Explanation Procedure used: Addition or Subtraction of Rational Numbers (Fraction) with unlike denominators: 1. Compute the LCD of the denominators. 2. Write fraction with the LCD as the denominator. 3. Add or subtract the numerators and write the result over the common denominator. 4. Write the result in lowest terms. “ Finding the lowest common denominator ( LCD ) : 1. Find the prime factors of each denominator. 2. For each different prime factor that appears, note the highest power to which it is raised in any one of the denominators. 3. Form the product of all the different prime factors, each raised to the power found in step 2. This product is called the lowest common denominator.” Calculation: The given expression is 3 x 2 x 2 − 2 − 2 4 x 2 − 5 x + 1 . Factor the denominators of the above expression as follows. 3 x 2 x 2 − 2 − 2 4 x 2 − 5 x + 1 = 3 x 2 ( x 2 − 1 ) − 2 4 x 2 − 4 x − x + 1 = 3 x 2 ( x + 1 ) ( x − 1 ) − 2 4 x ( x − 1 ) − ( x − 1 ) = 3 x 2 ( x + 1 ) ( x − 1 ) − 2 ( x − 1 ) ( 4 x − 1 ) Note that, the denominators of the above fractional expression are different. To make the denominators of the fractions equivalent, find the LCD of the denominators. The factors of 2 x 2 − 2 are 2 , ( x + 1 ) and ( x − 1 ) . The factors of 4 x 2 − 5 x + 1 are ( x − 1 ) and ( 4 x − 1 ) . Therefore, 2 , ( x − 1 ) , ( x + 1 ) and ( 4 x − 1 ) are the factors of the LCD. Thus, the LCD of the denominators of each fractions is the product of 2 , ( x − 1 ) , ( x + 1 ) and ( 4 x − 1 ) . That is, 2 ( x − 1 ) ( x + 1 ) ( 4 x − 1 ) . To make the resulting denominator in each fraction equal to the LCD, multiply and divide the fraction by the proper quantity

Basic Technical Mathematics with Calculus (11th Edition)

11th Edition

ISBN: 9780134437736

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 6, Problem 67RE

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To perform: The indicated operations in the expression 3x2x2−2−24x2−5x+1 and simplify it.

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Chapter 6, Problem 66RE

Chapter 6, Problem 68RE

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Chapter 6 Solutions

Basic Technical Mathematics with Calculus (11th Edition)

Ch. 6.1 - Prob. 1PE Ch. 6.1 - Prob. 2PE Ch. 6.1 - Prob. 3PE Ch. 6.1 - Prob. 4PE Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E

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To perform: The indicated operations in the expression 3 x 2 x 2 − 2 − 2 4 x 2 − 5 x + 1 and simplify it.

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To perform: The indicated operations in the expression 3x2x2−2−24x2−5x+1 and simplify it.

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Chapter 6 Solutions