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{align*}
24y+8 &= 50y \\
8 &= 26y \\
\boxed{\frac{4}{13}} &= y
\end{align*}
If $\frac{\sqrt{6y+2}}{\sqrt{2y}}=\frac52$, solve for $y$. Express your answer in simplest fractional form.
{'answer': '\\frac{4}{13}', 'subject': 'Algebra', 'level': 4, 'unique_id': 'test/algebra/2392.json'}