Rational Expressions and Functions - Anna Kuczynska
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246 | Section RT1
Rational Expressions and Functions
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.
RT1 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
Section RT1 | 247
Integral Exponents and Scientific Notation
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
248 | Section RT1
Rational Expressions and Functions
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
Section RT1 | 249
Integral Exponents and Scientific Notation
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
250 | Section RT1
Rational Expressions and Functions
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
Earth is approximately 1.5 β 108 kilometers from the Sun. Estimate the time in days needed for a space probe moving at an average rate of 2.4 β 104 km/h to reach the Sun? Assume that the probe moves along a straight line.
Solution
fill the empty places by zeros
Solution
fill the empty places by zeros
divide powers by subtracting exponents
Section RT1 | 251
Integral Exponents and Scientific Notation
To find time ππ needed for the space probe travelling at the rate π π = 2.4 β 104 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
2.4 β 104= 0.625 β 104 = 6.25 β 103
So, it will take 6.25 β 103 hours = 625024
days β ππππππ.ππ days for the space probe to reach the Sun.
RT.1 Exercises
True or false.
1. οΏ½34οΏ½β2
= οΏ½43οΏ½2 2. 10β4 = 0.00001 3. (0.25)β1 = 4
4. β45 = 145
5. (β2)β10 = 4β5 6. 2 β 2 β 2β1 = 18
7. 3π₯π₯β2 = 13π₯π₯2
8. β2β2 = β14 9. 510
5β12= 5β2
10. The number 0.68 β 10β5 is written in scientific notation. 11. 98.6 β 107 = 9.86 β 106
12. Match each expression in Row I with the equivalent expression(s) in Row II, if possible.
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
Evaluate each expression.
13. 4β6 β 43 14. β93 β 9β5 15. 2β3
26 16. 2β7
2β5
17. β3β4
5β3 18. βοΏ½3
2οΏ½β2
19. 2β2 + 2β3 20. (2β1 β 3β1)β1
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.
21. (β2π₯π₯β3)(7π₯π₯β8) 22. (5π₯π₯β2π¦π¦3)(β4π₯π₯β7π¦π¦β2) 23. (9π₯π₯β4ππ)(β4π₯π₯β8ππ)
24. (β3π¦π¦β4ππ)(β5π¦π¦β3ππ) 25. β4π₯π₯β3 26. π₯π₯β4ππ
π₯π₯6ππ
27. 3ππ5
ππππβ2 28. 14ππβ4ππβ3
β8ππ8ππβ5 29. β18π₯π₯β3π¦π¦3
β12π₯π₯β5π¦π¦5
Solution
252 | Section RT1
Rational Expressions and Functions
30. (2β1ππβ7ππ)β4 31. (β3ππ2ππβ5)β3 32. οΏ½5π₯π₯β2
π¦π¦3οΏ½β3
33. οΏ½2π₯π₯3π¦π¦β2
3π¦π¦β3οΏ½β3
34. οΏ½ β4π₯π₯β3
5π₯π₯β1π¦π¦4οΏ½β4
35. οΏ½125π₯π₯2π¦π¦β3
5π₯π₯4π¦π¦β2οΏ½β5
36. οΏ½β200π₯π₯3π¦π¦β5
8π₯π₯5π¦π¦β7οΏ½β4
37. [(β2π₯π₯β4π¦π¦β2)β3]β2 38. 12ππβ2οΏ½ππβ3οΏ½
β2
6ππ7
39. (β2ππ)2ππβ5
(ππππ)β3 40. οΏ½2ππππ2οΏ½3οΏ½3ππ
4
ππβ4οΏ½β1
41. οΏ½ β3π₯π₯4π¦π¦6
15π₯π₯β6π¦π¦7οΏ½β3
42. οΏ½ β4ππ3ππ2
12ππ6ππβ5οΏ½β3
43. οΏ½β9β2π₯π₯β4π¦π¦
3β3π₯π₯β3π¦π¦2οΏ½8 44. (4βπ₯π₯)2π¦π¦
45. (5ππ)βππ 46. π₯π₯πππ₯π₯βππ 47. 9ππ2βπ₯π₯
3ππ2β2π₯π₯
48. 12π₯π₯ππ+1
β4π₯π₯2βππ 49. οΏ½π₯π₯ππβ1οΏ½3οΏ½π₯π₯ππβ4οΏ½β2 50. 25π₯π₯ππ+πππ¦π¦ππβππ
β5π₯π₯ππβπππ¦π¦ππβππ
Convert each number to scientific notation.
51. 26,000,000,000 52. β0.000132 53. 0.0000000105 54. 705.6 Convert each number to decimal notation.
55. 6.7 β 108 56. 5.072 β 10β5 57. 2 β 1012 58. 9.05 β 10β9 59. One megabyte of computer memory equals 220 bytes. Using decimal notation, write the number of bytes in
1 megabyte. Then, using scientific notation, approximate this number by rounding the scientific notation coefficient to two decimals places.
Evaluate. State your answer in scientific notation.
60. (6.5 β 103)(5.2 β 10β8) 61. (2.34 β 10β5)(5.7 β 10β6)
62. (3.26 β 10β6)(5.2 β 10β8) 63. 4 β 10β7
8 β 10β3
64. 7.5 β 109
2.5 β 104 65. 4 β 10β7
8 β 10β3
66. 0.05 β 160000.0004
67. 0.003 β 40,0000.00012 β600
Solve each problem. State your answer in scientific notation.
68. A light-year is an astronomical unit measuring the distance that light travels in one year. If light travels approximately 3 β 105 kilometers per second, how long is a light-year in kilometers?
69. In 2018, the national debt in Canada was about 6.7 β 1011 dollars. If the Canadian population in 2018 was approximately 3.7 β 107, what was the share of this debt per person?
Section RT1 | 253
Integral Exponents and Scientific Notation
70. One of the brightest stars in the night sky, Vega, is about 2.365 β 1014 kilometers from Earth. If one light-year is approximately 9.46 β 1012 kilometers, how many light-years is it from Earth to Vega?
71. The Columbia River discharges its water to the Pacific Ocean at approximately 265,000 ft3/sec. What is the supply of water that comes from the Columbia River in one minute? in one day? State the answer in scientific notation.
72. Assuming the current trends continue, the population ππ of Canada, in millions, can be modelled by the equation ππ = 34(1.011)π₯π₯, where π₯π₯ is the number of years passed after the year 2010. According to this model, what is the predicted Canadian population for the years 2025 and 2030?
73. The mass of the Moon is 7.348 β 1022 kg while the mass of Earth is 5.976 β 1024 kg. How many times heavier is Earth than the Moon?
74. Most calculators cannot handle operations on numbers outside of the interval (10β100, 10100). How can we compute (5 β 10120)3 without the use of a calculator?
254 | Section RT2
Rational Expressions and Functions
RT2 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 RT5. Figure 1
ππ(π₯π₯)
π₯π₯
1
3
Section RT2 | 255
Rational Expressions and Functions; Multiplication and Division of Rational Expressions
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
256 | Section RT2
Rational Expressions and Functions
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
Section RT2 | 257
Rational Expressions and Functions; Multiplication and Division of Rational Expressions
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 involves 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
258 | Section RT2
Rational Expressions and Functions
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 nonzero 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₯
Section RT2 | 259
Rational Expressions and Functions; Multiplication and Division of Rational Expressions
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 nonzero 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.
260 | Section RT2
Rational Expressions and Functions
RT.2 Exercises
True or false.
1. ππ(π₯π₯) = 4βπ₯π₯β4
is a rational function. 2. The domain of ππ(π₯π₯) = π₯π₯β24
is the set of all real numbers.
3. π₯π₯β34βπ₯π₯
is equivalent to βπ₯π₯β3π₯π₯β4
. 4. ππ2+1ππ2β1
is equivalent to ππ+1ππβ1
.
Given the rational function f, find ππ(β1), ππ(0), and ππ(2).
5. ππ(π₯π₯) = π₯π₯π₯π₯β2
6. ππ(π₯π₯) = 5π₯π₯3π₯π₯βπ₯π₯2
7. ππ(π₯π₯) = π₯π₯β2π₯π₯2+π₯π₯β6
For each rational function, find all numbers that are not in the domain. Then give the domain, using both set notation and interval notation.
8. ππ(π₯π₯) = π₯π₯π₯π₯+2
9. ππ(π₯π₯) = π₯π₯π₯π₯β6
10. β(π₯π₯) = 2π₯π₯β13π₯π₯+7
11. ππ(π₯π₯) = 3π₯π₯+25π₯π₯β4
12. ππ(π₯π₯) = π₯π₯+2π₯π₯2β4
13. β(π₯π₯) = π₯π₯β2π₯π₯2+4
14. ππ(π₯π₯) = 53π₯π₯βπ₯π₯2
15. ππ(π₯π₯) = π₯π₯2+π₯π₯β6π₯π₯2+12π₯π₯+35
16. β(π₯π₯) = 7|4π₯π₯β3|
17. Which rational expressions are equivalent and what is their simplest form?
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
18. 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.
19. 24ππ3ππ3ππππ3
20. β18π₯π₯2π¦π¦3
8π₯π₯3π¦π¦ 21. 7βπ₯π₯
π₯π₯β7 22. π₯π₯+2
π₯π₯β2
23. ππβ5β5+ππ
24. (3βπ¦π¦)(π₯π₯+1)(π¦π¦β3)(π₯π₯β1) 25. 12π₯π₯β15
21 26. 18ππβ2
22
27. 4π¦π¦β124π¦π¦+12
28. 7π₯π₯+147π₯π₯β14
29. 6ππ+187ππ+21
30. 3π§π§2+π§π§18π§π§+6
31. ππ2β2520β4ππ
32. 9ππ2β34β12ππ2
33. π‘π‘2β25π‘π‘2β10π‘π‘+25
34. ππ2β36ππ2+12π‘π‘+36
Section RT2 | 261
Rational Expressions and Functions; Multiplication and Division of Rational Expressions
35. π₯π₯2β9π₯π₯+8π₯π₯2+3π₯π₯β4
36. ππ2+8ππβ9ππ2β5ππ+4
37. π₯π₯3βπ¦π¦3
π₯π₯2βπ¦π¦2 38. ππ2βππ2
ππ3βππ3
Perform operations and simplify. Assume nonzero denominators.
39. 18ππ4
5ππ2β 25ππ
4
9ππ3 40. 28
π₯π₯π¦π¦Γ· 63π₯π₯3
2π¦π¦2 41. 12π₯π₯
49(π₯π₯π¦π¦2)3 β(7π₯π₯π¦π¦)2
8
42. π₯π₯+12π₯π₯β3
β 2π₯π₯β32π₯π₯
43. 10ππ6ππβ12
β 20ππβ4030ππ3
44. ππ2β14ππ
β 21βππ
45. π¦π¦2β254π¦π¦
β 25βπ¦π¦
46. (8π₯π₯ β 16) Γ· 3π₯π₯β610 47. (π¦π¦2 β 4) Γ· 2βπ¦π¦
8π¦π¦
48. 3ππβ9ππ2β9
β (ππ3 + 27) 49. π₯π₯2β16π₯π₯2
β π₯π₯2β4π₯π₯π₯π₯2βπ₯π₯β12
50. π¦π¦2+10π¦π¦+25π¦π¦2β9
β π¦π¦2β3π¦π¦π¦π¦+5
51. ππβ3ππ2β4ππ+3
Γ· ππ2βππππβ1
52. π₯π₯2β6π₯π₯+9π₯π₯2+3π₯π₯
Γ· π₯π₯2β9π₯π₯
53. π₯π₯2β2π₯π₯3π₯π₯2β5π₯π₯β2
β 9π₯π₯2β49π₯π₯2β12π₯π₯+4
54. π‘π‘2β49π‘π‘2+4π‘π‘β21
β π‘π‘2+8π‘π‘+15π‘π‘2β2π‘π‘β35
55. ππ3βππ3
ππ2βππ2Γ· 2ππβ2ππ
2ππ+2ππ 56. 64π₯π₯3+1
4π₯π₯2β100β 4π₯π₯+2064π₯π₯2β16π₯π₯+4
57. 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 58. ππ3β27ππ3
ππ2+ππππβ12ππ2β ππ
2β2ππππβ24ππ2
ππ2β5ππππβ6ππ2
59. 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 60. 2π₯π₯2+π₯π₯β1
6π₯π₯2+π₯π₯β2Γ· 2π₯π₯2+5π₯π₯+3
6π₯π₯2+13π₯π₯+6
61. 6π₯π₯2β13π₯π₯+614π₯π₯2β25π₯π₯+6
Γ· 14β21π₯π₯49π₯π₯2+7π₯π₯β6
62. 4π¦π¦2β12π¦π¦+3627β3π¦π¦2
Γ· (π¦π¦3 + 27)
63. 3π¦π¦π₯π₯2
Γ· π¦π¦2
π₯π₯Γ· π¦π¦
5π₯π₯ 64. π₯π₯+1
π¦π¦β2Γ· 2π₯π₯+2
π¦π¦β2Γ· π₯π₯
π¦π¦
65. ππ2β4ππ2
ππ+2ππΓ· (ππ + 2ππ) β 2ππ
ππβ2ππ 66. 9π₯π₯2
π₯π₯2β16π¦π¦2Γ· 1
π₯π₯2+4π₯π₯π¦π¦β π₯π₯β4π¦π¦
3π₯π₯
67. π₯π₯2β25π₯π₯β4
Γ· π₯π₯2β2π₯π₯β15π₯π₯2β10π₯π₯+24
β π₯π₯+3π₯π₯2+10π₯π₯+25
68. 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₯).
69. ππ(π₯π₯) = π₯π₯β4π₯π₯2+π₯π₯
and ππ(π₯π₯) = 2π₯π₯π₯π₯+1
70. ππ(π₯π₯) = π₯π₯3β3π₯π₯2
π₯π₯+5 and ππ(π₯π₯) = 4π₯π₯2
π₯π₯β3
71. ππ(π₯π₯) = π₯π₯2β7π₯π₯+12π₯π₯+3
and ππ(π₯π₯) = 9βπ₯π₯2
π₯π₯β4 72. ππ(π₯π₯) = π₯π₯+6
4βπ₯π₯2 and ππ(π₯π₯) = 2βπ₯π₯
π₯π₯2+8π₯π₯+12
262 | Section RT3
Rational Expressions and Functions
RT3 Addition and Subtraction of Rational Expressions
Many real-world applications involve adding or subtracting algebraic fractions. Like 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
Section RT3 | 263
Addition and Subtraction of Rational Expressions
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
264 | Section RT3
Rational Expressions and Functions
To add or subtract algebraic fractions, follow the steps:
Factor the denominators of all algebraic fractions completely. Find the LCD of all the denominators. Convert each algebraic fraction to the lowest common denominator found in the
previous step and write the sum (or difference) as a single fraction. 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 10ππ. This can be done by multiplying the numerator and denominator of each expression by the factors of 10ππ 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.
Section RT3 | 265
Addition and Subtraction of Rational Expressions
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
Assume that a boat travels ππ kilometers up the river and then returns back to the starting point. If the water in the river flows with a constant current of ππ km/h, the total time for the round-trip can be calculated via the expression ππ
ππ+ππ+ ππ
ππβππ, where ππ is the speed of the boat
in still water in kilometers per hour. Write a single rational expression representing the total time of this trip.
multiply by the missing bracket
πΏπΏπΏπΏπ·π· = (π₯π₯β 2)(π₯π₯+ 2)
nothing to simplify this time
266 | Section RT3
Rational Expressions and Functions
To find a single rational expression representing the total time, we perform the addition using (ππ + ππ)(ππ β ππ) as the lowest common denominator. So,
ππππ + ππ
+ππ
ππ β ππ=ππ(ππ β ππ) + ππ(ππ + ππ)
(ππ + ππ)(ππ β ππ) =ππππ β ππππ + ππππ + ππππ
(ππ + ππ)(ππ β ππ) =ππππππ
ππππ β ππππ
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
1. a. What is the LCM for 6 and 9? b. What is the LCD for 1
6 and 1
9 ?
2. 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
? Find the LCD and then perform the indicated operations. Simplify the resulting fraction.
3. 512
+ 1318
4. 1130β 19
75 5. 3
4+ 7
30β 1
16 6. 5
8β 7
12+ 11
40
Solution
Solution a.
b.
Section RT3 | 267
Addition and Subtraction of Rational Expressions
Find the least common multiple (LCM) for each group of expressions.
7. 24ππ3ππ4, 18ππ5ππ2 8. 6π₯π₯2π¦π¦2, 9π₯π₯3π¦π¦, 15π¦π¦3 9. π₯π₯2 β 4, π₯π₯2 + 2π₯π₯
10. 10π₯π₯2, 25(π₯π₯2 β π₯π₯) 11. (π₯π₯ β 1)2, 1 β π₯π₯ 12. π¦π¦2 β 25, 5 β π¦π¦
13. π₯π₯2 β π¦π¦2, π₯π₯π¦π¦ + π¦π¦2 14. 5ππ β 15, ππ2 β 6ππ + 9 15. π₯π₯2 + 2π₯π₯ + 1, π₯π₯2 β 4π₯π₯ β 1
16. ππ2 β 7ππ + 10, ππ2 β 8ππ + 15 17. 2π₯π₯2 β 5π₯π₯ β 3, 2π₯π₯2 β π₯π₯ β 1, π₯π₯2 β 6π₯π₯ + 9
18. 1 β 2π₯π₯, 2π₯π₯ + 1, 4π₯π₯2 β 1 19. π₯π₯5 β 4π₯π₯4 + 4π₯π₯3, 12 β 3π₯π₯2, 2π₯π₯ + 4 True or false? If true, explain why. If false, correct it.
20. 12π₯π₯
+ 13π₯π₯
= 15π₯π₯
21. 1π₯π₯β3
+ 13βπ₯π₯
= 0 22. 1π₯π₯
+ 1π¦π¦
= 1π₯π₯+π¦π¦
23. 34
+ π₯π₯5
= 3+π₯π₯20
Perform the indicated operations and simplify if possible.
24. π₯π₯β2π¦π¦π₯π₯+π¦π¦
+ 3π¦π¦π₯π₯+π¦π¦
25. ππ+3ππ+1
β ππβ5ππ+1
26. 4ππ+3ππβ3
β 1
27. ππ+1ππβ2
+ 2 28. π₯π₯2
π₯π₯βπ¦π¦+ π¦π¦2
π¦π¦βπ₯π₯ 29. 4ππβ2
ππ2β49+ 5+3ππ
49βππ2
30. 2π¦π¦β3π¦π¦2β1
β 4βπ¦π¦1βπ¦π¦2
31. ππ3
ππβππ+ ππ3
ππβππ 32. 1
π₯π₯+ββ 1
π₯π₯
33. π₯π₯β2π₯π₯+3
+ π₯π₯+2π₯π₯β4
34. π₯π₯β13π₯π₯+1
+ 2π₯π₯β3
35. 4π₯π₯π¦π¦π₯π₯2βπ¦π¦2
+ π₯π₯βπ¦π¦π₯π₯+π¦π¦
36. π₯π₯β13π₯π₯+15
β π₯π₯+35π₯π₯+25
37. π¦π¦β24π¦π¦+8
β π¦π¦+65π¦π¦+10
38. 4π₯π₯π₯π₯β1
β 2π₯π₯+1
β 4π₯π₯2β1
39. β2π¦π¦+2
+ 5π¦π¦β2
+ π¦π¦+3π¦π¦2β4
40. π¦π¦π¦π¦2βπ¦π¦β20
+ 2π¦π¦+4
41. 5π₯π₯π₯π₯2β6π₯π₯+8
β 3π₯π₯π₯π₯2βπ₯π₯β12
42. 9π₯π₯+23π₯π₯2β2π₯π₯β8
+ 73π₯π₯2+π₯π₯β4
43. 3π¦π¦+22π¦π¦2βπ¦π¦β10
+ 82π¦π¦2β7π¦π¦+5
44. 6π¦π¦2+6π¦π¦+9
+ 5π¦π¦2β9
45. 3π₯π₯β1π₯π₯2+2π₯π₯β3
β π₯π₯+4π₯π₯2β9
46. 1π₯π₯+1
β π₯π₯π₯π₯β2
+ π₯π₯2+2π₯π₯2βπ₯π₯β2
47. 2π¦π¦+3
β π¦π¦π¦π¦β1
+ π¦π¦2+2π¦π¦2+2π¦π¦β3
48. 4π₯π₯π₯π₯2β1
+ 3π₯π₯1βπ₯π₯
β 4π₯π₯β1
49. 5π¦π¦1β2π¦π¦
β 2π¦π¦2π¦π¦+1
+ 34π¦π¦2β1
50. π₯π₯+5π₯π₯β3
β π₯π₯+2π₯π₯+1
β 6π₯π₯+10π₯π₯2β2π₯π₯β3
Perform the indicated operations and simplify if possible.
51. 2π₯π₯β3 + (3π₯π₯)β1 52. (π₯π₯2 β 9)β1 + 2(π₯π₯ β 3)β1 53. οΏ½π₯π₯+13οΏ½β1β οΏ½π₯π₯β4
2οΏ½β1
54. οΏ½ππβ3ππ2
β ππβ39οΏ½Γ· ππ2β9
3ππ 55. π₯π₯2β4π₯π₯+4
2π₯π₯+1β 2π₯π₯
2+π₯π₯π₯π₯3β4π₯π₯
β 3π₯π₯β2π₯π₯+1
56. 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.
57. ππ(π₯π₯) = π₯π₯π₯π₯+2
, ππ(π₯π₯) = 4π₯π₯β3
58. ππ(π₯π₯) = π₯π₯π₯π₯2β4
, ππ(π₯π₯) = 1π₯π₯2+4π₯π₯+4
59. ππ(π₯π₯) = 3π₯π₯π₯π₯2+2π₯π₯β3
, ππ(π₯π₯) = 1π₯π₯2β2π₯π₯+1
60. ππ(π₯π₯) = π₯π₯ + 1π₯π₯β1
, ππ(π₯π₯) = 1π₯π₯+1
268 | Section RT3
Rational Expressions and Functions
Solve each problem.
61. There are two part-time waitresses at a restaurant. One waitress works every fourth day, and the other one works every sixth day. Both waitresses were hired and start working on the same day. How often do they both work on the same day?
62. A cylindrical water tank is being filled and drained at the same time. To find the rate of change of the water level one could use the expression π»π»
ππππππβ π»π»
ππππππππ, where π»π» is the height of the water
in the full tank while ππππππ and πππππππ‘π‘ represent the time needed to fill and empty the tank, respectively. Write the rate of change of the water level as a single algebraic fraction.
63. To determine the Canadian population percent growth over the past year, one could use the expression
100 οΏ½ππ1ππ0β 1οΏ½, where ππ1 represents the current population and ππ0 represents the last yearβs population. Write
this expression as a single algebraic fraction.
64. A boat travels ππ kilometers against a ππ km/h current. Assuming the current remains constant, one could calculate the total time, in hours, needed for the entire trip via the expression ππ
π π βππ+ ππ
π π +ππ, where π π represents the speed of the
boat in calm water. Write this expression as a single algebraic fraction.
Section RT4 | 269
Complex Fractions
RT4 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
270 | Section RT4
Rational Expressions and Functions
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
Section RT4 | 271
Complex Fractions
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
272 | Section RT4
Rational Expressions and Functions
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
Simplify each complex fraction.
1. 2 β 133 + 73
2. 5 β 344 + 12
3. 38 β 523 + 6
4. 23 + 4534 β 12
Simplify each complex rational expression.
5. π₯π₯3
π¦π¦π₯π₯2
π¦π¦3 6.
ππ β 56ππππ β 58ππ2
7. 1 β 1ππ4 + 1ππ
8. 2ππ + 35ππ β 6
9. 9 β 3π₯π₯4π₯π₯ + 12π₯π₯ β 36π₯π₯ β 24
10. 9π¦π¦
15π¦π¦ β 6
11. 4π₯π₯ β 2π¦π¦4π₯π₯ + 2π¦π¦
12. 3ππ + 4ππ4ππ β 3ππ
13. ππ β 3ππππππ β ππππ
14. 1π₯π₯ β 1π¦π¦π₯π₯2βπ¦π¦2π₯π₯π¦π¦
15. 4π¦π¦ β π¦π¦
π₯π₯21π₯π₯ β 2π¦π¦
16. 5ππ β 1ππ1
5ππ2 β 5
ππ2
17. ππβ12ππ +ππ
ππ + 4 18. 2π‘π‘β1
3ππβ2ππ + 2π‘π‘
19. 1
ππββ β 1ππβ
20. 1
(π₯π₯+β)2 β 1π₯π₯2
β
21. 4 + 12
2π₯π₯β3
5 + 152π₯π₯β3
22. 1 + 3
π₯π₯+2
1 + 6π₯π₯β1
23. 1ππ2
β 1ππ2
1ππ β 1ππ
24. 1π₯π₯2
β 1π¦π¦2
1π₯π₯ + 1π¦π¦
25. π₯π₯+3π₯π₯ β 4
π₯π₯β1π₯π₯
π₯π₯β1 + 1π₯π₯ 26.
3π₯π₯2+6π₯π₯+9
+ 3π₯π₯+3
6π₯π₯2β9
+ 63βπ₯π₯
27. 1ππ2
β 1ππ2
1ππ3
+ 1ππ3
28.
4ππ2β12ππ+92ππ2+7ππβ152ππ2β15ππ+18ππ2βππβ30
29. Are the expressions π₯π₯β2+π¦π¦β2
π₯π₯β1+π¦π¦β1 and π₯π₯+π¦π¦
π₯π₯2+π¦π¦2 equivalent? Explain why or why not.
This bracket is essential!
keep the denominator in a factored form
Section RT4 | 273
Complex Fractions
Simplify each expression. Leave the answer with only positive exponents.
30. 1ππβ2 β ππβ2
31. π₯π₯β1 + π₯π₯β2
3π₯π₯β1 32. π₯π₯β2
π¦π¦β3 β π₯π₯β3 33. 1 β (2ππ+1)β1
1 + (2ππ+1)β1
Find and simplify the difference quotient ππ(ππ+β)βππ(ππ)
β for the given function.
34. ππ(π₯π₯) = 5π₯π₯ 35. ππ(π₯π₯) = 2
π₯π₯2 36. ππ(π₯π₯) = 1
1βπ₯π₯ 37. ππ(π₯π₯) = β 1
π₯π₯β2
Simplify each continued fraction.
38. ππ β ππ1 β ππ
1 β ππ 39. 3 β 2
1 β 2
3 β 2π₯π₯
40. ππ + ππ2+ 1
1 β 2ππ
274 | Section RT5
Rational Expressions and Functions
RT5 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.
Section RT5 | 275
Applications of Rational Equation
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
276 | Section RT5
Rational Expressions and Functions
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)
Section RT5 | 277
Applications of Rational Equation
(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
Determine the domain of the variable.
Clear all the fractions by multiplying both sides of the equation by the LCD of these fractions.
Find possible solutions by solving the resulting equation.
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.
278 | Section RT5
Rational Expressions and Functions
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
Section RT5 | 279
Applications of Rational Equation
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
280 | Section RT5
Rational Expressions and Functions
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
Section RT5 | 281
Applications of Rational Equation
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
Elevating the outer rail of a track allows for a safer turn of a train on a circular curve. The elevation depends on the allowable speed of the train and the radius of the curve. Suppose that a circular curve with a radius of ππ meters is being designed for a train travelling 100 kilometers per hour. Assume that the function ππ(ππ) = 3000
ππ can be used to calculate the
proper elevation π¦π¦ = ππ(ππ), in centimeters, 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
Solution
ππ(π₯π₯) = 12π₯π₯β3
π₯π₯
1
2
ππ(π₯π₯) = 2π₯π₯ β 3
elevation
282 | Section RT5
Rational Expressions and Functions
a. Evaluate ππ(300) and interpret the result.
b. Suppose that the outer rail for a curve is elevated 12 centimeters. Find the radius of this curve.
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) = 3000300
= 10. Thus, the outer rail on a curve with a 300-meter radius should be elevated 10 centimeters for a train to travel through it at 100 km/hr safely.
b. Since the elevation π¦π¦ = ππ(ππ) = 12 centimeters, to find the corresponding value of ππ, we need to solve the equation
12 = 3000ππ
.
After multiplying this equation by ππ and dividing it by 12, we obtain
ππ = 300012
= 250
So, the radius of this curve should be 250 meters. c. As the radius increases, the outer rail needs less elevation.
RT.5 Exercises
State the domain for each equation. There is no need to solve it.
1. π₯π₯+54β π₯π₯+3
3= π₯π₯
6 2. 5
6ππβ ππ
4= 8
2ππ
3. 3π₯π₯+4
= 2π₯π₯β9
4. 43π₯π₯β5
+ 2π₯π₯
= 94π₯π₯+7
5. 4π¦π¦2β25
β 1π¦π¦+5
= 2π¦π¦β7
6. π₯π₯2π₯π₯β6
β 3π₯π₯2β6π₯π₯+9
= π₯π₯β23π₯π₯β9
Solve each equation.
7. 38
+ 13
= π₯π₯12
8. 14β 5
6= 1
π¦π¦
9. π₯π₯ + 8π₯π₯
= β9 10. 43ππβ 3
ππ= 10
3
11. ππ8
+ ππβ412
= ππ24
12. ππβ22β ππ
6= 4ππ
9
Solution
radius (m)
elev
atio
n (c
m)
10
ππ(ππ)
ππ
20
300
40
30
500 100
50 ππ(ππ) =
3000ππ
Section RT5 | 283
Applications of Rational Equation
13. 5ππ+20
= 3ππ 14. 5
ππ+4= 3
ππβ2
15. π¦π¦+2π¦π¦
= 53 16. π₯π₯β4
π₯π₯+6= 2π₯π₯+3
2π₯π₯β1
17. π₯π₯π₯π₯β1
β π₯π₯2
π₯π₯β1= 5 18. 3 β 12
π₯π₯2= 5
π₯π₯
19. 13β π₯π₯β1
π₯π₯= π₯π₯
3 20. 1
π₯π₯+ 2
π₯π₯+10= π₯π₯
π₯π₯+10
21. 1π¦π¦β1
+ 512
= β23π¦π¦β3
22. 76π₯π₯+3
β 13
= 22π₯π₯+1
23. 83ππ+9
β 815
= 25ππ+15
24. 6ππβ4
+ 5ππ
= 2ππ2β4ππ
25. 3π¦π¦β2
+ 2π¦π¦4βπ¦π¦2
= 5π¦π¦+2
26. π₯π₯π₯π₯β2
+ π₯π₯π₯π₯2β4
= π₯π₯+3π₯π₯+2
27. 12π₯π₯+10
= 8π₯π₯2β25
β 2π₯π₯β5
28. 5π¦π¦+3
= 14π¦π¦2β36
+ 2π¦π¦β3
29. 6π₯π₯2β4π₯π₯+3
β 1π₯π₯β3
= 14π₯π₯β4
30. 7π₯π₯β2
β 8π₯π₯+5
= 12π₯π₯2+6π₯π₯β20
31. 5π₯π₯β4
β 3π₯π₯β1
= π₯π₯2β1π₯π₯2β5π₯π₯+4
32. π¦π¦π¦π¦+1
+ 3π¦π¦+5π¦π¦2+4π¦π¦+3
= 2π¦π¦+3
33. 3π₯π₯π₯π₯+2
+ 72π₯π₯3+8
= 24π₯π₯2β2π₯π₯+4
34. 4π₯π₯+3
+ 7π₯π₯2β3π₯π₯+9
= 108π₯π₯3+27
35. π₯π₯2π₯π₯β9
β 3π₯π₯ = 109β2π₯π₯
36. 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.
37. ππ(π₯π₯) = 2π₯π₯ β 15π₯π₯
; ππ(π₯π₯) = 1 38. ππ(π₯π₯) = π₯π₯β5π₯π₯+1
; ππ(π₯π₯) = 35
39. ππ(π₯π₯) = β3π₯π₯π₯π₯+3
+ π₯π₯; ππ(π₯π₯) = 4 40. ππ(π₯π₯) = 4π₯π₯
+ 1π₯π₯β2
; ππ(π₯π₯) = 3
Graph each rational function. State its domain, range and the equations of the vertical and horizontal asymptotes.
41. ππ(π₯π₯) = 2π₯π₯ 42. ππ(π₯π₯) = β 1
π₯π₯ 43. β(π₯π₯) = 2
π₯π₯β3
44. ππ(π₯π₯) = β1π₯π₯+1
45. ππ(π₯π₯) = π₯π₯β1π₯π₯+2
46. β(π₯π₯) = π₯π₯+2π₯π₯β3
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.
47. ππ(π₯π₯) = 12π₯π₯ + 1 48. ππ(π₯π₯) = βπ₯π₯ + 2 49. ππ(π₯π₯) = β2π₯π₯ β 3
284 | Section RT5
Rational Expressions and Functions
Solve each equation.
50. π₯π₯1 + 1
π₯π₯+1= π₯π₯ β 3 51.
2 β 1π₯π₯4 β 1
π₯π₯2= 1
Solve each problem.
52. Suppose that the number of vehicles searching for a parking place at UFV parking lot is modelled by the function
ππ(π₯π₯) = π₯π₯2
2(1βπ₯π₯),
where 0 β€ π₯π₯ < 1 is a quantity known as traffic intensity.
a. For each traffic intensity, find the number of vehicles searching for a parking place. Round your answer to the nearest one.
i. 0.2 ii. 0.8 iii. 0.98
b. Observing answers to part (a), conclude how does the number of vehicles searching for a parking place changes when the traffic intensity get closer to 1.
53. Suppose that the percent of deaths caused by smoking, called the incidence rate, is modelled by the rational function
π·π·(π₯π₯) =π₯π₯ β 1π₯π₯
,
where π₯π₯ tells us how many times a smoker is more likely to die of lung cancer than a non-smoker.
a. Find π·π·(10) and interpret it in the context of the problem. b. Find the π₯π₯-value corresponding to the incidence rate of 0.5. c. Under what condition would the incidence rate equal to 0?
Section RT6 | 285
Applications of Rational Equation
RT6 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
286 | Section RT6
Rational Expressions and Functions
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)
Section RT6 | 287
Applications of Rational Equation
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 of the same distance ππ.
a. Given the speed π£π£1 for the first part of the trip and π£π£2 for the second part of the trip, find a formula for the average speed π£π£ for the whole trip. (Make sure to leave this formula in the simplified form.)
b. Find the average speed π£π£ for the whole trip, if the speed for the first part of the trip was 75 km/h and the speed for the second part of the trip was 105 km/h.
c. How does the π£π£-value from (b) compare to the average of π£π£1 and π£π£2?
a. The total distance, π·π·, for the whole trip is ππ + ππ = 2ππ. 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‘π‘π‘ππ = πππ‘π‘π π ππππππππππ, we have
ππ1 = πππ£π£1
and ππ2 = πππ£π£2
. Therefore,
ππ =πππ£π£1
+πππ£π£2
,
which after substituting to the formula for the average speed, ππ = π·π·ππ, gives us
ππ =2ππ
πππ£π£1
+ πππ£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
ππ =2ππ
πππ£π£1
+ πππ£π£2
βπ£π£1π£π£2π£π£1π£π£2
ππ =2πππ£π£1π£π£2
πππ£π£1π£π£2π£π£1
+ πππ£π£1π£π£2π£π£2
Solution
/+πΏπΏππ
/Γ· (π π + πΏπΏ)
288 | Section RT6
Rational Expressions and Functions
ππ =2πππ£π£1π£π£2πππ£π£2 + πππ£π£1
ππ =2πππ£π£1π£π£2
ππ(π£π£2 + π£π£1)
π½π½ =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 = 75 km/h and π£π£2 = 105 km/h, using the formula developed in Example 2a, we calculate
π£π£ =2 β 75 β 10575 + 105
=15750
180= ππππ.ππ π€π€π€π€/π‘π‘
c. The average speed for the whole trip, π£π£ = 87.5 km/h, is lower than the average of the
speeds for each part of the trip, which is 75+1052
= 90 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 is 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
.
factor the ππ
Section RT6 | 289
Applications of Rational Equation
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 a particular area in Nevada, a forest ranger catches 452 wild horses, tags them, and releases them. In a week, he catches 95 horses out of which 10 are found to be tagged. Assuming that the horses mix freely when they are released, estimate the number of wild horses in this region. Round your answer to the nearest hundreds. Suppose there are π₯π₯ wild horses in region. 452 of them were tagged, so the ratio of the tagged horses to the whole population of the wild horses there is
452π₯π₯
The ratio of the tagged horses found in the sample of 95 horses caught in the later time is
1095
So, we form the proportion:
452π₯π₯
=1095
After solving for π₯π₯, we have
π₯π₯ =452 β 95
10= 4294 β 4300
So, we can say that approximatly 4300 wild horses live in this region.
Solution
452
95
10
x wild horses
tagged horses
all horses
population sample
290 | Section RT6
Rational Expressions and Functions
πππ‘π‘π‘π‘ππ 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 6.1a. Find the size of the largest cubic box fitting in this room when placed with its base on the floor. 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‘π‘π‘ππ 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
Section RT6 | 291
Applications of Rational Equation
Solving a Motion Problem Where Times are the Same
Two bikers participate in a Cross-Mountain Crusher. One biker is 2 km/h faster than the other. The faster biker travels 35 km in the same amount of time that it takes the slower biker to cover only 30 km. Find the average speed of each biker. Let ππ represent the average speed of the slower biker. Then ππ + 2 represents the average speed of the faster biker. The slower biker travels 30 km, while the faster biker travels 35 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
=35ππ + 2
6(ππ + 2) = 7ππ
6ππ + 12 = 7ππ
ππ = 12
Thus, the average speed of the slower biker is ππ = ππππ km/h and the average speed of the faster biker is ππ + 2 = ππππ km/h.
Solving a Motion Problem Where the Total Time is Given
Judy and Nathan drive from Abbotsford to Kelowna, a distance of 322 km. Judyβs average driving rate is 5 km/h faster than Nathanβs. Judy got tired after driving the first 154 kilometers, so Nathan drove the remaining part of the trip. If the total driving time was 3 hours, what was the average rate of each driver? Let ππ represent Nathanβs average rate. Then ππ + 5 represents Judyβs average rate. Since Judy drove 154 km, Nathan drove 322β 154 = 168 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 + 2 35ππ + 2
35
π π β π»π» = π«π«
Judy ππ + 5 154ππ + 5
154
Nathan ππ 168ππ
168
total 3 322
Solution
Solution
To complete the Time column, we divide the Distance by the Rate.
/Γ· 5 and cross-multiply
/β6ππ
292 | Section RT6
Rational Expressions and Functions
The equation to solve comes from the Time column.
154ππ + 5
+168ππ
= 3
154ππ + 168(ππ + 5) = 3ππ(ππ + 5)
154ππ + 168ππ + 840 = 3ππ2 + 15ππ
0 = 3ππ2 β 307ππ β 840
(3ππ + 8)(ππ β 105) = 0
ππ = β83 ππππ ππ = 105
Since a rate cannot be negative, we discard the solution ππ = β8
3. Therefore, Nathanβs average rate was ππ = ππππππ km/h and Judyβs average rate was ππ + 5 = ππππππ km/h.
Work Problems
When solving work problems, refer to the formula
π π πππΉπΉππ ππππ π€π€ππππππ β π»π»ππππππ = πππ‘π‘ππππππππ ππππ π±π±π±π±ππ πππππ‘π‘ππππππππππππ 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
Adam can trim the shrubs at Centralia College in 8 hr. Bruce can do the same job in 6 hr. To the nearest minute, 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 Adam and Bruce work together. Since trimming the shrubs at Centralia 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‘π‘π‘ππ
=ππ
π»π»π‘π‘π‘π‘ππ.
To organize the information, we can complete the table below.
π π β π»π» = J worker I worker II together
/β ππ(ππ + 5)
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
Section RT6 | 293
Applications of Rational Equation
Since the rate of work when both Adam and Bruce trim the shrubs is the sum of rates of individual workers, we form and solve the equation
18
+16
=1ππ
3ππ + 4ππ = 24
7ππ = 24
ππ =247β 3.43
So, if both Adam and Bruce work together, the amount of time needed to complete the job if approximately 3.43 hours β 3 hours 26 minutes. Note: The time needed for both workers is shorter than either of the individual times.
Solving a Work Problem Involving Subtraction of Rates
The inlet pipe can fill a swimming pool in 4 hours, while the outlet pipe can empty the pool in 5 hours. If both pipes were left open, how long would it take to fill the pool? 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.
π π β π»π» = π±π±
Adam ππππ
8 1
Bruce ππππ
6 1
together πππΉπΉ
ππ 1
π π β π»π» = π±π±
inlet pipe ππππ
4 1
outlet pipe βππππ
5 β1
both pipes πππΉπΉ
ππ 1
/β 24ππ
/Γ· 7
Solution
To complete the Rate column, we divide the
Job by the Time.
The job column is often equal to 1, although
sometimes other values might need to be used.
294 | Section RT6
Rational Expressions and Functions
The equation to solve comes from the Rate column.
14β
15
=1ππ
5ππ β 4ππ = 20
ππ = 20
So, it will take 20 hours to fill the pool when both pipes are left open.
Inverse and Combined Variation
When two quantities vary in such a way that their product remains constant, we say that they are inversely proportional. For example, consider rate π π and time ππ of a moving object that covers a constant distance π·π·. In particular, if π·π· = 100 km, we have
π π = 100ππ
= 100 β 1ππ
This relation tells us that the rate is 100 times larger than the reciprocal of time. Observe though that when the time doubles, the rate is half as large. When the time triples, the rate is three times smaller, and so on. One can observe that the rate decreases proportionally to the increase of time. Such a reciprocal relation between the two quantities is called an inverse variation.
Definition 6.1 Two quantities, ππ and ππ, are inversely proportional to each other (there is an inverse variation between them) iff there is a real constant ππ β ππ, such that
ππ =ππππ
.
We say that ππ varies inversely as ππ with the variation constant ππ. (or equivalently: ππ is inversely proportional to ππ with the proportionality constant ππ.)
Solving Inverse Variation Problems
The volume ππ of a gas is inversely proportional to the pressure ππ of the gas. If a pressure of 30 kg/cm2 corresponds to a volume of 240 cm3, find the following: a. The equation that relates ππ and ππ, b. The pressure needed to produce a volume of 150 cm3. a. To find the inverse variation equation that relates ππ and ππ, we need to find the variation
constant ππ first. This can be done by substituting ππ = 240 and ππ = 30 into the equation ππ = ππ
ππ. So, we obtain
240 =ππ
30
ππ = 7200.
Therefore, our equation is π½π½ = πππππππππ·π·
.
/β 20ππ
Solution
/ β 30
Time in hours
Rate
in k
m/h
50
π π (ππ)
ππ
100
3
200
150
5 1
250 π π (ππ) =
100ππ
Section RT6 | 295
Applications of Rational Equation
βswapβ 150 and ππ
b. The required pressure can be found by substituting ππ = 150 into the inverse variation
equation,
150 =7200ππ
.
This gives us
ππ =7200150
= 48.
So, the pressure of the gas that assumes the volume of 150 cm3 is 48 kg/cm2.
Extension: We say that ππ varies inversely as the ππ-th power of ππ iff ππ = ππππππ
, for some nonzero constant ππ.
Solving an Inverse Variation Problem Involving the Square of a Variable
The intensity of light varies inversely as the square of the distance from the light source. If 4 meters from the source the intensity of light is 9 candelas, what is the intensity of this light 3 meters from the source? Let πΌπΌ represents the intensity of the light and ππ the distance from the source of this light. Since πΌπΌ varies inversely as ππ2, we set the equation
πΌπΌ =ππππ2
After substituting the data given in the problem, we find the value of ππ:
9 =ππ42
ππ = 9 β 16 = 144
So, the inverse variation equation is πΌπΌ = 144ππ2
. Hence, the light intensity at 3 meters from the
source is πΌπΌ = 14432
= ππππ candelas.
Recall from Section L2 that two variables, say ππ and ππ, vary directly with a proportionality constant ππ β ππ if ππ = ππππ. Also, we say that one variable, say ππ, varies jointly as other variables, say ππ and ππ, with a proportionality constant ππ β ππ if ππ = ππππππ.
Definition 6.2 A combination of the direct or joint variation with the inverse variation is called a combined variation.
Example: ππ may vary jointly as ππ and ππ and inversely as the square of ππ. This means that there is a
real constant ππ β ππ, such that
ππ =ππππππππππ
.
Solution
/ β 16
/ β ππ, Γ· 150
296 | Section RT6
Rational Expressions and Functions
Solving Combined Variation Problems
The resistance of a cable varies directly as its length and inversely as the square of its diameter. A 20-meter cable with a diameter of 1.2 cm has a resistance of 0.2 ohms. A 50-meter cable with a diameter of 0.6 cm is made out of the same material. What would be its resistance? Let π π , ππ, and ππ represent respectively the resistance, length, and diameter of a cable. Since π π varies directly as ππ and inversely as ππ2, we set the combined variation equation
π π =ππππππ2
.
Substituting the data given in the problem, we have
0.2 =ππ β 201.22
,
which gives us
ππ =0.2 β 1.44
20= 0.0144
So, the combined variation equation is π π = 0.0144ππππ2
. Therefore, the resistance of a 50-meter
cable with the diameter of 0.6 cm is π π = 0.0144β500.62
= ππ ohms.
RT.6 Exercises
1. Using the formula 1ππ
= 1ππ
+ 1ππ
, find ππ if ππ = 6 and ππ = 10.
2. The gravitational force between two masses is given by the formula πΉπΉ = πΊπΊπΊπΊππππ2
. Find πΏπΏ if πΉπΉ = 20, πΊπΊ = 6.67 β 10β11, π‘π‘ = 1, and ππ = 4 β 10β6. Round your answer to one decimal
place.
3. What is the first step in solving the formula ππππ + ππππ = ππ β ππ for ππ?
4. What is the first step in solving the formula π΄π΄ = ππππππβππ
for ππ?
Solve each formula for the specified variable.
5. π‘π‘ = πΉπΉππ for ππ 6. πΌπΌ = πΈπΈ
π π for π π 7. ππ1
ππ2= ππ1
ππ2 for ππ1
8. πΉπΉ = πΊπΊπΊπΊππππ2
for π‘π‘ 9. π π = (π£π£1+π£π£2)π‘π‘2
for ππ 10. π π = (π£π£1+π£π£2)π‘π‘2
for π£π£1
Solution
/ β 1.44, Γ· 20
Section RT6 | 297
Applications of Rational Equation
11. 1π π
= 1ππ1
+ 1ππ2
for π π 12. 1π π
= 1ππ1
+ 1ππ2
for ππ1 13. 1ππ
+ 1ππ
= 1ππ for ππ
14. π‘π‘ππ
+ π‘π‘ππ
= 1 for ππ 15. ππππππ
= πππ£π£π‘π‘
for π£π£ 16. ππππππ
= πππ£π£π‘π‘
for ππ
17. π΄π΄ = β(ππ+ππ)2
for ππ 18. ππ = ππβπ£π£π‘π‘
for ππ 19. π π = πππ π ππ+π π
for π π
20. πΌπΌ = 2ππππ+2ππ
for ππ 21. πΌπΌ = πππΈπΈπΈπΈ+ππππ
for ππ 22. πΈπΈππ
= π π +ππππ
for ππ
23. πΈπΈππ
= π π +ππππ
for ππ 24. ππ = π»π»ππ(π‘π‘1βπ‘π‘2) for ππ1 25. ππ = ππβ2(3π π ββ)
3 for π π
26. ππ = π΄π΄1+ππ
for ππ 27. ππ2
π π 2= 2ππ
π π +β for β 28. π£π£ = ππ2βππ1
π‘π‘2βπ‘π‘1 for ππ2
Solve each problem.
29. The ratio of the weight of an object on Earth to the weight of an object on the moon is 200 to 33. What would be the weight of a 75-kg astronaut on the moon?
30. A 30-meter long ribbon is cut into two sections. How long are the two sections if the ratio of their lengths is 5 to 7?
31. Assume that burning 7700 calories causes a decrease of 1 kilogram in body mass. If walking 7 kilometers in 2 hours burns 700 calories, how many kilometers would a person need to walk at the same rate to lose 1 kg?
32. On a map of Canada, the linear distance between Vancouver and Calgary is 1.8 cm. The flight 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 flight distance between the two cities is approximately 3000 kilometers?
33. To estimate the population of Cape Mountain Zebra in South Africa, biologists caught, tagged, and then released 68 Cape Mountain Zebras. In a month, they caught a random sample of 84 of this type of zebras. It turned out that 5 of them were tagged. Assuming that zebras mixed freely, approximately how many Cape Mountain Zebras lived in South Africa?
34. To estimate the number of white bass fish in a particular lake, biologists caught, tagged, and then released 300 of this fish. In two weeks, they returned and collected a random sample of 196 white bass fish. This sample contained 12 previously tagged fish. Approximately how many white bass fish does the lake have?
35. Eighteen white-tailed eagles are tagged and released into the wilderness. In a few weeks, a sample of 43 white-tailed eagles was examined, and 5 of them were tagged. Estimate the white-tailed eagle population in this wilderness area.
36. A meter stick casts a 64 cm long shadow. At the same time, a 15-year old cottonwood tree casts an 18-meter long shadow. To the nearest meter, how tall is the tree?
37. The ratio of corresponding sides of similar triangles is 5 to 3. The two shorter sides of the larger triangle are 5 and 7 units long, correspondingly. Find the length of each side of the smaller triangle if its longest side is 4 units shorter than the corresponding side of the larger triangle. P
R S
π₯π₯ β 4
7
A
B C 5
π₯π₯
298 | Section RT6
Rational Expressions and Functions
38. The width of a rectangle is the same as the length of a similar rectangle. If the dimensions of the smaller rectangle are 7 cm by 12 cm, what are the dimensions of the larger rectangle?
39. Justin runs twice around a park. He averages 20 kilometers per hour during the first round and only 16 kilometers per hour during the second round. What is his average speed for the whole run? Round your answer to one decimal place.
40. Robert runs twice around a stadium. He averages 18 km/h during the first round. What should his average speed be during the second round to have an overall average of 20 km/h for the whole run?
41. Jimβs boat moves at 20 km/h in still water. Suppose it takes the same amount of time for Jim to travel by his boat either 15 km downriver or 10 km upriver. Find the rate of the current.
42. The average speed of a plane flying west was 880 km/h. On the return trip, the same plane averaged only 620 km/h. If the total flying time in both directions was 6 hours, what was the one-way distance?
43. A plane flies 3800 kilometers with the wind, while only 3400 kilometers against the same wind. If the airplane speed in still air is 900 km/h, find the speed of the wind.
44. Walking on a moving sidewalk, Sarah could travel 40 meters forward in the same time it would take her to travel 15 meters in the opposite direction. If the rate of the moving sidewalk was 35 m/min, what was Sarah's rate of walking?
45. Arthur travelled by car from Madrid to Paris. He usually averages 100 km/h on such trips. This time, due to heavier traffic and few stops, he averaged only 85 km/h, and he reached his destination 2 hours 15 minutes later than expected. How far did Arthur travel?
46. Tony averaged 100 km/h on the first part of his trip to Lillooet, BC. The second part of his trip was 20 kilometers longer than the first, and his average speed was only 80 km/h. If the second part of the trip took him 30 minutes longer than the first part, what was the overall distance travelled by Tony?
47. Page is a college student who lives in a near-campus apartment. When she rides her bike to campus, she gets there 24 min faster than when she walks. If her average walking rate is 4 km/h and her average biking rate is 20 km/h, how far does she live from the campus?
48. Sonia can respond to all the daily e-mails in 2 hours. Betty needs 3 hours to do the same job. If they both work on responding to e-mails, what portion of this daily job can be done in 1 hour? How much more time would they need to complete the job?
49. Brenda can paint a deck in π₯π₯ hours, while Tony can do the same job in π¦π¦ hours. Write a rational expression that represents the portion of the deck that can be painted by both of them in 4 hours.
50. Aaron and Ben plan to paint a house. Aaron needs 24 hours to paint the house by himself. Ben needs 18 to do the same job. To the nearest minute, how long would it take them to paint the house if they work together?
51. When working together, Adam and Brian can paint a house in 6 hours. Brian could paint this house on his own in 10 hours. How long would it take Adam to paint the house working alone?
Section RT6 | 299
Applications of Rational Equation
52. An experienced floor installer can install a parquet floor twice as fast as an apprentice. Working together, it takes the two workers 2 days to install the floor in a particular house. How long would it take the apprentice to do the same job on his own?
53. A pool can be filled in 8 hr and drained in 12 hr. On one occasion, when filling the pool, the drain was accidentally left open. How long did it take to fill this pool?
54. One inlet pipe can fill a hot tub in 15 minutes. Another inlet pipe can fill the tub in 10 minutes. An outlet pipe can drain the hot tub in 18 minutes. How long would it take to fill the hot tub if all three pipes are left open?
55. Two different width escalators can empty a 1470-people auditorium in 12 min. If the wider escalator can move twice as many people as the narrower one, how many people per hour can the narrower escalator move?
56. At what times between 3:00 and 4:00 are the minute and hour hands
perfectly lined up?
57. If Miranda drives to work at an average speed of 60 km/h, she is 1 min late. When she drives at an average speed of 75 km/h, she is 3 min early. How far is Miranda's workplace from her home?
58. The current in an electrical circuit at a constant potential varies inversely as the resistance of the circuit. Suppose that the current πΌπΌ is 9 amperes when the resistance π π is 10 ohms. Find the current when the resistance is 6 ohms.
59. Assuming the same rate of work for all workers, the number of workers needed for a job varies inversely as the time required to complete the job. If it takes 3 hours for 8 workers to build a deck, how long would it take two workers to build the same deck?
60. The length of a guitar string is inversely proportional to the frequency of the string vibrations. Suppose a 60-cm long string vibrates at a frequency of 500 Hz (1 hertz = one cycle per second). What is the frequency of the same string when it is shortened to 50 centimeters?
61. A musical toneβs pitch is inversely proportional to its wavelength. If a wavelength of 2.2 meters corresponds to a pitch of 420 vibrations per second, find the wavelength of a tone with a pitch of 660 vibrations per second.
62. The intensity, πΌπΌ, of a television signal is inversely proportional to the square of the distance, ππ, from a transmitter. If 2 km away from the transmitter the intensity is 25 W/m2 (watts per square meter), how far from the transmitter is a TV set that receives a signal with the intensity of 2.56 W/m2?
63. The weight ππ of an object is inversely proportional to the square of the distance π·π· from the center of Earth. To the nearest kilometer, how high above the surface of Earth must a 60-kg astronaut be to weigh half as much? Assume the radius of Earth to be 6400 km.
64. The number of long-distance phone calls between two cities during a specified period in time varies jointly as the populations of the cities, ππ1 and ππ2, and inversely as the distance between them. Suppose 80,000 calls are made between two cities that are 400 km apart and have populations of 70,000 and 100,000. How many calls are made between Vancouver and Abbotsford that are 70 km apart and have populations of 630,000 and 140,000, respectively?
300 | Section RT6
Rational Expressions and Functions
65. The force that keeps a car from skidding on a curve is inversely proportional to the radius of the curve and jointly proportional to the weight of the car and the square of its speed. Knowing that a force of 880 N (Newtons) keeps an 800-kg car moving at 50 km/h from skidding on a curve of radius 160 m, estimate the force that would keep the same car moving at 80 km/h from skidding on a curve of radius 200 meters.
66. Suppose that the renovation time is inversely proportional to the number of workers hired for the job. Will
the renovation time decrease more when hiring additional 2 workers in a 4-worker company or a 6-worker company? Justify your answer.
Attributions
p.246 Venice Beach at Sunset by Austin Dixon / Unsplash Licence p.250 Earth to Sun by LucasVB / public domain p.266 Aerial view of Columbia River and Bonneville Dam by U.S. Army Corps of Engineers / Public Domain; Close-up of Moon by
Martin Adams / Unsplash Licence p.269 Architecture with Round Balcomies by Chuttersnap / Unsplash Licence
p.274 Snowflake Hunting by Aaron Burden / Unsplash Licence p.281 Railway superelevation at Dunbar, July 2012 by Calvinps / Public Domain p.284 Blur Cigar Cigarette by Irina Iriser / Pexels Licence p.285 Fiddlehead by lisaleo / Morguefile Licence p.292 Buxus by Ellen26 / Pixabay Licence p.293 FTF Demo by Greg Leaman / Unsplash Licence p.295 Candle by Gadini / CC0 1.0 Universal (CC0 1.0) Public Domain Dedication p.298 A Close-up Picture of Boat on Water by Tamara Mills on Pixnio/ public domain (CC0); Three Persons Standing on Escalator by
Negative Space / Pexels Licence; Staining by - Jaco - Jahluka / CC BY-ND 2.0 p.299 Escalation by Christian DeKnock /Unsplash Licence; Big Ben by Michael Jin / Unsplash Licence
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