Explanation Given information: The provided equation is, x x 2 + 3 x − 4 + x + 1 x 2 + 6 x + 8 = 2 x x 2 + x − 2 Formula used: To find the least common multiple or LCM, factor each term completely and then use each factor the greatest number of times it occurred. The difference of squares of two numbers: For any real number a and b , a 2 − b 2 = ( a + b ) ( a − b ) Calculation: Consider the provided equation x x 2 + 3 x − 4 + x + 1 x 2 + 6 x + 8 = 2 x x 2 + x − 2 . The equation to solve as below, x x 2 + 3 x − 4 + x + 1 x 2 + 6 x + 8 = 2 x x 2 + x − 2 Factorize all the denominators by splitting the terms as below, x x 2 + 3 x − 4 + x + 1 x 2 + 6 x + 8 = 2 x x 2 + x − 2 x x 2 + 4 x − x − 4 + x + 1 x 2 + 4 x + 2 x + 8 = 2 x x 2 + 2 x − x − 2 x x ( x + 4 ) − 1 ( x + 4 ) + x + 1 x ( x + 4 ) + 2 ( x + 4 ) = 2 x x ( x + 2 ) − 1 ( x + 2 ) x ( x + 4 ) ( x − 1 ) + x + 1 ( x + 4 ) ( x + 2 ) = 2 x ( x + 2 ) ( x − 1 ) Since the denominators in the equation are different, make them same by taking least common multiple. The least common multiple of the denominators is: ( x + 4 ) ( x − 1 ) ( x + 2 ) Multiply by least common multiple = ( x + 4 ) ( x − 1 ) ( x + 2 ) on both sides and solve as below, ( x ( x + 4 ) ( x − 1 ) + x + 1 ( x + 4 ) ( x + 2 ) ) = 2 x ( x + 2 ) ( x − 1 ) ( x ( x + 4 ) ( x − 1 ) + x + 1 ( x + 4 ) ( x + 2 ) ) × ( x + 4 ) ( x − 1 ) ( x + 2 ) = 2 x ( x + 2 ) ( x − 1 ) × ( x + 4 ) ( x − 1 ) ( x + 2 ) Multiply with each term by opening the bracket and simplify by using the property the difference of squares of two numbers a 2 − b 2 = ( a + b ) ( a − b ) then further simplify, x ( x + 4 ) ( x − 1

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ISBN: 9780321997173

Author: Marvin L. Bittinger, Judith A. Beecher

Publisher: PEARSON

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Chapter 12.7, Problem 57ES

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To calculate: The value of x form the equation xx2+3x−4+x+1x2+6x+8=2xx2+x−2.

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Chapter 12.7, Problem 56ES

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Chapter 12 Solutions

Developmental Mathematics (9th Edition)

Ch. 12.1 - Objective 11.3a: Factor trinomials of the type...Ch. 12.1 - Prob. 2SR Ch. 12.1 - Prob. 1DE Ch. 12.1 - Prob. 2DE Ch. 12.1 - Prob. 3DE Ch. 12.1 - Prob. 4DE Ch. 12.1 - Prob. 5DE Ch. 12.1 - Prob. 6DE Ch. 12.1 - Prob. 7DE Ch. 12.1 - Prob. 8DE

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The value of x form the equation x x 2 + 3 x − 4 + x + 1 x 2 + 6 x + 8 = 2 x x 2 + x − 2 .

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To calculate: The value of x form the equation xx2+3x−4+x+1x2+6x+8=2xx2+x−2.

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