Explanation Given: 5 ( x + 1 ) ( x + 5 ) − 2 ( x + 5 ) 2 Concept Used: To add/subtract rational expressions with unlike denominator 1. find LCD(least common divisor) of denominators 2. convert each rational expressions as an equivalent fraction whose denominators the LCD 3. Add/subtract the like rational expressions To simplify a given expression, we have to find the common factor of the numerator and denominator and then divide them by the common factor. Calculation: To solve 5 ( x + 1 ) ( x + 5 ) − 2 ( x + 5 ) 2 , first find the LCD (least common denominator) for each rational expressions, now as the LCD of denominator ( ( x + 1 ) ( x + 5 ) and ( x + 5 ) 2 ) is ( x + 5 ) 2 ( x + 1 ) , thus write each rational expression as an equivalent rational expression whose denominator is ( x + 5 ) 2 ( x + 1 ) , thus 5 ( x + 1 ) ( x + 5 ) ⋅ x + 5 x + 5 = 5 ( x + 5 ) ( x + 1 ) ( x + 5 ) 2 , 2 ( x + 5 ) 2 ⋅ x + 1 x + 1 = 2 ( x + 1 ) ( x + 1 ) ( x + 5 ) 2

Algebra Foundations: Prealgebra, Introductory Algebra, & Intermediate Algebra - 18 Week Standalone Access Card

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ISBN: 9780134582788

Author: Elayn Martin-Gay

Publisher: PEARSON

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Chapter 14.4, Problem 42ES

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Solve 5(x+1)(x+5)−2( x+5)2 , and simplify the result if possible

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Chapter 14.4, Problem 41ES

Chapter 14.4, Problem 43ES

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Chapter 14 Solutions

Algebra Foundations: Prealgebra, Introductory Algebra, & Intermediate Algebra - 18 Week Standalone Access Card

Ch. 14.1 - Use the choices below to fill in each blank. Some...Ch. 14.1 - Prob. 2VRVC Ch. 14.1 - Prob. 3VRVC Ch. 14.1 - Prob. 4VRVC Ch. 14.1 - Prob. 5VRVC Ch. 14.1 - Prob. 6VRVC Ch. 14.1 - Prob. 7VRVC Ch. 14.1 - Prob. 8VRVC Ch. 14.1 - Prob. 1ES Ch. 14.1 - Prob. 2ES

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Solve 5 ( x + 1 ) ( x + 5 ) − 2 ( x + 5 ) 2 , and simplify the result if possible

Solution Summary: The author explains how to add/subtract rational expressions with unlike denominator.

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Solve 5(x+1)(x+5)−2( x+5)2 , and simplify the result if possible

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Chapter 14 Solutions