## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 7 - Review: 109

#### Answer

$\dfrac{3\sqrt[]{y}+6}{y-4}$

#### Work Step by Step

Multiplying by the conjugate of the denominator, the rationalized-denominator form of the given expression, $\dfrac{3}{\sqrt[]{y}-2} ,$ is \begin{array}{l}\require{cancel} \dfrac{3}{\sqrt[]{y}-2}\cdot\dfrac{\sqrt[]{y}+2}{\sqrt[]{y}+2} \\\\= \dfrac{3\sqrt[]{y}+6}{(\sqrt[]{y})^2-(2)^2} \\\\= \dfrac{3\sqrt[]{y}+6}{y-4} .\end{array} Note that all variables are assumed to have positive real numbers.

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