Chapter 10 - Section 10.6 - Radical Equations and Problem Solving - Exercise Set - Page 729: 38

$3=y$

$\sqrt {3y+6}=\sqrt {7y-6}$ $(\sqrt {3y+6})^2=(\sqrt {7y-6})^2$ $3y+6=7y-6$ $3y+12=7y$ $3y+12-3y=7y-3y$ $12=4y$ $12/4=4y/4$ $3=y$ $\sqrt {3y+6}=\sqrt {7y-6}$ $\sqrt {3*3+6}=\sqrt {7*3-6}$ $\sqrt {9+6}=\sqrt {21-6}$ $\sqrt {15}=\sqrt {15}$ (true)