Explanation Given: Consider the equation 2 3 x − 1 + 2 3 x + 1 = 4 x 9 x 2 − 1 . Calculation: Given equation is 2 3 x − 1 + 2 3 x + 1 = 4 x 9 x 2 − 1 Take LCM in the denominator of the LHS. 2 ( 3 x + 1 ) + 2 ( 3 x − 1 ) ( 3 x + 1 ) ( 3 x − 1 ) = 4 x 9 x 2 − 1 6 x + 2 + 6 x − 2 9 x 2 − 1 = 4 x 9 x 2 − 1 12 x 9 x 2 − 1 = 4 x 9 x 2 − 1 Multiply both sides by 9 x 2 − 1

Pearson eText for Intermediate Algebra -- Instant Access (Pearson+)

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ISBN: 9780137553464

Author: John Tobey, Jeffrey Slater

Publisher: PEARSON+

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Chapter 6.4, Problem 18E

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To calculate: The solution for the equation 23x−1+23x+1=4x9x2−1 and check the solution.

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Chapter 6.4, Problem 17E

Chapter 6.4, Problem 19E

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Pearson eText for Intermediate Algebra -- Instant Access (Pearson+)

Ch. 6.1 - Find the domain of each of the following rational...Ch. 6.1 - Find the domain of each of the following rational...Ch. 6.1 - Find the domain of each of the following rational...Ch. 6.1 - Prob. 4E Ch. 6.1 - Simplify completely. − 18 x 4 y 12 x 2 y 6 Ch. 6.1 - Simplify completely. − 7 x y 2 28 x 5 y Ch. 6.1 - Prob. 7E Ch. 6.1 - Simplify completely. 24 x 2 − 4 x 4 x 2 − 8 x Ch. 6.1 - Simplify completely. 8 x 2 28 x 2 + 4 x Ch. 6.1 - Simplify completely. 30 x 2 6 x 2 + 18 x

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The solution for the equation 2 3 x − 1 + 2 3 x + 1 = 4 x 9 x 2 − 1 and check the solution.

Solution Summary: The author calculates the solution for the equation 23x-1+1= x=0.

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To calculate: The solution for the equation 23x−1+23x+1=4x9x2−1 and check the solution.

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