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<para>This module is from<link document="col10614">Elementary Algebra</link>by Denny Burzynski and Wade Ellis, Jr.</para><para>A detailed study of arithmetic operations with rational expressions is presented in this chapter, beginning with the definition of a rational expression and then proceeding immediately to a discussion of the domain. The process of reducing a rational expression and illustrations of multiplying, dividing, adding, and subtracting rational expressions are also included. Since the operations of addition and subtraction can cause the most difficulty, they are given particular attention. We have tried to make the written explanation of the examples clearer by using a "freeze frame" approach, which walks the student through the operation step by step.</para><para>The five-step method of solving applied problems is included in this chapter to show the problem-solving approach to number problems, work problems, and geometry problems. The chapter also illustrates simplification of complex rational expressions, using the combine-divide method and the LCD-multiply-divide method.</para><para>Objectives of this module: be able to use the five-step method to solve various applied problems.</para>

Overview

  • The Five-Step Method

The five-step method

We are now in a position to study some applications of rational equations. Some of these problems will have practical applications while others are intended as logic developers.

We will apply the five-step method for solving word problems.

Five-step method

  1. Represent all unknown quantities in terms of x or some other letter.
  2. Translate the verbal phrases to mathematical symbols and form an equation.
  3. Solve this equation.
  4. Check the solution by substituting the result into the original statement of the problem.
  5. Write the conclusion.

Remember, step 1 is very important: always

Introduce a variable.

Sample set a

When the same number is added to the numerator and denominator of the fraction 3 5 , the result is 7 9 . What is the number that is added?

Step 1: Let  x = the number being added. Step 2: 3 + x 5 + x = 7 9 . Step 3: 3 + x 5 + x = 7 9 . An excluded value is  5 . The LCD is 9 ( 5 + x ) .  Multiply each term by 9 ( 5 + x ) . 9 ( 5 + x ) · 3 + x 5 + x = 9 ( 5 + x ) · 7 9 9 ( 3 + x ) = 7 ( 5 + x ) 27 + 9 x = 35 + 7 x 2 x = 8 x = 4 Check this potential solution . Step 4: 3 + 4 5 + 4 = 7 9 . Yes, this is correct. Step 5: The number added is 4 .

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Practice set a

The same number is added to the numerator and denominator of the fraction 4 9 . The result is 2 3 . What is the number that is added?

Step 1:   Let x =

Step 2:


Step 3:





Step 4:



Step 5:   The number added is .

The number added is 6.

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Sample set b

Two thirds of a number added to the reciprocal of the number yields 25 6 . What is the number?

Step 1:    Let x = the number.

Step 2:    Recall that the reciprocal of a number x is the number 1 x .

       2 3 · x + 1 x = 25 6

Step 3: 2 3 · x + 1 x = 25 6 The LCD is 6 x . Multiply each term by 6 x . 6 x · 2 3 x + 6 x · 1 x = 6 x · 25 6 4 x 2 + 6 = 25 x Solve this nonfractional quadratic equation to obtain the potential solutions . (Use the zero-factor property .) 4 x 2 25 x + 6 = 0 ( 4 x 1 ) ( x 6 ) = 0 x = 1 4 , 6 Check these potential solutions .

Step 4:    Substituting into the original equation, it can be that both solutions check.

Step 5:    There are two solutions: 1 4 and 6.

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Source:  OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3
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