π¦π¦+9π¦π¦2 66. 2π₯π₯2+π₯π₯β1
6π₯π₯2+π₯π₯β2Γ· 2π₯π₯2+5π₯π₯+3
6π₯π₯2+13π₯π₯+6
67. 6π₯π₯2β13π₯π₯+614π₯π₯2β25π₯π₯+6
Γ· 14β21π₯π₯49π₯π₯2+7π₯π₯β6
68. 4π¦π¦2β12π¦π¦+3627β3π¦π¦2
Γ· (π¦π¦3 + 27)
69. 3π¦π¦π₯π₯2
Γ· π¦π¦2
π₯π₯Γ· π¦π¦
5π₯π₯ 70. π₯π₯+1
π¦π¦β2Γ· 2π₯π₯+2
π¦π¦β2Γ· π₯π₯
π¦π¦
71. ππ2β4ππ2
ππ+2ππΓ· (ππ + 2ππ) β 2ππ
ππβ2ππ 72. 9π₯π₯2
π₯π₯2β16π¦π¦2Γ· 1
π₯π₯2+4π₯π₯π¦π¦β π₯π₯β4π¦π¦
3π₯π₯
73. π₯π₯2β25π₯π₯β4
Γ· π₯π₯2β2π₯π₯β15π₯π₯2β10π₯π₯+24
β π₯π₯+3π₯π₯2+10π₯π₯+25
74. π¦π¦β3π¦π¦2β8π¦π¦+16
β π¦π¦2β16π¦π¦+4
Γ· π¦π¦2+3π¦π¦β18π¦π¦2+11π¦π¦+30
Given ππ(π₯π₯) and ππ(π₯π₯), find ππ(π₯π₯) β ππ(π₯π₯) and ππ(π₯π₯) Γ· ππ(π₯π₯).
75. ππ(π₯π₯) = π₯π₯β4π₯π₯2+π₯π₯
and ππ(π₯π₯) = 2π₯π₯π₯π₯+1
76. ππ(π₯π₯) = π₯π₯3β3π₯π₯2
π₯π₯+5 and ππ(π₯π₯) = 4π₯π₯2
π₯π₯β3
77. ππ(π₯π₯) = π₯π₯2β7π₯π₯+12π₯π₯+3
and ππ(π₯π₯) = 9βπ₯π₯2
π₯π₯β4 78. ππ(π₯π₯) = π₯π₯+6
4βπ₯π₯2 and ππ(π₯π₯) = 2βπ₯π₯
π₯π₯2+8π₯π₯+12
269
RT.3 Addition and Subtraction of Rational Expressions
Many real-world applications involve adding or subtracting algebraic fractions. Similarly as in the case of common fractions, to add or subtract algebraic fractions, we first need to change them equivalently to fractions with the same denominator. Thus, we begin by discussing the techniques of finding the least common denominator.
Least Common Denominator
The least common denominator (LCD) for fractions with given denominators is the same as the least common multiple (LCM) of these denominators. The methods of finding the LCD for fractions with numerical denominators were reviewed in section R3. For example,
πΏπΏπΏπΏπ·π·(4,6,8) = 24,
because 24 is a multiple of 4, 6, and 8, and there is no smaller natural number that would be divisible by all three numbers, 4, 6, and 8.
Suppose the denominators of three algebraic fractions are 4(π₯π₯2 β π¦π¦2), β6(π₯π₯ + π¦π¦)2, and 8π₯π₯. The numerical factor of the least common multiple is 24. The variable part of the LCM is built by taking the product of all the different variable factors from each expression, with each factor raised to the greatest exponent that occurs in any of the expressions. In our example, since 4(π₯π₯2 β π¦π¦2) = 4(π₯π₯ + π¦π¦)(π₯π₯ β π¦π¦), then
πΏπΏπΏπΏπ·π·( 4(π₯π₯ + π¦π¦)(π₯π₯ β π¦π¦) , β 6(π₯π₯ + π¦π¦)2, 8π₯π₯ ) = ππππππ(ππ + ππ)ππ(ππ β ππ) Notice that we do not worry about the negative sign of the middle expression. This is because a negative sign can always be written in front of a fraction or in the numerator rather than in the denominator. For example,
1β6(π₯π₯ + π¦π¦)2 = β
16(π₯π₯ + π¦π¦)2 =
β16(π₯π₯ + π¦π¦)2
In summary, to find the LCD for algebraic fractions, follow the steps:
Factor each denominator completely. Build the LCD for the denominators by including the following as factors:
o LCD of all numerical coefficients, o all of the different factors from each denominator, with each factor raised to the
greatest exponent that occurs in any of the denominators. Note: Disregard any factor of β1.
Determining the LCM for the Given Expressions
Find the LCM for the given expressions.
a. 12π₯π₯3π¦π¦ and 15π₯π₯π¦π¦2(π₯π₯ β 1) b. π₯π₯2 β 2π₯π₯ β 8 and π₯π₯2 + 3π₯π₯ + 2
c. π¦π¦2 β π₯π₯2, 2π₯π₯2 β 2π₯π₯π¦π¦, and π₯π₯2 + 2π₯π₯π¦π¦ + π¦π¦2
270
a. Notice that both expressions, 12π₯π₯3π¦π¦ and 15π₯π₯π¦π¦2(π₯π₯ β 1), are already in factored form. The πΏπΏπΏπΏπΏπΏ(12,15) = 60, as
ππ β
12 4
15 β 5
= ππππ
The highest power of π₯π₯ is 3, the highest power of π¦π¦ is 2, and (π₯π₯ β 1) appears in the first power. Therefore,
πΏπΏπΏπΏπΏπΏοΏ½12π₯π₯3π¦π¦, 15π₯π₯π¦π¦2(π₯π₯ β 1)οΏ½ = ππππππππππππ(ππ β ππ)
b. To find the LCM of π₯π₯2 β 2π₯π₯ β 8 and π₯π₯2 + 3π₯π₯ + 2, we factor each expression first:
π₯π₯2 β 2π₯π₯ β 8 = (π₯π₯ β 4)(π₯π₯ + 2) π₯π₯2 + 3π₯π₯ + 2 = (π₯π₯ + 1)(π₯π₯ + 2)
There are three different factors in these expressions, (π₯π₯ β 4), (π₯π₯ + 2), and (π₯π₯ + 1). All of these factors appear in the first power, so
πΏπΏπΏπΏπΏπΏ( π₯π₯2 β 2π₯π₯ β 8, π₯π₯2 + 3π₯π₯ + 2 ) = (ππ β ππ)(ππ + ππ)(ππ + ππ)
c. As before, to find the LCM of π¦π¦2 β π₯π₯2, 2π₯π₯2 β 2π₯π₯π¦π¦, and π₯π₯2 + 2π₯π₯π¦π¦ + π¦π¦2, we factor
each expression first:
π¦π¦2 β π₯π₯2 = (π¦π¦ + π₯π₯)(π¦π¦ β π₯π₯) = β(π₯π₯ + π¦π¦)(π₯π₯ β π¦π¦) 2π₯π₯2 β 2π₯π₯π¦π¦ = 2π₯π₯(π₯π₯ β π¦π¦) π₯π₯2 + 2π₯π₯π¦π¦ + π¦π¦2 = (π₯π₯ + π¦π¦)2
Since the factor of β1 can be disregarded when finding the LCM, the opposite factors can be treated as the same by factoring the β1 out of one of the expressions. So, there are four different factors to consider, 2, π₯π₯, (π₯π₯ + π¦π¦), and (π₯π₯ β π¦π¦). The highest power of (π₯π₯ + π¦π¦) is 2 and the other factors appear in the first power. Therefore,
πΏπΏπΏπΏπΏπΏ( π¦π¦2 β π₯π₯2, 2π₯π₯2 β 2π₯π₯π¦π¦, π₯π₯2 + 2π₯π₯π¦π¦ + π¦π¦2 ) = ππππ(ππ β ππ)(ππ + ππ)ππ
Addition and Subtraction of Rational Expressions
Observe addition and subtraction of common fractions, as review in section R3.
12
+23β
56
=1 β 3 + 2 β 2 β 5
6=
3 + 4 β 56
=26
=ππππ
Solution
divide by 3
no more common factors, so we multiply the numbers in the letter L
notice that (π₯π₯ + 2) is taken only ones!
as ππ β ππ = β(ππ β ππ) and ππ + ππ = ππ + ππ
convert fractions to the lowest common denominator
work out the numerator
simplify, if possible
271
To add or subtract algebraic fractions, follow the steps:
Step 1: Factor the denominators of all algebraic fractions completely. Step 2: Find the LCD of all the denominators. Step 3: Convert each algebraic fraction to the lowest common denominator found in
Step 2 and write the sum (or difference) as a single fraction. Step 4: Simplify the numerator and the whole fraction, if possible.
Adding and Subtracting Rational Expressions
Perform the operations and simplify if possible.
a. ππ5β 3ππ
2ππ b. π₯π₯
π₯π₯βπ¦π¦+ π¦π¦
π¦π¦βπ₯π₯
c. 3π₯π₯2+3π₯π₯π¦π¦π₯π₯2βπ¦π¦2
β 2β3π₯π₯π₯π₯βπ¦π¦
d. π¦π¦+1π¦π¦2β7π¦π¦+6
+ π¦π¦+2π¦π¦2β5π¦π¦β6
e. 2π₯π₯π₯π₯2β4
+ 52βπ₯π₯
β 12+π₯π₯
f. (2π₯π₯ β 1)β2 + (2π₯π₯ β 1)β1 a. Since πΏπΏπΏπΏπΏπΏ(5, 2ππ) = 10ππ, we would like to rewrite expressions, ππ
5 and 3ππ
2ππ, so that they
have a denominator of 10a. This can be done by multiplying the numerator and denominator of each expression by the factors of 10a that are missing in each denominator. So, we obtain
ππ5β
3ππ2ππ
=ππ5β
2ππ2ππ
β3ππ2ππ
β55
=ππππππ β ππππππ
ππππππ
b. Notice that the two denominators, π₯π₯ β π¦π¦ and π¦π¦ β π₯π₯, are opposite expressions. If we
write π¦π¦ β π₯π₯ as β(π₯π₯ β π¦π¦), then π₯π₯
π₯π₯ β π¦π¦+
π¦π¦π¦π¦ β π₯π₯
=π₯π₯
π₯π₯ β π¦π¦+
π¦π¦β (π₯π₯ β π¦π¦) =
π₯π₯π₯π₯ β π¦π¦
βπ¦π¦
π₯π₯ β π¦π¦=π₯π₯ β π¦π¦π₯π₯ β π¦π¦
= ππ
c. To find the LCD, we begin by factoring π₯π₯2 β π¦π¦2 = (π₯π₯ β π¦π¦)(π₯π₯ + π¦π¦). Since this expression includes the second denominator as a factor, the LCD of the two fractions is (π₯π₯ β π¦π¦)(π₯π₯ + π¦π¦). So, we calculate
3π₯π₯2 + 3π₯π₯π¦π¦π₯π₯2 β π¦π¦2
β2 β 3π₯π₯π₯π₯ β π¦π¦
=(3π₯π₯2 + 3π₯π₯π¦π¦) β 1 + (2 + 3π₯π₯) β (π₯π₯ + π¦π¦)
(π₯π₯ β π¦π¦)(π₯π₯ + π¦π¦) =
3π₯π₯2 + 3π₯π₯π¦π¦ β (2π₯π₯ + 2π¦π¦ + 3π₯π₯2 + 3π₯π₯π¦π¦)
(π₯π₯ β π¦π¦)(π₯π₯ + π¦π¦) =3π₯π₯2 + 3π₯π₯π¦π¦ β 2π₯π₯ β 2π¦π¦ β 3π₯π₯2 β 3π₯π₯π¦π¦
(π₯π₯ β π¦π¦)(π₯π₯ + π¦π¦) =
β2π₯π₯ β 2π¦π¦
(π₯π₯ β π¦π¦)(π₯π₯ + π¦π¦) =β2(π₯π₯ + π¦π¦)
(π₯π₯ β π¦π¦)(π₯π₯ + π¦π¦) =βππ
(ππ β ππ)
Solution
keep the bracket after a βββ sign
combine the signs
Multiplying the numerator and denominator of a fraction
by the same factor is equivalent to multiplying the whole fraction by 1, which
does not change the value of the fraction.
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d. To find the LCD, we first factor each denominator. Since
π¦π¦2 β 7π¦π¦ + 6 = (π¦π¦ β 6)(π¦π¦ β 1) and π¦π¦2 β 5π¦π¦ β 6 = (π¦π¦ β 6)(π¦π¦ + 1),
then πΏπΏπΏπΏπ·π· = (π¦π¦ β 6)(π¦π¦ β 1)(π¦π¦ + 1) and we calculate
π¦π¦ + 1π¦π¦2 β 7π¦π¦ + 6
+π¦π¦ β 1
π¦π¦2 β 5π¦π¦ β 6=
π¦π¦ + 1(π¦π¦ β 6)(π¦π¦ β 1) +
π¦π¦ β 1(π¦π¦ β 6)(π¦π¦ + 1) =
(π¦π¦ + 1) β (π¦π¦ + 1) + (π¦π¦ β 1) β (π¦π¦ β 1)(π¦π¦ β 6)(π¦π¦ β 1)(π¦π¦ + 1) =
π¦π¦2 + 2π¦π¦ + 1 + (π¦π¦2 β 1)(π¦π¦ β 6)(π¦π¦ β 1)(π¦π¦ + 1) =
2π¦π¦2 + 2π¦π¦(π¦π¦ β 6)(π¦π¦ β 1)(π¦π¦ + 1) =
2π¦π¦(π¦π¦ + 1)(π¦π¦ β 6)(π¦π¦ β 1)(π¦π¦ + 1) =
ππππ(ππ β ππ)(ππ β ππ)
e. As in the previous examples, we first factor the denominators, including factoring out
a negative from any opposite expression. So,
2π₯π₯π₯π₯2 β 4
+5
2 β π₯π₯β
12 + π₯π₯
=2π₯π₯
(π₯π₯ β 2)(π₯π₯ + 2) +5
β (π₯π₯ β 2) β1
π₯π₯ + 2=
2π₯π₯ β 5(π₯π₯ + 2) β 1(π₯π₯ β 2)(π₯π₯ β 2)(π₯π₯ + 2) =
2π₯π₯ β 5π₯π₯ β 10 β π₯π₯ + 2(π₯π₯ β 2)(π₯π₯ + 2) =
β4π₯π₯ β 8
(π₯π₯ β 2)(π₯π₯ + 2) =β4(π₯π₯ + 2)
(π₯π₯ β 2)(π₯π₯ + 2) =βππ
(ππ β ππ)
e. Recall that a negative exponent really represents a hidden fraction. So, we may choose
to rewrite the negative powers as fractions, and then add them using techniques as shown in previous examples.
3(2π₯π₯ β 1)β2 + (2π₯π₯ β 1)β1 =1
(2π₯π₯ β 1)2 +1
2π₯π₯ β 1=
1 + 1 β (2π₯π₯ β 1)(2π₯π₯ β 1)2 =
3 + 2π₯π₯ β 1(2π₯π₯ β 1)2 =
2π₯π₯ + 2(2π₯π₯ β 1)2 =
ππ(ππ + ππ)(ππππ β ππ)ππ
Note: Since addition (or subtraction) of rational expressions results in a rational expression, from now on the term βrational expressionβ will include sums of rational expressions as well.
Adding Rational Expressions in Application Problems
An airplane flies ππ miles with a windspeed of π€π€ mph. On the return flight, the airplane flies against the same wind. The expression ππ
π π +π€π€+ ππ
π π βπ€π€, where π π is the speed of the airplane
in still air, represents the total time in hours that it takes to make the round-trip flight. Write a single rational expression representing the total time.
multiply by the missing bracket
πΏπΏπΏπΏπ·π· = (π₯π₯β 2)(π₯π₯+ 2)
nothing to simplify this time
273
To find a single rational expression representing the total time, we perform the addition using (π π + π€π€)(π π β π€π€) as the lowest common denominator. So,
πππ π + π€π€
+ππ
π π β π€π€=ππ(π π β π€π€) + ππ(π π + π€π€)
(π π + π€π€)(π π β π€π€) =πππ π βπππ€π€ + πππ π + πππ€π€
(π π + π€π€)(π π β π€π€) =ππππππ
ππππ βππππ
Adding and Subtracting Rational Functions
Given ππ(π₯π₯) = 1π₯π₯2+10π₯π₯+24
and ππ(π₯π₯) = 2π₯π₯2+4π₯π₯
, find
a. (ππ + ππ)(π₯π₯) b. (ππ β ππ)(π₯π₯).
(ππ + ππ)(ππ) = ππ(π₯π₯) + ππ(π₯π₯) =1
π₯π₯2 + 10π₯π₯ + 24+
2π₯π₯2 + 4π₯π₯
=1
(π₯π₯ + 6)(π₯π₯ + 4) +2
π₯π₯(π₯π₯ + 4) =1 β π₯π₯ + 2(π₯π₯ + 6)π₯π₯(π₯π₯ + 6)(π₯π₯ + 4) =
π₯π₯ + 2π₯π₯ + 12π₯π₯(π₯π₯ + 6)(π₯π₯ + 4)
=3π₯π₯ + 12
π₯π₯(π₯π₯ + 6)(π₯π₯ + 4) =3(π₯π₯ + 4)
π₯π₯(π₯π₯ + 6)(π₯π₯ + 4) =ππ
ππ(ππ + ππ)
((ππ β ππ)(ππ) = ππ(π₯π₯) β ππ(π₯π₯) =1
π₯π₯2 + 10π₯π₯ + 24β
2π₯π₯2 + 4π₯π₯
=1
(π₯π₯ + 6)(π₯π₯ + 4) β2
π₯π₯(π₯π₯ + 4) =1 β π₯π₯ β 2(π₯π₯ + 6)π₯π₯(π₯π₯ + 6)(π₯π₯ + 4) =
π₯π₯ β 2π₯π₯ β 12π₯π₯(π₯π₯ + 6)(π₯π₯ + 4)
=βππβ ππππ
ππ(ππ + ππ)(ππ + ππ)
RT.3 Exercises
Vocabulary Check Complete each blank with the most appropriate term from the given list: denominator,
different, factor, highest, rational.
1. To add (or subtract) rational expressions with the same denominator, add (or subtract) the numerators and keep the same ________________.
2. To find the LCD of rational expressions, first ____________ each denominator completely.
3. To add (or subtract) rational expressions with _____________ denominators, first find the LCD for all the involved expressions.
Solution
Solution a.
b.
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4. The LCD of rational expressions is the product of the different factors in the denominators, where the power of each factor is the __________ number of times that it occurs in any single denominator.
5. A polynomial expression raised to a negative exponent represents a ______________ expression. Concept Check
6. a. What is the LCM for 6 and 9? b. What is the LCD for 16 and 1
9 ?
7. a. What is the LCM for π₯π₯2 β 25 and π₯π₯ + 5? b. What is the LCD for 1π₯π₯2β25
and 1π₯π₯+5
? Concept Check Find the LCD and then perform the indicated operations. Simplify the resulting fraction.
8. 512
+ 1318
9. 1130β 19
75 10. 3
4+ 7
30β 1
16 11. 5
8β 7
12+ 11
40
Concept Check Find the least common multiple (LCM) for each group of expressions.
12. 24ππ3ππ4, 18ππ5ππ2 13. 6π₯π₯2π¦π¦2, 9π₯π₯3π¦π¦, 15π¦π¦3 14. π₯π₯2 β 4, π₯π₯2 + 2π₯π₯
15. 10π₯π₯2, 25(π₯π₯2 β π₯π₯) 16. (π₯π₯ β 1)2, 1 β π₯π₯ 17. π¦π¦2 β 25, 5 β π¦π¦
18. π₯π₯2 β π¦π¦2, π₯π₯π¦π¦ + π¦π¦2 19. 5ππ β 15, ππ2 β 6ππ + 9 20. π₯π₯2 + 2π₯π₯ + 1, π₯π₯2 β 4π₯π₯ β 1
21. ππ2 β 7ππ + 10, ππ2 β 8ππ + 15 22. 2π₯π₯2 β 5π₯π₯ β 3, 2π₯π₯2 β π₯π₯ β 1, π₯π₯2 β 6π₯π₯ + 9
23. 1 β 2π₯π₯, 2π₯π₯ + 1, 4π₯π₯2 β 1 24. π₯π₯5 β 4π₯π₯4 + 4π₯π₯3, 12 β 3π₯π₯2, 2π₯π₯ + 4 Concept Check True or false? If true, explain why. If false, correct it.
25. 12π₯π₯
+ 13π₯π₯
= 15π₯π₯
26. 1π₯π₯β3
+ 13βπ₯π₯
= 0 27. 1π₯π₯
+ 1π¦π¦
= 1π₯π₯+π¦π¦
28. 34
+ π₯π₯5
= 3+π₯π₯20
Perform the indicated operations and simplify if possible.
29. π₯π₯β2π¦π¦π₯π₯+π¦π¦
+ 3π¦π¦π₯π₯+π¦π¦
30. ππ+3ππ+1
β ππβ5ππ+1
31. 4ππ+3ππβ3
β 1
32. ππ+1ππβ2
+ 2 33. π₯π₯2
π₯π₯βπ¦π¦+ π¦π¦2
π¦π¦βπ₯π₯ 34. 4ππβ2
ππ2β49+ 5+3ππ
49βππ2
35. 2π¦π¦β3π¦π¦2β1
β 4βπ¦π¦1βπ¦π¦2
36. ππ3
ππβππ+ ππ3
ππβππ 37. 1
π₯π₯+ββ 1
β
38. π₯π₯β2π₯π₯+3
+ π₯π₯+2π₯π₯β4
39. π₯π₯β13π₯π₯+1
+ 2π₯π₯β3
40. 4π₯π₯π¦π¦π₯π₯2βπ¦π¦2
+ π₯π₯βπ¦π¦π₯π₯+π¦π¦
41. π₯π₯β13π₯π₯+15
β π₯π₯+35π₯π₯+25
42. π¦π¦β24π¦π¦+8
β π¦π¦+65π¦π¦+10
43. 4π₯π₯π₯π₯β1
β 2π₯π₯+1
β 4π₯π₯2β1
44. β2π¦π¦+2
+ 5π¦π¦β2
+ π¦π¦+3π¦π¦2β4
45. π¦π¦π¦π¦2βπ¦π¦β20
+ 2π¦π¦+4
46. 5π₯π₯π₯π₯2β6π₯π₯+8
β 3π₯π₯π₯π₯2βπ₯π₯β12
47. 9π₯π₯+23π₯π₯2β2π₯π₯β8
+ 73π₯π₯2+π₯π₯β4
48. 3π¦π¦+22π¦π¦2βπ¦π¦β10
+ 82π¦π¦2β7π¦π¦+5
49. 6π¦π¦2+6π¦π¦+9
+ 5π¦π¦2β9
50. 3π₯π₯β1π₯π₯2+2π₯π₯β3
β π₯π₯+4π₯π₯2β9
51. 1π₯π₯+1
β π₯π₯π₯π₯β2
+ π₯π₯2+2π₯π₯2βπ₯π₯β2
52. 2π¦π¦+3
β π¦π¦π¦π¦β1
+ π¦π¦2+2π¦π¦2+2π¦π¦β3
275
53. 4π₯π₯π₯π₯2β1
+ 3π₯π₯1βπ₯π₯
β 4π₯π₯β1
54. 5π¦π¦1β2π¦π¦
β 2π¦π¦2π¦π¦+1
+ 34π¦π¦2β1
55. π₯π₯+5π₯π₯β3
β π₯π₯+2π₯π₯+1
β 6π₯π₯+10π₯π₯2β2π₯π₯β3
Discussion Point
56. Consider the following calculation
π₯π₯π₯π₯ β 2
β4π₯π₯ β 1π₯π₯2 β 4
=π₯π₯(π₯π₯ + 2) β 4π₯π₯ β 1
π₯π₯2 β 4=π₯π₯2 + 2π₯π₯ β 4π₯π₯ β 1
π₯π₯2 β 4=π₯π₯2 β 2π₯π₯ β 1π₯π₯2 β 4
Is this correct? If yes, check if the result can be simplified. If no, correct it. Perform the indicated operations and simplify if possible.
57. 2π₯π₯β3 + (3π₯π₯)β1 58. (π₯π₯2 β 9)β1 + 2(π₯π₯ β 3)β1 59. οΏ½π₯π₯+13οΏ½β1β οΏ½π₯π₯β4
2οΏ½β1
60. οΏ½ππβ3ππ2
β ππβ39οΏ½Γ· ππ2β9
3ππ 61. π₯π₯2β4π₯π₯+4
2π₯π₯+1β 2π₯π₯
2+π₯π₯π₯π₯3β4π₯π₯
β 3π₯π₯β2π₯π₯+1
62. 2π₯π₯β3
β π₯π₯π₯π₯2βπ₯π₯β6
β π₯π₯2β2π₯π₯β3π₯π₯2βπ₯π₯
Given ππ(π₯π₯) and ππ(π₯π₯), find (ππ + ππ)(π₯π₯) and (ππ β ππ)(π₯π₯). Leave the answer in simplified single fraction form.
63. ππ(π₯π₯) = π₯π₯π₯π₯+2
, ππ(π₯π₯) = 4π₯π₯β3
64. ππ(π₯π₯) = π₯π₯π₯π₯2β4
, ππ(π₯π₯) = 1π₯π₯2+4π₯π₯+4
65. ππ(π₯π₯) = 3π₯π₯π₯π₯2+2π₯π₯β3
, ππ(π₯π₯) = 1π₯π₯2β2π₯π₯+1
66. ππ(π₯π₯) = π₯π₯ + 1π₯π₯β1
, ππ(π₯π₯) = 1π₯π₯+1
Solve each problem.
67. Two friends work part-time at a store. The first person works every sixth day and the second person works every tenth day. If they are both working today, how many days pass before they both work on the same day again?
68. Suppose a cylindrical water tank is being drained. The change in the water level can be found using the expression ππ1
ππππ2β ππ2
ππππ2, where ππ1 and ππ2 represent the original and new volume,
respectively, and ππ is the radius of the tank. Write the change in water level as a single algebraic fraction.
69. To determine the percent growth in sales from the previous year, the owner of a company uses the expression
100 οΏ½ππ1ππ0β 1οΏ½, where ππ1 represents the current yearβs sales and ππ0 represents last yearβs sales. Write this
expression as a single algebraic fraction.
70. A boat travels ππ mi against the current whose rate is ππ mph. On the return trip, the boat travels with the same current. The expression ππ
ππβππ+ ππ
ππ+ππ , where ππ is the speed
of the boat in calm water, represents the total amount of time in hours it takes for the entire boating trip. Represent this amount of time as a single algebraic fraction.
276
RT.4 Complex Fractions
When working with algebraic expressions, sometimes we come across needing to simplify expressions like these:
π₯π₯2 β 9π₯π₯ + 1π₯π₯ + 3π₯π₯2 β 1
, 1 + 1
π₯π₯1 β 1
π¦π¦,
1π₯π₯ + 2 β
1π₯π₯ + β + 2β
, 1
1ππ β
1ππ
A complex fraction is a quotient of rational expressions (including sums of rational expressions) where at least one of these expressions contains a fraction itself. In this section, we will examine two methods of simplifying such fractions.
Simplifying Complex Fractions
Definition 4.1 A complex fraction is a quotient of rational expressions (including their sums) that result
in a fraction with more than two levels. For example, 123
has three levels while 12π₯π₯34π₯π₯
has four
levels. Such fractions can be simplified to a single fraction with only two levels. For example,
12 3
=12β
13
=16
, ππππ 1
2π₯π₯3
4π₯π₯2=
12π₯π₯
β4π₯π₯2
3=
2π₯π₯3
There are two common methods of simplifying complex fractions.
Method I (multiplying by the reciprocal of the denominator)
Replace the main division in the complex fraction with a multiplication of the numerator fraction by the reciprocal of the denominator fraction. We then simplify the resulting fraction if possible. Both examples given in Definition 4.1 were simplified using this strategy.
Method I is the most convenient to use when both the numerator and the denominator of a complex fraction consist of single fractions. However, if either the numerator or the denominator of a complex fraction contains addition or subtraction of fractions, it is usually easier to use the method shown below.
Method II (multiplying by LCD)
Multiply the numerator and denominator of a complex fraction by the least common denominator of all the fractions appearing in the numerator or in the denominator of the complex fraction. Then, simplify the resulting fraction if possible. For example, to simplify π¦π¦+1π₯π₯π₯π₯+1π¦π¦
, multiply the numerator π¦π¦ + 1π₯π₯ and the denominator π₯π₯ + 1
π¦π¦ by the πΏπΏπΏπΏπ·π· οΏ½1
π₯π₯, 1π¦π¦οΏ½ = π₯π₯π¦π¦. So,
οΏ½π¦π¦ + 1π₯π₯οΏ½
οΏ½π₯π₯ + 1π¦π¦οΏ½
βπ₯π₯π¦π¦π₯π₯π¦π¦
=π₯π₯π¦π¦2 + π¦π¦π₯π₯2π¦π¦ + π₯π₯
=π¦π¦(π₯π₯π¦π¦ + 1)π₯π₯(π₯π₯π¦π¦ + 1) =
ππππ
1
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Simplifying Complex Fractions
Use a method of your choice to simplify each complex fraction.
a. π₯π₯2βπ₯π₯β12π₯π₯2β2π₯π₯β15π₯π₯2+8π₯π₯+12π₯π₯2β5π₯π₯β14
b. ππ+ππ1ππ3 + 1ππ3
c. π₯π₯ + 15π₯π₯ β 13
d. 6
π₯π₯2β4 β 5
π₯π₯+27
π₯π₯2β4 β 4
π₯π₯β2
a. Since the expression π₯π₯2βπ₯π₯β12π₯π₯2β2π₯π₯β15π₯π₯2+8π₯π₯+12π₯π₯2β5π₯π₯β14
contains a single fraction in both the numerator and
denominator, we will simplify it using method I, as below.
π₯π₯2 β 2π₯π₯ β 8π₯π₯2 β 2π₯π₯ β 15π₯π₯2 + 8π₯π₯ + 12π₯π₯2 β 4π₯π₯ β 21
=(π₯π₯ β 4)(π₯π₯ + 2)(π₯π₯ β 5)(π₯π₯ + 3) β
(π₯π₯ β 7)(π₯π₯ + 3)(π₯π₯ + 6)(π₯π₯ + 2) =
(ππ β ππ)(ππ β ππ)(ππ β ππ)(ππ + ππ)
b. ππ+ππ
1ππ3 + 1ππ3
can be simplified in the following two ways:
Method I Method II
ππ+ππ1ππ3
+ 1ππ3
= ππ+ππππ3+ππ3
ππ3ππ3 = (ππ+ππ)ππ3ππ3
ππ3+ππ3 ππ+ππ
1ππ3
+ 1ππ3β ππ
3ππ3
ππ3ππ3= (ππ+ππ)ππ3ππ3
ππ3+ππ3
= (ππ+ππ)ππ3ππ3
(ππ+ππ)(ππ2βππππ+ππ2) = ππππππππ
ππππβππππ+ππππ = (ππ+ππ)ππ3ππ3
(ππ+ππ)(ππ2βππππ+ππ2) = ππππππππ
ππππβππππ+ππππ
Caution: In Method II, the factor that we multiply the complex fraction by must be equal to 1. This means that the numerator and denominator of this factor must be exactly the same.
c. To simplify π₯π₯ + 15π₯π₯ β 13
, we will use method II. Multiplying the numerator and denominator
by the πΏπΏπΏπΏπ·π· οΏ½15
, 13οΏ½ = 15, we obtain
π₯π₯ + 15
π₯π₯ β 13β
1515
=ππππππ + ππππππππ β ππ
Solution
factor and multiply by the reciprocal
278
d. Again, to simplify 6
π₯π₯2β4 β 5
π₯π₯+27
π₯π₯2β4 β 4
π₯π₯β2, we will use method II. Notice that the lowest common
multiple of the denominators in blue is (π₯π₯ + 2)(π₯π₯ β 2). So, after multiplying the numerator and denominator of the whole expression by the LCD, we obtain
6
π₯π₯2 β 4 β 5π₯π₯ + 2
7π₯π₯2 β 4 β 4
π₯π₯ β 2β
(π₯π₯+ 2)(π₯π₯ β 2)(π₯π₯+ 2)(π₯π₯ β 2) =
6 β 5(π₯π₯ β 2)7 β 4(π₯π₯ + 2) =
6 β 5π₯π₯ + 107 β 4π₯π₯ β 8
=β5π₯π₯ + 16β4π₯π₯ β 1
=ππππ β ππππππππ + ππ
Simplifying Rational Expressions with Negative Exponents
Simplify each expression. Leave the answer with only positive exponents.
a. π₯π₯β2 β π¦π¦β1
π¦π¦ βπ₯π₯ b. ππβ3
ππβ1βππβ1
a. If we write the expression with no negative exponents, it becomes a complex fraction,
which can be simplified as in Example 1. So,
π₯π₯β2 β π¦π¦β1
π¦π¦ β π₯π₯=
1π₯π₯ β 1
π¦π¦π¦π¦ β π₯π₯
βπ₯π₯π¦π¦π₯π₯π¦π¦
=π¦π¦ β π₯π₯
π₯π₯π¦π¦(π¦π¦ β π₯π₯) =ππππππ
b. As above, first, we rewrite the expression with only positive exponents and then
simplify as any other complex fraction.
ππβ3
ππβ1 β ππβ1=
1ππ3
1ππ β
1ππβππ3ππππ3ππ
=ππ
ππ2ππ β ππ3=
ππππππ(ππ β ππ)
Simplifying the Difference Quotient for a Rational Function
Find and simplify the expression ππ(ππ+β)βππ(ππ)β
for the function ππ(π₯π₯) = 1π₯π₯+1
.
Since ππ(ππ + β) = 1
ππ+β+1 and ππ(ππ) = 1
ππ+1, then
ππ(ππ + β) β ππ(ππ)β
=1
ππ + β + 1 β1
ππ + 1β
Solution
Solution
Remember! This factor must be = 1
279
To simplify this expression, we can multiply the numerator and denominator by the lowest common denominator, which is (ππ + β + 1)(ππ + 1). Thus,
1ππ + β + 1 β
1ππ + 1
ββ
(ππ + β + 1)(ππ + 1)(ππ + β + 1)(ππ + 1) =
ππ + 1 β (ππ + β + 1)β(ππ + β + 1)(ππ + 1)
=ππ + 1 β ππ β β β 1β(ππ + β + 1)(ππ + 1) =
βββ(ππ + β + 1)(ππ + 1) =
βππ(ππ + ππ + ππ)(ππ + ππ)
RT.4 Exercises
Vocabulary Check Complete each blank with the most appropriate term from the given list: complex, LCD,
multiply, reciprocal, same, single.
1. A quotient of two rational expressions that results in a fraction with more than two levels is called a _______________ algebraic fraction.
2. To simplify a complex fraction means to find an equivalent __________ fraction ππππ
, where ππ and ππ are
polynomials with no essential common factors.
3. To simplify a complex rational expression using the ____________ method, first write both the numerator and the denominator as single fractions in simplified form.
4. To simplify a complex fraction using the ______ method, first find the LCD of all rational expressions within the complex fraction. Then, ______________ the numerator and denominator by the _________ LCD expression.
Concept Check Simplify each complex fraction.
5. 2 β 133 + 73
6. 5 β 344 + 12
7. 38 β 523 + 6
8. 23 + 4534 β 12
Simplify each complex rational expression.
9. π₯π₯3
π¦π¦π₯π₯2
π¦π¦3 10.
ππ β 56ππππ β 58ππ2
11. 1 β 1ππ4 + 1ππ
12. 2ππ + 35ππ β 6
13. 9 β 3π₯π₯4π₯π₯ + 12π₯π₯ β 36π₯π₯ β 24
14. 9π¦π¦
15π¦π¦ β 6
15. 4π₯π₯ β 2π¦π¦4π₯π₯ + 2π¦π¦
16. 3ππ + 4ππ4ππ β 3ππ
This bracket is essential!
keep the denominator in a factored form
280
17. ππ β 3ππππππ β ππππ
18. 1π₯π₯ β 1π¦π¦π₯π₯2βπ¦π¦2π₯π₯π¦π¦
19. 4π¦π¦ β π¦π¦
π₯π₯21π₯π₯ β 2π¦π¦
20. 5ππ β 1ππ1
5ππ2 β 5
ππ2
21. ππβ12ππ +ππ
ππ + 4 22. 2π‘π‘β1
3π‘π‘β2π‘π‘ + 2π‘π‘
23. 1
ππββ β 1ππβ
24. 1
(π₯π₯+β)2 β 1π₯π₯2
β
25. 4 + 12
2π₯π₯β3
5 + 152π₯π₯β3
26. 1 + 3
π₯π₯+2
1 + 6π₯π₯β1
27. 1ππ2
β 1ππ2
1ππ β 1ππ
28. 1π₯π₯2
β 1π¦π¦2
1π₯π₯ + 1π¦π¦
29. π₯π₯+3π₯π₯ β 4
π₯π₯β1π₯π₯
π₯π₯β1 + 1π₯π₯ 30.
3π₯π₯2+6π₯π₯+9
+ 3π₯π₯+3
6π₯π₯2β9
+ 63βπ₯π₯
31. 1ππ2
β 1ππ2
1ππ3
+ 1ππ3
32.
4ππ2β12ππ+92ππ2+7ππβ152ππ2β15ππ+18ππ2βππβ30
Discussion Point
33. Are the expressions π₯π₯β1+π¦π¦β1
π₯π₯β2+π¦π¦β2 and
π₯π₯2+π¦π¦2
π₯π₯+π¦π¦ equivalent? Explain why or why not.
Simplify each expression. Leave your answer with only positive exponents.
34. 1ππβ2 β ππβ2
35. π₯π₯β1 + π₯π₯β2
3π₯π₯β1 36. π₯π₯β2
π¦π¦β3 β π₯π₯β3 37. 1 β (2ππ+1)β1
1 + (2ππ+1)β1
Analytic Skills Find and simplify the difference quotient ππ(ππ+β)βππ(ππ)
β for the given function.
38. ππ(π₯π₯) = 5π₯π₯ 39. ππ(π₯π₯) = 2
π₯π₯2 40. ππ(π₯π₯) = 1
1βπ₯π₯ 41. ππ(π₯π₯) = β 1
π₯π₯β2
Analytic Skills Simplify each continued fraction.
42. ππ β ππ1 β ππ
1 β ππ 43. 3 β 2
1 β 2
3 β 2π₯π₯
44. ππ + ππ2+ 1
1 β 2ππ
281
RT.5 Rational Equations and Graphs
In previous sections of this chapter, we worked with rational expressions. If two rational expressions are equated, a rational equation arises. Such equations often appear when solving application problems that involve rates of work or amounts of time considered in motion problems. In this section, we will discuss how to solve rational equations, with close attention to their domains. We will also take a look at the graphs of reciprocal functions, their properties and transformations.
Rational Equations
Definition 5.1 A rational equation is an equation involving only rational expressions and containing at least one fractional expression.
Here are some examples of rational equations:
π₯π₯2β
12π₯π₯
= β1, π₯π₯2
π₯π₯ β 5=
25π₯π₯ β 5
, 2π₯π₯π₯π₯ β 3
β6π₯π₯
=18
π₯π₯2 β 3π₯π₯
Attention! A rational equation contains an equals sign, while a rational expression does not. An equation can be solved for a given variable, while an expression can only be simplified or evaluated. For example, ππ
ππβ ππππ
ππ is an expression to
simplify, while ππππ
= ππππππ
is an equation to solve. When working with algebraic structures, it is essential to identify whether they
are equations or expressions before applying appropriate strategies.
By Definition 5.1, rational equations contain one or more denominators. Since division by zero is not allowed, we need to pay special attention to the variable values that would make any of these denominators equal to zero. Such values would have to be excluded from the set of possible solutions. For example, neither 0 nor 3 can be solutions to the equation
2π₯π₯π₯π₯ β 3
β6π₯π₯
=18
π₯π₯2 β 3π₯π₯,
as it is impossible to evaluate either of its sides for π₯π₯ = 0 or 3. So, when solving a rational equation, it is important to find its domain first.
Definition 5.2 The domain of the variable(s) of a rational equation (in short, the domain of a rational equation) is the intersection of the domains of all rational expressions within the equation.
As stated in Definition 2.1, the domain of each single algebraic fraction is the set of all real numbers except for the zeros of the denominator (the variable values that would make the denominator equal to zero). Therefore, the domain of a rational equation is the set of all real numbers except for the zeros of all the denominators appearing in this equation.
282
Determining Domains of Rational Equations
Find the domain of the variable in each of the given equations.
a. π₯π₯2β 12
π₯π₯= β1 b. 2π₯π₯
π₯π₯β2= β3
π₯π₯+ 4
π₯π₯β2
c. 2π¦π¦2β2π¦π¦β8
β 4π¦π¦2+6π¦π¦+8
= 2π¦π¦2β16
a. The equation π₯π₯2β 12
π₯π₯= β1 contains two denominators, 2 and π₯π₯. 2 is never equal to
zero and π₯π₯ becomes zero when π₯π₯ = 0. Thus, the domain of this equation is β β {ππ}. b. The equation 2π₯π₯
π₯π₯β2= β3
π₯π₯+ 4
π₯π₯β2 contains two types of denominators, π₯π₯ β 2 and π₯π₯. The
π₯π₯ β 2 becomes zero when π₯π₯ = 2, and π₯π₯ becomes zero when π₯π₯ = 0. Thus, the domain of this equation is β β {ππ,ππ}.
c. The equation 2
π¦π¦2β2π¦π¦β8β 4
π¦π¦2+6π¦π¦+8= 2
π¦π¦2β16 contains three different denominators.
To find the zeros of these denominators, we solve the following equations by factoring:
π¦π¦2 β 2π¦π¦ β 8 = 0 π¦π¦2 + 6π¦π¦ + 8 = 0 π¦π¦2 β 16 = 0
(π¦π¦ β 4)(π¦π¦ + 2) = 0 (π¦π¦ + 4)(π¦π¦ + 2) = 0 (π¦π¦ β 4)(π¦π¦ + 4) = 0
π¦π¦ = 4 or π¦π¦ = β2 π¦π¦ = β4 or π¦π¦ = β2 π¦π¦ = 4 or π¦π¦ = β4
So, β4, β2, and 4 must be excluded from the domain of this equation. Therefore, the domain π·π· = β β {βππ,βππ,ππ}.
To solve a rational equation, it is convenient to clear all the fractions first and then solve the resulting polynomial equation. This can be achieved by multiplying all the terms of the equation by the least common denominator.
Caution! Only equations, not expressions, can be changed equivalently by multiplying both of their sides by the LCD.
Multiplying expressions by any number other than 1 creates expressions that are NOT equivalent to the original ones. So, avoid multiplying rational expressions by the LCD.
Solving Rational Equations
Solve each equation.
a. π₯π₯2β 12
π₯π₯= β1 b. 2π₯π₯
π₯π₯β2= β3
π₯π₯+ 4
π₯π₯β2
c. 2π¦π¦2β2π¦π¦β8
β 4π¦π¦2+6π¦π¦+8
= 2π¦π¦2β16
d. π₯π₯β1π₯π₯β3
= 2π₯π₯β3
Solution
283
a. The domain of the equation π₯π₯2β 12
π₯π₯= β1 is the set β β {0}, as discussed in Example
1a. The πΏπΏπΏπΏπΏπΏ(2, π₯π₯) = 2π₯π₯, so we calculate
π₯π₯2β
12π₯π₯
= β1
2π₯π₯ β
π₯π₯2β 2π₯π₯
β12π₯π₯
= β1 β 2π₯π₯
π₯π₯2 β 24 = β2π₯π₯
π₯π₯2 + 2π₯π₯ β 24 = 0
(π₯π₯ + 6)(π₯π₯ β 4) = 0
π₯π₯ = β6 or π₯π₯ = 4
Since both of these numbers belong to the domain, the solution set of the original equation is {βππ,ππ}.
b. The domain of the equation 2π₯π₯
π₯π₯β2= β3
π₯π₯+ 4
π₯π₯β2 is the set β β {0, 2}, as discussed in
Example 1b. The πΏπΏπΏπΏπΏπΏ(π₯π₯ β 2, π₯π₯) = π₯π₯(π₯π₯ β 2), so we calculate
2π₯π₯π₯π₯ β 2
=β3π₯π₯
+4
π₯π₯ β 2
π₯π₯(π₯π₯ β 2)
β2π₯π₯π₯π₯ β 2
=β3π₯π₯β π₯π₯(π₯π₯ β 2)
+
4π₯π₯ β 2
β π₯π₯(π₯π₯ β 2)
2π₯π₯2 = β3(π₯π₯ β 2) + 4π₯π₯
2π₯π₯2 = β3π₯π₯ + 6 + 4π₯π₯
2π₯π₯2 β π₯π₯ + 6 = 0
(2π₯π₯ + 3)(π₯π₯ β 2) = 0
π₯π₯ = β32 or π₯π₯ = 2
Since 2 is excluded from the domain, there is only one solution to the original equation, π₯π₯ = βππ
ππ.
c. The domain of the equation 2
π¦π¦2β2π¦π¦β8β 4
π¦π¦2+6π¦π¦+8= 2
π¦π¦2β16 is the set β β {β4,β2, 4},
as discussed in Example 1c. To find the LCD, it is useful to factor the denominators first. Since
π¦π¦2 β 2π¦π¦ β 8 = (π¦π¦ β 4)(π¦π¦ + 2), π¦π¦2 + 6π¦π¦ + 8 = (π¦π¦ + 4)(π¦π¦ + 2), and π¦π¦2 β 16 = (π¦π¦ β 4)(π¦π¦ + 4), then the LCD needed to clear the fractions in the original
equation is (π¦π¦ β 4)(π¦π¦ + 4)(π¦π¦ + 2). So, we calculate
2(π¦π¦ β 4)(π¦π¦ + 2) β
4(π¦π¦ + 4)(π¦π¦ + 2) =
2(π¦π¦ β 4)(π¦π¦ + 4)
Solution
multiply each term by the LCD
expand the bracket, collect like terms, and
bring the terms over to one side
/ β 2π₯π₯
factor to find the possible roots
/ β π₯π₯(π₯π₯ β 2)
factor to find the possible roots
/ β (π¦π¦ β 4)(π¦π¦ + 4)(π¦π¦ + 2)
284
(π¦π¦ β 4)(π¦π¦ + 4)(π¦π¦ + 2)
β
2(π¦π¦ β 4)(π¦π¦ + 2) β
(π¦π¦ β 4)(π¦π¦ + 4)(π¦π¦ + 2)
β4
(π¦π¦ + 4)(π¦π¦ + 2)
=2
(π¦π¦ β 4)(π¦π¦ + 4) β(π¦π¦ β 4)(π¦π¦ + 4)(π¦π¦ + 2)
2(π¦π¦ + 4) β 4(π¦π¦ β 4) = 2(π¦π¦ + 2)
2π¦π¦ + 8 β 4π¦π¦ + 16 = 2π¦π¦ + 4
20 = 4π¦π¦
π¦π¦ = 5
Since 5 is in the domain, this is the true solution. d. First, we notice that the domain of the equation π₯π₯β1
π₯π₯β3= 2
π₯π₯β3 is the set β β {3}. To solve
this equation, we can multiply it by the πΏπΏπΏπΏπ·π· = π₯π₯ β 3, as in the previous examples, or we can apply the method of cross-multiplication, as the equation is a proportion. Here, we show both methods.
Multiplication by LCD: Cross-multiplication: π₯π₯β1
π₯π₯β3= 2
π₯π₯β3 π₯π₯β1
π₯π₯β3= 2
π₯π₯β3
π₯π₯ β 1 = 2 (π₯π₯ β 1)(π₯π₯ β 3) = 2(π₯π₯ β 3)
π₯π₯ = 3 π₯π₯ β 1 = 2
π₯π₯ = 3
Since 3 is excluded from the domain, there is no solution to the original equation.
Summary of Solving Rational Equations in One Variable
1. Determine the domain of the variable.
2. Clear all the fractions by multiplying both sides of the equation by the LCD of these fractions.
3. Find possible solutions by solving the resulting equation.
4. Check the possible solutions against the domain. The solution set consists of only these possible solutions that belong to the domain.
Graphs of Basic Rational Functions
So far, we discussed operations on rational expressions and solving rational equations. Now, we will look at rational functions, such as
/ β (π₯π₯ β 3)
/ Γ· (π₯π₯ β 3)
this division is permitted as π₯π₯ β 3 β 0
this multiplication
is permitted as π₯π₯ β 3 β 0
Use the method of your choice β either one is
fine.
285
ππ(π₯π₯) =1π₯π₯
, ππ(π₯π₯) =β2π₯π₯ + 3
, ππππ β(π₯π₯) =π₯π₯ β 3π₯π₯ β 2
.
Definition 5.3 A rational function is any function that can be written in the form
ππ(ππ) =π·π·(ππ)πΈπΈ(ππ)
,
where ππ and ππ are polynomials and ππ is not a zero polynomial.
The domain π«π«ππ of such function ππ includes all π₯π₯-values for which ππ(π₯π₯) β 0.
Finding the Domain of a Rational Function Find the domain of each function.
a. ππ(π₯π₯) = β2π₯π₯+3
b. β(π₯π₯) = π₯π₯β3π₯π₯β2
a. Since π₯π₯ + 3 = 0 for π₯π₯ = β3, the domain of ππ is the set of all real numbers except for
β3. So, the domain π«π«ππ = β β {βππ}. b. Since π₯π₯ β 2 = 0 for π₯π₯ = 2, the domain of β is the set of all real numbers except for 2.
So, the domain π«π«ππ = β β {ππ}.
Note: The subindex ππ in the notation π·π·ππ indicates that the domain is of function ππ.
To graph a rational function, we usually start by making a table of values. Because the graphs of rational functions are typically nonlinear, it is a good idea to plot at least 3 points on each side of each π₯π₯-value where the function is undefined. For example, to graph the
basic rational function, ππ(π₯π₯) = 1π₯π₯, called the reciprocal function, we
evaluate ππ for a few points to the right of zero and to the left of zero. This is because ππ is undefined at π₯π₯ = 0, which means that the graph of ππ does not cross the π¦π¦-axis. After plotting the obtained points, we connect them within each group, to the right of zero and to the left of zero, creating two disjoint curves. To see the shape of each curve clearly, we might need to evaluate ππ at some additional points.
The domain of the reciprocal function ππ(π₯π₯) = 1
π₯π₯ is β β {ππ}, as the denominator π₯π₯ must be different than zero. Projecting the graph
of this function onto the π¦π¦-axis helps us determine the range, which is also β β {ππ}.
Solution
ππ ππ(ππ) ππππ 2 ππ 1 2 1
2
ππ undefined
β ππππ β2
βππ β1 βππ β1
2
ππ(π₯π₯)
π₯π₯
1
1
286
There is another interesting feature of the graph of the reciprocal function ππ(π₯π₯) = 1π₯π₯.
Observe that the graph approaches two lines, π¦π¦ = 0, the π₯π₯-axis, and π₯π₯ = 0, the π¦π¦-axis. These lines are called asymptotes. They effect the shape of the graph, but they themselves do not belong to the graph. To indicate the fact that asymptotes do not belong to the graph, we use a dashed line when graphing them.
In general, if the π¦π¦-values of a rational function approach β or ββ as the π₯π₯-values approach a real number ππ, the vertical line π₯π₯ = ππ is a vertical asymptote of the graph. This can be recorded with the use of arrows, as follows:
π₯π₯ = ππ is a vertical asymptote β π¦π¦ β β (or ββ) when π₯π₯ β ππ.
Also, if the π¦π¦-values approach a real number ππ as π₯π₯-values approach β or ββ, the horizontal line π¦π¦ = ππ is a horizontal asymptote of the graph. Again, using arrows, we can record this statement as:
π¦π¦ = ππ is a horizontal asymptote β π¦π¦ β ππ when π₯π₯ β β (or ββ).
Graphing and Analysing the Graphs of Basic Rational Functions
For each function, state its domain and the equation of the vertical asymptote, graph it, and then state its range and the equation of the horizontal asymptote.
a. ππ(π₯π₯) = β2π₯π₯+3
b. β(π₯π₯) = π₯π₯β3π₯π₯β2
a. The domain of function ππ(π₯π₯) = β2π₯π₯+3
is π«π«ππ = β β {βππ}, as discussed in Example 3a. Since β3 is excluded from the domain, we expect the vertical asymptote to be at ππ =
βππ.
To graph function ππ, we evaluate it at some points to the right and to the left of β3. The reader is encouraged to check the values given in the table. Then, we draw the vertical asymptote π₯π₯ = β3 and plot and join the obtained points on each side of this asymptote. The graph suggests that the horizontal asymptote is the π₯π₯-axis. Indeed, the value of zero cannot be attained by the function ππ(π₯π₯) = β2
π₯π₯+3, as in order
for a fraction to become zero, its numerator would have to be zero. So, the range of function ππ is β β {ππ} and ππ = ππ is the equation of the horizontal asymptote.
b. The domain of function β(π₯π₯) = π₯π₯β3π₯π₯β2
is π«π«ππ = β β {ππ}, as discussed in Example 3b. Since 2 is excluded from the domain, we expect the vertical asymptote to be at ππ = ππ.
ππ ππ(ππ)
β ππππ β4
βππ β2 βππ β1 ππ β 1
2
βππ undefined
β ππππ 4
βππ 2 βππ 1 βππ 2
3
Solution
read: approaches
ππ(π₯π₯)
π₯π₯
ππ
Horizontal Asymptote
ππ(π₯π₯)
π₯π₯ ππ
Vertical A
symptote
ππ(π₯π₯)
π₯π₯ 1
β3
287
As before, to graph function β, we evaluate it at some points to the right and to the left of 2. Then, we draw the vertical asymptote π₯π₯ = 2 and plot and join the obtained points on each side of this asymptote. The graph suggests that the horizontal asymptote is the line ππ = ππ. Thus, the range of function β is β β {ππ}.
Notice that π₯π₯β3
π₯π₯β2= π₯π₯β2β1
π₯π₯β2= π₯π₯β2
π₯π₯β2β 1
π₯π₯β2= 1 β 1
π₯π₯β2. Since 1
π₯π₯β2
is never equal to zero than 1 β 1π₯π₯β2
is never equal to 1. This confirms the range and the horizontal asymptote stated above.
Connecting the Algebraic and Graphical Solutions of Rational Equations
Given that ππ(π₯π₯) = π₯π₯+2π₯π₯β1
, find all the π₯π₯-values for which ππ(π₯π₯) = 2. Illustrate the situation with a graph. To find all the π₯π₯-values for which ππ(π₯π₯) = 2, we replace ππ(π₯π₯) in the equation ππ(π₯π₯) = π₯π₯+2
π₯π₯β1
with 2 and solve the resulting equation. So, we have
2 =π₯π₯ + 2π₯π₯ β 1
2π₯π₯ β 2 = π₯π₯ + 2
π₯π₯ = 4
Thus, ππ(π₯π₯) = 2 for ππ = ππ.
The geometrical connection can be observed by graphing the function ππ(π₯π₯) = π₯π₯+2
π₯π₯β1= π₯π₯β1+3
π₯π₯β1= 1 + 3
π₯π₯β1 and the line π¦π¦ = 2
on the same grid, as illustrated by the accompanying graph. The π₯π₯-coordinate of the intersection of the two graphs is the solution to the equation 2 = π₯π₯+2
π₯π₯β1. This also means that
ππ(4) = 4+24β1
= 2. So, we can say that ππ(4) = 2.
Graphing the Reciprocal of a Linear Function
Suppose ππ(π₯π₯) = 2π₯π₯ β 3. a. Determine the reciprocal function ππ(π₯π₯) = 1
ππ(π₯π₯) and its domain π·π·ππ.
ππ ππ(ππ)
βππ 43
ππ 32
ππ 2 ππππ 3
ππ undefined ππππ β1 ππ 0 ππ 1
2
ππ 34
β(π₯π₯)
π₯π₯
1
2
Solution
/ β (π₯π₯ β 1)
/ βπ₯π₯, + 2
ππ(π₯π₯)
π₯π₯ 1 ππ
2 π¦π¦ = 2
ππ(π₯π₯) =π₯π₯ + 2π₯π₯ β 1
288
b. Determine the equation of the vertical asymptote of the reciprocal function ππ.
c. Graph the function ππ and its reciprocal function ππ on the same grid. Then, describe the relations between the two graphs.
a. The reciprocal of ππ(π₯π₯) = 2π₯π₯ β 3 is the function ππ(ππ) = ππ
ππππβππ. Since 2π₯π₯ β 3 = 0 for
π₯π₯ = 32, then the domain π«π«ππ = β β οΏ½ππ
πποΏ½.
b. A vertical asymptote of a rational function in simplified form is a vertical line passing
through any of the π₯π₯-values that are excluded from the domain of such a function. So, the equation of the vertical asymptote of function ππ(π₯π₯) = 1
2π₯π₯β3 is ππ = ππ
ππ.
c. To graph functions ππ and ππ, we can use a table of values as below.
Notice that the vertical asymptote of the reciprocal function comes through the zero of the linear function. Also, the values of both functions are positive to the right of 3
2 and
negative to the left of 32. In addition, ππ(2) = ππ(2) = 1 and ππ(1) = ππ(1) = β1. This
is because the reciprocal of 1 is 1 and the reciprocal of β1 is β1. For the rest of the values, observe that the values of the linear function that are very close to zero become very large in the reciprocal function and conversely, the values of the linear function that are very far from zero become very close to zero in the reciprocal function. This suggests the horizontal asymptote at zero.
Using Properties of a Rational Function in an Application Problem
When curves are designed for train tracks, the outer rail is usually elevated so that a locomotive and its cars can safely take the curve at a higher speed than if the tracks were at the same level. Suppose that a circular curve with a radius of ππ feet is being designed for a train traveling 60 miles per hour. The function ππ(ππ) = 2540
ππ calculates the proper elevation
π¦π¦ = ππ(ππ), in inches, for the outer rail.
ππ ππ(ππ) ππ(ππ)
β ππππ β4 β1
4
ππππ β2 β1
2
ππ β1 β1 ππππ 1
2 2
ππππ 0 undefined ππππ β1
2 β2
ππ 1 1 ππππ 2 1
2
ππππ 4 1
4
elevation
Solution
ππ(π₯π₯) = 12π₯π₯β3
π₯π₯
1
2
ππ(π₯π₯) = 2π₯π₯ β 3
289
a. Evaluate ππ(300) and interpret the result. b. Suppose that the outer rail for a curve is elevated 6 inches.
What should the radius of the curve be? c. Observe the accompanying graph of the function ππ and
discuss how the elevation of the outer rail changes as the radius ππ increases.
a. ππ(300) = 2540
300β 8.5. Thus, the outer rail on a curve with
radius 300 ft should be elevated about 8.5 inches for a train to safely travel through it at 60 miles per hour.
b. Since the elevation π¦π¦ = ππ(ππ) = 6 inches, to find the corresponding value of ππ, we need to solve the equation
6 = 2540ππ
.
After multiplying this equation by ππ and dividing it by 6, we obtain
ππ = 25406
β 423
So, the radius of the curve should be about 423 feet.
c. As the radius increases, the outer rail needs less elevation.
RT.5 Exercises
Vocabulary Check Complete each blank with one of the suggested words, or with the most appropriate term
from the given list: asymptote, cross-product, domain, equations, LCD, numerator, rational.
1. A ______________ equation involves fractional expressions.
2. To solve a rational equation, first find the ________ of all the rational expressions.
3. Multiplication by any quantity other than zero can be used to create equivalent ______________.
4. When simplifying rational expressions, we ___________ππππππ / ππππππππππππ
multiply by LCD.
5. The ____________ of a rational function is the set of all real numbers except for the variable values that would make the denominator of this function equal to zero.
6. An _____________ is a line that a given graph approaches in a long run.
7. A fractional expression is equal to zero when its _____________ is equal to zero.
8. To solve a rational proportion we can use either the LCD method or the ___________________ method.
Solution radius (feet)
elev
atio
n (in
ches
)
10
ππ(ππ)
ππ
20
300
40
30
500 100
50 ππ(ππ) =
2540ππ
290
Concept Check State the domain for each equation. There is no need to solve it.
9. π₯π₯+54β π₯π₯+3
3= π₯π₯
6 10. 5
6ππβ ππ
4= 8
2ππ
11. 3π₯π₯+4
= 2π₯π₯β9
12. 43π₯π₯β5
+ 2π₯π₯
= 94π₯π₯+7
13. 4π¦π¦2β25
β 1π¦π¦+5
= 2π¦π¦β7
14. π₯π₯2π₯π₯β6
β 3π₯π₯2β6π₯π₯+9
= π₯π₯β23π₯π₯β9
Solve each equation.
15. 38
+ 13
= π₯π₯12
16. 14β 5
6= 1
π¦π¦
17. π₯π₯ + 8π₯π₯
= β9 18. 43ππβ 3
ππ= 10
3
19. ππ8
+ ππβ412
= ππ24
20. ππβ22β ππ
6= 4ππ
9
21. 5ππ+20
= 3ππ 22. 5
ππ+4= 3
ππβ2
23. π¦π¦+2π¦π¦
= 53 24. π₯π₯β4
π₯π₯+6= 2π₯π₯+3
2π₯π₯β1
25. π₯π₯π₯π₯β1
β π₯π₯2
π₯π₯β1= 5 26. 3 β 12
π₯π₯2= 5
π₯π₯
27. 13β π₯π₯β1
π₯π₯= π₯π₯
3 28. 1
π₯π₯+ 2
π₯π₯+10= π₯π₯
π₯π₯+10
29. 1π¦π¦β1
+ 512
= β23π¦π¦β3
30. 76π₯π₯+3
β 13
= 22π₯π₯+1
31. 83ππ+9
β 815
= 25ππ+15
32. 6ππβ4
+ 5ππ
= β20ππ2β4ππ
33. 3π¦π¦β2
+ 2π¦π¦4βπ¦π¦2
= 5π¦π¦+2
34. π₯π₯π₯π₯β2
+ π₯π₯π₯π₯2β4
= π₯π₯+3π₯π₯+2
35. 12π₯π₯+10
= 8π₯π₯2β25
β 2π₯π₯β5
36. 5π¦π¦+3
= 14π¦π¦2β36
+ 2π¦π¦β3
37. 6π₯π₯2β4π₯π₯+3
β 1π₯π₯β3
= 14π₯π₯β4
38. 7π₯π₯β2
β 8π₯π₯+5
= 12π₯π₯2+6π₯π₯β20
39. 5π₯π₯β4
β 3π₯π₯β1
= π₯π₯2β1π₯π₯2β5π₯π₯+4
40. π¦π¦π¦π¦+1
+ 3π¦π¦+5π¦π¦2+4π¦π¦+3
= 2π¦π¦+3
41. 3π₯π₯π₯π₯+2
+ 72π₯π₯3+8
= 24π₯π₯2β2π₯π₯+4
42. 4π₯π₯+3
+ 7π₯π₯2β3π₯π₯+9
= 108π₯π₯3+27
43. π₯π₯2π₯π₯β9
β 3π₯π₯ = 109β2π₯π₯
44. β2π₯π₯2+2π₯π₯β3
β 53β3π₯π₯
= 43π₯π₯+9
For the given rational function ππ, find all values of π₯π₯ for which ππ(π₯π₯) has the indicated value.
45. ππ(π₯π₯) = 2π₯π₯ β 15π₯π₯
; ππ(π₯π₯) = 1 46. ππ(π₯π₯) = π₯π₯β5π₯π₯+1
; ππ(π₯π₯) = 35
47. ππ(π₯π₯) = β3π₯π₯π₯π₯+3
+ π₯π₯; ππ(π₯π₯) = 4 48. ππ(π₯π₯) = 4π₯π₯
+ 1π₯π₯β2
; ππ(π₯π₯) = 3
291
Graph each rational function. State its domain, range and the equations of the vertical and horizontal asymptotes.
49. ππ(π₯π₯) = 2π₯π₯ 50. ππ(π₯π₯) = β 1
π₯π₯ 51. β(π₯π₯) = 2
π₯π₯β3
52. ππ(π₯π₯) = β1π₯π₯+1
53. ππ(π₯π₯) = π₯π₯β1π₯π₯+2
54. β(π₯π₯) = π₯π₯+2π₯π₯β3
Analytic Skills For each function ππ, find its reciprocal function ππ(π₯π₯) = 1ππ(π₯π₯)
and graph both functions on the
same grid. Then, state the equations of the vertical and horizontal asymptotes of function ππ.
55. ππ(π₯π₯) = 12π₯π₯ + 1 56. ππ(π₯π₯) = βπ₯π₯ + 2 57. ππ(π₯π₯) = β2π₯π₯ β 3
Analytic Skills Solve each equation.
58. π₯π₯1 + 1
π₯π₯+1= π₯π₯ β 3 59.
2 β 1π₯π₯4 β 1
π₯π₯2= 1
Solve each problem.
60. The average number of vehicles waiting in line to enter a parking area is modeled by the function defined by
π€π€(π₯π₯) = π₯π₯2
2(1βπ₯π₯),
where π₯π₯ is a quantity between 0 and 1 known as the traffic intensity.
a. For each traffic intensity, find the average number of vehicles waiting. Round the answer to the nearest one. i. 0.1 ii. 0.8 iii. 0.9
b. What happens to the number of vehicles waiting in line as traffic intensity increases?
61. The percent of deaths caused by smoking, called the incidence rate, is modeled by the rational function
ππ(π₯π₯) =π₯π₯ β 1π₯π₯
,
where π₯π₯ is the number of times a smoker is more likely to die of lung cancer than a non-smoker is. For example, π₯π₯ = 10 means that a smoker is 10 times more likely than a non-smoker to die from lung cancer.
a. Find ππ(π₯π₯) if π₯π₯ is 10. b. For what values of π₯π₯ is ππ(π₯π₯) = 80%? (Hint: Change 80% to a decimal.) c. Can the incidence rate equal 0? Explain. d. Can the incidence rate equal 1? Explain.
292
RT.6 Applications of Rational Equations
In previous sections of this chapter, we studied operations on rational expressions, simplifying complex fractions, and solving rational equations. These skills are needed when working with real-world problems that lead to a rational equation. The common types of such problems are motion or work problems. In this section, we first discuss how to solve a rational formula for a given variable, and then present several examples of application problems involving rational equations.
Formulas Containing Rational Expressions
Solving application problems often involves working with formulas. We might need to form a formula, evaluate it, or solve it for a desired variable. The basic strategies used to solve a formula for a variable were shown in section L2 and F4. Recall the guidelines that we used to isolate the desired variable:
Reverse operations to clear unwanted factors or addends; Example: To solve π΄π΄+π΅π΅
2= πΏπΏ for π΄π΄, we multiply by 2 and then subtract π΅π΅.
Multiply by the LCD to keep the desired variable in the numerator;
Example: To solve π΄π΄1+ππ
= ππ for ππ, first, we multiply by (1 + ππ).
Take the reciprocal of both sides of the equation to keep the desired variable in the numerator (this applies to proportions only); Example: To solve 1
πΆπΆ= π΄π΄+π΅π΅
π΄π΄π΅π΅ for πΏπΏ, we can take the reciprocal of both sides to
obtain πΏπΏ = π΄π΄π΅π΅π΄π΄+π΅π΅
.
Factor to keep the desired variable in one place. Example: To solve ππ + ππππππ = π΄π΄ for ππ, we first factor ππ out.
Below we show how to solve formulas containing rational expressions, using a combination of the above strategies.
Solving Rational Formulas for a Given Variable
Solve each formula for the indicated variable. a. 1
ππ= 1
ππ+ 1
ππ, for ππ b. πΏπΏ = πππ π
π·π·βππ, for π·π· c. πΏπΏ = πππ π
π·π·βππ, for ππ
a. Solution I: First, we isolate the term containing ππ, by βmovingβ 1
ππ to the other side
of the equation. So, 1ππ
=1ππ
+1ππ
1ππβ
1ππ
=1ππ
1ππ
=ππ β ππππππ
Solution
rewrite from the right to the left,
and perform the subtraction to
leave this side as a single fraction
/ β 1ππ
293
Then, to bring ππ to the numerator, we can take the reciprocal of both sides of the equation, obtaining
ππ =ππππππ β ππ
Caution! This method can be applied only to a proportion (an equation with a single
fraction on each side).
Solution II: The same result can be achieved by multiplying the original equation by the πΏπΏπΏπΏπ·π· = ππππππ, as shown below
1ππ
=1ππ
+1ππ
ππππ = ππππ + ππππ
ππππ β ππππ = ππππ
ππ(ππ β ππ) = ππππ
ππ =ππππππ β ππ
b. To solve πΏπΏ = πππ π π·π·βππ
for π·π·, we may start with multiplying the equation by the denominator to bring the variable π·π· to the numerator. So,
πΏπΏ =πππ π
π·π· β ππ
πΏπΏ(π·π· β ππ) = ππ
π·π· β ππ =πππ π πΏπΏ
π«π« =πππ π π³π³
+ ππ =πππ π + πππ³π³
π³π³
c. When solving πΏπΏ = πππ π π·π·βππ
for ππ, we first observe that the variable ππ appears in both the numerator and denominator. Similarly as in the previous example, we bring the ππ from the denominator to the numerator by multiplying the formula by the denominator π·π· βππ. Thus,
πΏπΏ =πππ π
π·π· β ππ
πΏπΏ(π·π· β ππ) = πππ π .
Then, to keep the ππ in one place, we need to expand the bracket, collect terms with ππ, and finally factor the ππ out. So, we have
factor ππ out
/β ππππππ
/βππππ
/Γ· (ππ β ππ)
/β (π·π· β ππ)
/Γ· πΏπΏ
/+ππ
This can be done in one step by interchanging πΏπΏ with π·π· β ππ.
The movement of the expressions resembles that of a
teeter-totter.
Both forms are correct answers.
/β (π·π· β ππ)
294
πΏπΏπ·π· β πΏπΏππ = πππ π
πΏπΏπ·π· = πππ π + πΏπΏππ
πΏπΏπ·π· = ππ(π π + πΏπΏ)
πΏπΏπ·π·π π + πΏπΏ
= ππ
Obviously, the final formula can be written starting with ππ,
ππ =π³π³π«π«π π + π³π³
.
Forming and Evaluating a Rational Formula
Suppose a trip consists of two parts, each of length π₯π₯.
a. Find a formula for determining the average speed π£π£ for the whole trip, given the speed π£π£1 for the first part of the trip and π£π£2 for the second part of the trip.
b. Find the average speed π£π£ for the whole trip, if the speed for the first part of the trip was 60 km/h and the speed for the second part of the trip was 90 km/h.
a. The total distance, ππ, for the whole trip is π₯π₯ + π₯π₯ = 2π₯π₯. The total time, ππ, for the whole
trip is the sum of the times for the two parts of the trip, ππ1 and ππ2. From the relation ππππππππ β πππ‘π‘ππππ = πππ‘π‘π π ππππππππππ, we have
ππ1 = π₯π₯π£π£1
and ππ2 = π₯π₯π£π£2
. Therefore,
ππ =π₯π₯π£π£1
+π₯π₯π£π£2
,
which after substituting to the formula for the average speed, π£π£ = πππ‘π‘, gives us
π£π£ =2π₯π₯
π₯π₯π£π£1
+ π₯π₯π£π£2
.
Since the formula involves a complex fraction, it should be simplified. We can do this by multiplying the numerator and denominator by the πΏπΏπΏπΏπ·π· = π£π£1π£π£2. So, we have
π£π£ =2π₯π₯
π₯π₯π£π£1
+ π₯π₯π£π£2βπ£π£1π£π£2π£π£1π£π£2
π£π£ =2π₯π₯π£π£1π£π£2
π₯π₯π£π£1π£π£2π£π£1
+ π₯π₯π£π£1π£π£2π£π£2
π£π£ =2π₯π₯π£π£1π£π£2π₯π₯π£π£2 + π₯π₯π£π£1
Solution
factor the π₯π₯
/+πΏπΏππ
/Γ· (π π + πΏπΏ)
295
π£π£ =2π₯π₯π£π£1π£π£2
π₯π₯(π£π£2 + π£π£1)
ππ =ππππππππππππππ + ππππ
Note 1: The average speed in this formula does not depend on the distance travelled.
Note 2: The average speed for the total trip is not the average (arithmetic mean) of the speeds for each part of the trip. In fact, this formula represents the harmonic mean of the two speeds.
b. Since π£π£1 = 60 km/h and π£π£2 = 90km/h, using the formula developed in Example 2a, we calculate
π£π£ =2 β 60 β 9060 + 90
=10800
150= ππππ π€π€π€π€/π‘π‘
Observation: Notice that the average speed for the whole trip is lower than the average of the speeds for each part of the trip, which is 60+90
2= 75 km/h.
Applied Problems
Many types of application problems were already introduced in sections L3 and E2. Some of these types, for example motion problems, may involve solving rational equations. Below we show examples of proportion and motion problems as well as introduce another type of problems, work problems.
Proportion Problems
When forming a proportion,
πππππππππππππππ¦π¦ πΌπΌ πππππππππππππππππππππππππππ¦π¦ πΌπΌπΌπΌ ππππππππππππ
=πππππππππππππππ¦π¦ πΌπΌ πππππππππππππππππππππππππ¦π¦ πΌπΌπΌπΌ ππππππππππ
,
it is essential that the same type of data are placed in the same row or the same column.
Recall: To solve a proportion ππππ
=ππππ
,
for example, for ππ, it is enough to multiply the equation by ππ. This gives us
ππ =ππππππ
.
296
Similarly, to solve ππππ
=ππππ
for ππ, we can use the cross-multiplication method, which eventually (we encourage the reader to check this) leads us to
ππ =ππππππ
.
Notice that in both cases the desired variable equals the product of the blue variables lying across each other, divided by the remaining purple variable. This is often referred to as the βcross multiply and divideβ approach to solving a proportion.
In statistics, proportions are often used to estimate the population by analysing its sample in situations where the exact count of the population is too costly or not possible to obtain.
Estimating Numbers of Wild Animals To estimate the number of wild horses in Utah, a forest ranger catches 620 wild horses, tags them, and releases them. Later, 122 horses are caught and it is found that 31 of them are tagged. Assuming that the horses mix freely when they are released, estimate how many wild horses there are in Utah.
Suppose there are π₯π₯ wild horses in Utah. 620 of them were tagged, so the ratio of the tagged horses in the whole population of the wild horses in Utah is
620π₯π₯
The ratio of the tagged horses found in the sample of 122 horses caught in the later time is
31122
So, we form the proportion:
620π₯π₯
=31
122
After solving for π₯π₯, we have
π₯π₯ =620 β 122
31= ππππππππ
So, we can estimate that over 2400 wild horses live in Utah.
Solution
620
122
31
x wild horses
tagged horses
all horses
population sample
297
πππ‘π‘ππππ π»π» =πππ‘π‘π π ππππππππππ π«π«ππππππππ π π
In geometry, proportions are the defining properties of similar figures. One frequently used theorem that involves proportions is the theorem about similar triangles, attributed to the Greek mathematician Thales.
Thalesβ Theorem Two triangles are similar iff the ratios of the corresponding sides are the same.
βΏπ¨π¨π¨π¨π¨π¨ βΌ βΏπ¨π¨π¨π¨β²π¨π¨β² β π¨π¨π¨π¨π¨π¨π¨π¨β²
=π¨π¨π¨π¨π¨π¨π¨π¨β²
=π¨π¨π¨π¨π¨π¨β²π¨π¨β²
Using Similar Triangles in an Application Problem
A cross-section of a small storage room is in the shape of a right triangle with a height of 2 meters and a base of 1.2 meters, as shown in Figure 6.1. What is the largest cubic box that can fit in this room? Assume that the base of the box is positioned on the floor of the storage room. Suppose that the height of the box is π₯π₯ meters. Since the height of the storage room is 2 meters, the expression 2 β π₯π₯ represents the height of the wall above the box, as shown in Figure 6.1b. Since the blue and brown triangles are similar, we can use the Thalesβ Theorem to form the proportion
2 β π₯π₯2
=π₯π₯
1.2.
Employing cross-multiplication, we obtain
2.4 β 1.2π₯π₯ = 2π₯π₯
2.4 = 3.2π₯π₯
π₯π₯ =2.43.2
= ππ.ππππ
So, the dimensions of the largest cubic box fitting in this storage room are 75 cm by 75 cm by 75 cm.
Motion Problems
Motion problems in which we compare times usually involve solving rational equations. This is because when solving the motion formula ππππππππ π π β πππ‘π‘ππππ π»π» = πππ‘π‘π π ππππππππππ π«π« for time, we create a fraction
Solution
π΄π΄ π΅π΅
πΏπΏ πΏπΏβ
π΅π΅β
ππ
ππ
ππ.ππ
ππ
Figure 6.1a
π₯π₯ 2
1.2
2βπ₯π₯
Figure 6.1b
298
Solving a Motion Problem Where Times are the Same
The speed of one mountain biker is 3 km/h faster than the speed of another biker. The first biker travels 40 km in the same amount of time that it takes the second to travel 30 km. Find the speed of each biker.
Let ππ represent the speed of the slower biker. Then ππ + 3 represents the speed of the faster biker. The slower biker travels 30 km, while the faster biker travels 40 km. Now, we can complete the table
Since the time of travel is the same for both bikers, we form and then solve the equation:
30ππ
=40ππ + 3
3(ππ + 3) = 4ππ
3ππ + 9 = 4ππ
ππ = 9
Thus, the speed of the slower biker is ππ = ππ km/h and the speed of the faster biker is ππ + 3 = ππππ km/h.
Solving a Motion Problem Where the Total Time is Given
Kris and Pat are driving from Vancouver to Princeton, a distance of 297 km. Kris, whose average rate is 6 mph faster than Patβs, will drive the first 153 km of the trip, and then Pat will drive the rest of the way to their destination. If the total driving time is 3 hours, determine the average rate of each driver. Let ππ represent Patβs average rate. Then ππ + 6 represents Krisβ average rate. Kris travelled 153 km, while Pat travelled 297β 153 = 144 km. Now, we can complete the table:
Note: In motion problems we may add times or distances but we usually do not add rates!
π π β π»π» = π«π«
slower biker ππ 30ππ
30
faster biker ππ + 3 40ππ + 3
40
π π β π»π» = π«π«
Kris ππ + 6 153ππ + 6
153
Pat ππ 144ππ
144
total 3 297
Solution
Solution
To complete the Time column, we divide the Distance by the Rate.
/Γ· 10 and cross-multiply
/β3ππ
299
The equation to solve comes from the Time column.
153ππ + 6
+144ππ
= 3
153ππ + 144(ππ + 6) = 3ππ(ππ + 6)
51ππ + 48ππ + 288 = ππ2 + 6ππ
0 = ππ2 β 93ππ β 288
(ππ β 96)(ππ + 3) = 0
ππ = 96 ππππ ππ = β3 Since a rate cannot be negative, we discard the solution ππ = β3. Therefore, Patβs average rate was ππ = ππππ km/h and Krisβ average rate was ππ + 6 = ππππππ km/h.
Work Problems
When solving work problems, refer to the formula
π π πππΉπΉππ ππππ π€π€πππππ€π€ β π»π»ππππππ = ππππππππππππ ππππ π±π±π±π±ππ ππππππππππππππππππ and organize data in a table like this:
Note: In work problems we usually add rates but do not add times!
Solving a Work Problem Involving Addition of Rates
Alex can trim the shrubs at Beecher Community College in 6 hr. Bruce can do the same job in 4 hr. How long would it take them to complete the same trimming job if they work together?
Let ππ be the time needed to trim the shrubs when Alex and Bruce work together. Since trimming the shrubs at Beecher Community College is considered to be the whole one job to complete, then the rate π π in which this work is done equals
π π =π±π±πππππ»π»π‘π‘ππππ
=ππ
π»π»π‘π‘ππππ.
To organize the information, we can complete the table below.
π π β π»π» = J worker I worker II together
/β ππ(ππ + 6)
/Γ· 3 and distribute; then collect like terms on one side factor
Notice the similarity to the
formula π π β π»π» = π«π« used in motion
problems.
Solution
The job column is often equal to 1, although
sometimes other values might need to be used.
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Since the rate of work when both Alex and Bruce trim the shrubs is the sum of rates of individual workers, we form and solve the equation
16
+14
=1ππ
2ππ + 3ππ = 12
5ππ = 12
ππ =125
= 2.4
So, if both Alex and Bruce work together, the amount of time needed to complete the job if 2.4 hours = 2 hours 24 minutes. Note: The time needed for both workers is shorter than either of the individual times.
Solving a Work Problem Involving Subtraction of Rates
One pipe can fill a swimming pool in 10 hours, while another pipe can empty the pool in 15 hours. How long would it take to fill the pool if both pipes were left open? Suppose ππ is the time needed to fill the pool when both pipes are left open. If filling the pool is the whole one job to complete, then emptying the pool corresponds to β1 job. This is because when emptying the pool, we reverse the filling job.
To organize the information given in the problem, we complete the following table.
The equation to solve comes from the Rate column.
π π β π»π» = π±π±
Alex ππππ
6 1
Bruce ππππ
4 1
together πππΉπΉ
ππ 1
π π β π»π» = π±π±
pipe I ππππππ
10 1
pipe II βππππππ
15 β1
both pipes πππΉπΉ
ππ 1
/β 12ππ
/Γ· 5
Solution
/β 30ππ
To complete the Rate column, we divide the
Job by the Time.
301
110
β1
15=
1ππ
3ππ β 2ππ = 30
ππ = 30
So, it will take 30 hours to fill the pool when both pipes are left open.
RT.6 Exercises
Vocabulary Check Complete each blank with the most appropriate term or phrase from the given list: column,
fraction, motion, numerator, proportions, row, similar, work.
1. When solving a formula for a specified variable, we want to keep the variable in the _______________.
2. The strategy of taking reciprocal of each side of an equation is applicable only in solving _________________. This means that each side of such equation must be in the form of a single __________.
3. When forming a proportion, it is essential that the same type of data are placed in the same _________ or in the same ____________.
4. The Thalesβ Theorem states that for any two ____________ triangles the ratios of the corresponding sides are the same.
5. In ___________ problems, we usually add times or distances but not the rates.
6. In ___________ problems, we usually add rates but not times.
Concept Check
7. Using the formula 1ππ
= 1ππ
+ 1ππ, find ππ if ππ = 8 and ππ = 12.
8. The gravitational force between two masses is given by the formula πΉπΉ = πΊπΊπΊπΊππππ2
. Find πΏπΏ if πΉπΉ = 10, πΊπΊ = 6.67 β 10β11, ππ = 1, and ππ = 3 β 10β6. Round your answer to two decimal
places.
Concept Check
9. What is the first step in solving the formula ππππ β ππππ = ππ + ππ for ππ?
10. What is the first step in solving the formula ππ = ππππππβππ
for ππ? Solve each formula for the specified variable.
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11. ππ = πΉπΉππ for ππ 12. πΌπΌ = πΈπΈ
π π for π π 13. ππ1
ππ2= ππ1
ππ2 for ππ1
14. πΉπΉ = πΊπΊπΊπΊππππ2
for ππ 15. π π = (π£π£1+π£π£2)π‘π‘2
for ππ 16. π π = (π£π£1+π£π£2)π‘π‘2
for π£π£1
17. 1π π
= 1ππ1
+ 1ππ2
for π π 18. 1π π
= 1ππ1
+ 1ππ2
for ππ1 19. 1ππ
+ 1ππ
= 1ππ for ππ
20. π‘π‘ππ
+ π‘π‘ππ
= 1 for ππ 21. ππππππ
= πππ£π£π‘π‘
for π£π£ 22. ππππππ
= πππ£π£π‘π‘
for ππ
23. π΄π΄ = β(ππ+ππ)2
for ππ 24. ππ = ππβπ£π£π‘π‘
for ππ 25. π π = πππ π ππ+π π
for π π
26. πΌπΌ = 2ππππ+2ππ
for ππ 27. πΌπΌ = πππΈπΈπΈπΈ+ππππ
for ππ 28. πΈπΈππ
= π π +ππππ
for ππ
29. πΈπΈππ
= π π +ππππ
for ππ 30. ππ = π»π»ππ(π‘π‘1βπ‘π‘2) for ππ1 31. ππ = ππβ2(3π π ββ)
3 for π π
32. ππ = π΄π΄1+ππ
for ππ 33. ππ2
π π 2= 2ππ
π π +β for β 34. π£π£ = ππ2βππ1
π‘π‘2βπ‘π‘1 for ππ2
Analytic Skills Solve each problem
35. The ratio of the weight of an object on the moon to the weight of an object on Earth is 0.16 to 1. How much will an 80-kg astronaut weigh on the moon?
36. A rope is 24 meters long. How can the rope be cut in such a way that the ratio of the resulting two segments is 3 to 5?
37. Walking 4 miles in 2 hours will use up 650 calories. Walking at the same rate, how many miles would a person need to walk to lose 1 lb? (Burning 3500 calories is equivalent to losing 1 pound.) Round to the nearest hundredth.
38. On a map of Canada, the linear distance between Vancouver and Calgary is 1.8 cm. The airline distance between the two cities is about 675 kilometers. On this same map, what would be the linear distance between Calgary and Montreal if the airline distance between the two cities is approximately 3000 kilometers?
39. To estimate the deer population of a forest preserve, wildlife biologists caught, tagged, and then released 42 deer. A month later, they returned and caught a sample of 75 deer and found that 15 of them were tagged. Based on this experiment, approximately how many deer lived in the forest preserve?
40. Biologists tagged 500 fish in a lake on January 1. On February 1, they returned and collected a random sample of 400 fish, 8 of which had been previously tagged. On the basis of this experiment, approximately how many fish does the lake have?
41. Twenty-two bald eagles are tagged and released into the wilderness. Later, an observed sample of 56 bald eagles contains 7 eagles that are tagged. Estimate the bald eagle population in this wilderness area.
42. A 6βfoot tall person casts a 4βfoot long shadow. If a nearby tree casts a 44βfoot long shadow, estimate the height of the tree.
43. Suppose the following triangles are similar. Find π¦π¦ and the lengths of the unknown sides of each triangle.
P
Q R
2π¦π¦ + 2 3
A
B C
4 5π¦π¦ β 2
5
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44. A rectangle has sides of 9 cm and 14 cm. In a similar rectangle the longer side is 8 cm. What is the length of
the shorter side?
45. What is the average speed if Shane runs 18 kilometers per hour for the first half of a race and 22 kilometers per hour for the second half of the race?
46. If you average π₯π₯ kilometers per hour during the first half of a trip, find the speed π¦π¦ in kilometers per hour needed during the second half of the trip to reach 80 kilometers per hour as an overall average.
47. Kellenβs boat goes 12 mph. Find the rate of the current of the river if she can go 6 mi upstream in the same amount of time she can go 10 mi downstream.
48. A plane averaged 500 mph on a trip going east, but only 350 mph on the return trip. The total flying time in both directions was 8.5 hr. What was the one-way distance?
49. A Boeing 747 flies 2420 mi with the wind. In the same amount of time, it can fly 2140 mi against the wind. The cruising speed (in still air) is 570 mph. Find the speed of the wind.
50. A moving sidewalk moves at a rate of 1.7 ft/sec. Walking on the moving sidewalk, Hunter can travel 120 ft forward in the same time it takes to travel 52 ft in the opposite direction. How fast would Hunter be walking on a non-moving sidewalk?
51. On his drive from Montpelier, Vermont, to Columbia, South Carolina, Victor Samuels averaged 51 mph. If he had been able to average 60 mph, he would have reached his destination 3 hr earlier. What is the driving distance between Montpelier and Columbia?
52. On the first part of a trip to Carmel traveling on the freeway, Marge averaged 60 mph. On the rest of the trip, which was 10 mi longer than the first part, she averaged 50 mph. Find the total distance to Carmel if the second part of the trip took 30 min more than the first part.
53. Cathy is a college professor who lives in an off-campus apartment. On days when she rides her bike to campus, she gets to her first class 36 min faster than when she walks. If her average walking rate is 3 mph and her average biking rate is 12 mph, how far is it from her apartment to her first class?
54. Gina can file all of the daily invoices in 4 hr and Bert can do the same job in 6 hr. If they work together, then what portion of the invoices can they file in 1 hr?
55. Melanie can paint the entire house in π₯π₯ hours and Timothy can do the same job in π¦π¦ hours. Write a rational expression that represents the portion of the house that they can paint in 2 hr working together.
56. Walter and Helen are asked to paint a house. Walter can paint the house by himself in 12 hours and Helen can paint the house by herself in 16 hours. How long would it take to paint the house if they worked together?
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57. Brian, Mark, and Jeff are painting a house. Working together they can paint the house is 6 hours. Working alone Brain can paint the house in 15 hours and Jeff can paint the house in 20 hours. How long would it take Mark to paint the house working alone?
58. An experienced carpenter can frame a house twice as fast as an apprentice. Working together, it takes the carpenters 2 days. How long would it take the apprentice working alone?
59. A tank can be filled in 9 hr and drained in 11 hr. How long will it take to fill the tank if the drain is left open?
60. A cold water faucet can fill the bath tub in 12 minutes, and a hot water faucet can fill the bath tub in 18 minutes. The drain can empty the bath tub in 24 minutes. If both faucets are on and the drain is open, how long would it take to fill the bath tub?
61. Together, a 100-cm wide escalator and a 60-cm wide escalator can empty a 1575-person auditorium in 14 min. The wider escalator moves twice as many people as the narrower one does. How many people per hour does the 60-cm wide escalator move?
Discussion Point
62. At what time after 4:00 will the minute hand overlap the hour hand of a clock for the first time?
63. Michelle drives to work at 50 mph and arrives 1 min late. When she drives to work at 60 mph, she arrives 5 min early. How far does Michelle live from work?