jmc

To begin, we observe that the expression $\frac{\sqrt{6y+2}}{\sqrt{2y}}$ is defined if and only if $y>0$. In this case, it is equal to $\sqrt{\frac{6y+2}{2y}}$. Since the quantity under the radical is always nonnegative for $y>0$, we may safely square both sides of our equation without creating fake solutions: $$\frac{6y+2}{2y}=\frac{25}{4}.$$Now we cross-multiply to obtain $$4(6y+2)=25(2y),$$and solve this linear equation: $$\begin{array}{rl}24y+8& =50y\\ 8& =26y\\ \boxed{\frac{4}{13}}& =y\end{array}$$