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

The rational equation consists of three fractions. The lowest common denominator for these three fractions is ( 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 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} } } {}

We can simplify things on a term by term basis

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

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

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.

Our answer is therefore 3.

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.

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Source:  OpenStax, Math 1508 (laboratory) engineering applications of precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11337/1.3
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