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# Rational Expressions and Equations

Module by: Scott Starks. E-mail the author

Summary: This module is part of collection of modules developed for use by students enrolled in a special section of MATH 1508 (PreCalculus) for preengineers. This module introduces several engineering applications which are based upon rational expressions and equations.

## Rational Expressions and Equations

### Introduction

We have seen that much of the analysis that it takes to simplify rational expressions and to solve rational equations is an outgrowth of the mathematics associated with fractions. In most cases, one can begin to solve a problem involving a rational equation by factoring the polynomials that are constituents of the rational equation. Then, one may seek to identify if there are any common factors in the numerator and denominator that cancel with one another. If there are, cancelling these terms will often make our job of solving the equation simpler.

The fields of science and engineering are filled with rational equations that describe many different application areas. In this section of notes, we will focus on how rational expressions and equations are solved in several different applications. Before we do so, let us examine an example that involves the manipulation of rational expressions to solve a numerical problem involving positive integers.

### Numerical Problem Involving Rational Expressions

Question: One positive integer is 3 more than another positive integer. When the reciprocal of the smaller integer is added to the reciprocal of the larger integer, the resulting sum is ½. Find the positive integer.

As a reminder the reciprocal of an integer is 1 divided by the integer. For example, the reciprocal of 10 is 1/10.

Solution: We begin the solution by defining the variable x.

Let x = smaller positive integer

With this definition of x, we know that the larger positive integer can be expressed algebraically as (x + 3).

Using the definition of the reciprocal, we can translate the problem statement into a rational equation

1 x + 1 x + 3 = 1 2 1 x + 1 x + 3 = 1 2 size 12{ left ( { {1} over {x} } right )+ left ( { {1} over {x+3} } right )= { {1} over {2} } } {}
(1)

The rational equation consists of three fractions. The lowest common denominator for these three fractions is (x)(x+3)(2).(x)(x+3)(2). size 12{ $$x$$  $$x+3$$  $$2$$ "." } {} We can multiply each side of the rational equation by the lowest common denominator to obtain the equation

( 2 ) ( x ) ( x + 3 ) 1 x + 1 x + 3 = ( 2 ) ( x ) ( x + 3 ) 1 2 ( 2 ) ( x ) ( x + 3 ) 1 x + 1 x + 3 = ( 2 ) ( x ) ( x + 3 ) 1 2 size 12{ $$2$$ $$x$$ $$x+3$$  left lbrace left ( { {1} over {x} } right )+ left ( { {1} over {x+3} } right ) right rbrace = $$2$$ $$x$$ $$x+3$$  { {1} over {2} } } {}
(2)

We can simplify things on a term by term basis

2 ( x + 3 ) + 2 ( x ) = x ( x + 3 ) 2 ( x + 3 ) + 2 ( x ) = x ( x + 3 ) size 12{2 $$x+3$$ +2 $$x$$ =x $$x+3$$ } {}
(3)
2x + 6 + 2x = x 2 + 3x 2x + 6 + 2x = x 2 + 3x size 12{2x+6+2x=x rSup { size 8{2} } +3x} {}
(4)
4x + 6 = x 2 + 3x 4x + 6 = x 2 + 3x size 12{4x+6=x rSup { size 8{2} } +3x} {}
(5)
x 2 x 6 = 0 x 2 x 6 = 0 size 12{x rSup { size 8{2} } - x - 6=0} {}
(6)

This quadratic equation can be solved a variety of ways. One simple way to do so is to employ factoring.

( x 3 ) ( x + 2 ) = 0 ( x 3 ) ( x + 2 ) = 0 size 12{ $$x - 3$$ $$x+2$$ =0} {}
(7)

The roots are therefore x = 3 and x = -2.

From the problem statement, we know that the solution for x must be positive. We therefore can ignore the root of -2.

### Rate Applications

Engineering and science abound with problems that make use of rates. Examples of rate include concepts such as speed which is a measure of the rate of change of distance per unit of time. Flow rates are common in fluid mechanics problems. The flow rate of liquid in a pipe can be expressed as a number of liters of liquid per unit of time.

An important consideration in working rate problems is to recognize that a rate can be written as a ratio of the quantity of a particular entity over time. The following two examples illustrate the use of rates to solve engineering problems.

### Computing the Time It Takes to Finish a Construction Project

Civil engineers perform important duties in the management of construction projects. Suppose that a civil engineer is charged with the the completion of a construction project. The civil engineer knows that if he (she) hires Subcontractor A to complete the construction project, it will take 10 months. On the other hand, if Subcontractor B is hired, the construction project can be completed in 15 months

Instead of hiring just one of the subcontractor, the civil engineer decides to hire both Subcontractor A and Subcontractor B. By doing so, the civil engineer expects the project to be completed more quickly.

Question: How long will it take both subcontractors working together to complete the construction project?

Let us begin by considering Subcontractor A. Working alone, it would take Subcontractor A 10 months to complete the project. If we consider the completion of the construction project as a job, then Subcontractor A can complete (1/10) of the job in one month. Another way to look at this is that the rate at which Subcontractor A can complete the job is as (1/10) job/month.

Now, let us look at Subcontractor B. Working alone, it would take Subcontractor B 15 months to complete the construction project. Thus Subcontractor B can complete (1/15) of job in one month. The rate of Subcontractor B can be stated as (1/15) job/month

Let us define the variable t

t = the time that it takes to complete the entire job using both of the subcontractors,

where t is measured in months.

In a length of time equal to t, the fraction of the job that Subcontractor A can complete is equal to the product

1 10 t = t 10 1 10 t = t 10 size 12{ left ( { {1} over {"10"} } right )t= { {t} over {"10"} } } {}
(8)

Similarly the fraction of the job that Subcontractor B can complete in the same length of time is

1 15 t = t 15 1 15 t = t 15 size 12{ left ( { {1} over {"15"} } right )t= { {t} over {"15"} } } {}
(9)

We may obtain the fraction of the job that can be completed in an interval of time equal to t by adding the fraction associated with the Subcontractor A to that of Subcontractor B, which is t10+t15t10+t15 size 12{ { {t} over {"10"} } + { {t} over {"15"} } } {}{}

If we set this sum to one, we obtain a rational equation that can be solved to produce the amount of time that it will take both subcontractors working together to complete the entire job

t 10 + t 15 = 1 t 10 + t 15 = 1 size 12{ { {t} over {"10"} } + { {t} over {"15"} } =1} {}
(10)

The solution follows

15 t 150 + 10 t 150 = 1 15 t 150 + 10 t 150 = 1 size 12{ { {"15"t} over {"150"} } + { {"10"t} over {"150"} } =1} {}
(11)
150 15 t + 10 t 150 = 150 ( 1 ) 150 15 t + 10 t 150 = 150 ( 1 ) size 12{"150" left ( { {"15"t+"10"t} over {"150"} } right )="150" $$1$$ } {}
(12)

Simplification of this equation yields

25 t = 150 25 t = 150 size 12{"25"t="150"} {}
(13)

From this, we conclude that the two subcontractors working together can complete the job in 6 months.

### Motion of an Automobile Using Two Different Speeds

Question: Suppose that you are interested in taking an automobile trip to visit a friend in Van Horn, Texas. On your journey, you will travel 40 miles inside the City of El Paso. Once you leave the city limits of El Paso, you will have to travel an additional 90 miles to reach Van Horn. Because of traffic, you anticipate that your average speed within the city limits of El Paso will be only half of the average speed once you leave the city limits. You also project that the total time of your trip from El Paso to Van Horn will be 2.5 hours,

What is your average speed within the city limits of El Paso? What would be your average velocity outside the city limits?

Solution: We begin by defining the following variable

x = average speed (miles/hr) within in the city limits of El Paso.

The speed at which you would drive outside the city limits would be twice that of the velocity within the city limits. Thus

2 x = average velocity (miles/hr) outside the city limits of El Paso.

The amount of time that you would spend driving in El Paso would be

40 miles x 40 miles x size 12{ { {"40" ital "miles"} over {x} } } {}
(14)

The amount of time that you would spend driving outside the city limits would be

90 miles 2 x 90 miles 2 x size 12{ { {"90" ital "miles"} over {2x} } } {}
(15)

If we sum these together we obtain the total time of the trip

40 miles x + 90 miles 2 x = 2 . 5 hr 40 miles x + 90 miles 2 x = 2 . 5 hr size 12{ { {"40" ital "miles"} over {x} } + { {"90" ital "miles"} over {2x} } =2 "." 5 ital "hr"} {}
(16)

This rational equation can be solved as follows

2 ( 40 ) miles 2 ( x ) + 90 miles 2 x = 2 . 5 hr 2 ( 40 ) miles 2 ( x ) + 90 miles 2 x = 2 . 5 hr size 12{ { {2 $$"40"$$  ital "miles"} over {2 $$x$$ } } + { {"90" ital "miles"} over {2x} } =2 "." 5 ital "hr"} {}
(17)

which can be simplified as

170 miles = 5 x hr 170 miles = 5 x hr size 12{"170" ital "miles"=5x ital "hr"} {}
(18)

We solve this for x to obtain the result of 34 miles/hr. This represents the average speed within the city limits of El Paso. Outside the city limits would be twice that within the city limits or 68 miles/hr.

### Exercises

1. The difference between the reciprocals of two consecutive positive odd integers is 2/15. Find the two positive odd integers.

2. Two pipes (A and B) attach to the top of a holding tank that is part of a waste water processing system. The pipes are of different size. If liquid is supplied to the tank by Pipe A only, it takes 12 minutes to fill the tank. Pipe B has a smaller cross-sectional area than that of pipe A. If liquid is supplied to the tank using Pipe B only, it takes 15 minutes to fill the tank.

A civil engineer is tasked with managing this process. In order to minimize the time needed to fill the tank, the civil engineer decides to supply liquid to the tank via both Pipe A and Pipe B. If this is done, how long will it take to fill the tank?

3. The first leg of Mary’s Spring Break trip consisted of 120 miles of traffic. When the traffic cleared she was able to drive twice as fast for 300 miles. If the total trip took 9 hours, how long was Mary stuck in traffic?

4. Three resistors are connected in parallel. There individual values are 10 Ω, 20 Ω and 50 Ω. What is the equivalent resistance of the parallel combination?

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