## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 1 - Review: 93

#### Answer

$\frac{2}{3}$ $\ne$ $2(n+\frac{1}{4})$

#### Work Step by Step

From the problem: "Two-thirds:" $\frac{2}{3}$ "is not equal to:" $\ne$ "twice the sum of:" $2(addition)$ "the sum of $n$ and one-fourth": $(n+\frac{1}{4})$ Combine: $\frac{2}{3}$ $\ne$ $2(n+\frac{1}{4})$

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