## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 10 - Exponents and Radicals - Review Exercises: Chapter 10 - Page 693: 43

#### Answer

$y=19$

#### Work Step by Step

Using the properties of equality to isolate the radical expression results to \begin{array}{l}\require{cancel} \sqrt{y+6}-2=3 \\\\ \sqrt{y+6}=3+2 \\\\ \sqrt{y+6}=5 .\end{array} Squaring both sides and isolating the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} \sqrt{y+6}=5 \\\\ \left( \sqrt{y+6} \right)^2=(5)^2 \\\\ y+6=25 \\\\ y=25-6 \\\\ y=19 .\end{array} Upon checking, $y=19$ satisfies the original equation.

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