If $\frac{\sqrt{2x}}{\sqrt{3x-1}}=\frac32$, solve for $x$. Express your answer in simplest fractional form.
We can start by cross-multiplying: \begin{align*} 3\sqrt{3x-1}&=2\sqrt{2x}
\\\Rightarrow \qquad (3\sqrt{3x-1})^2 &=(2\sqrt{2x})^2
\\\Rightarrow \qquad 9(3x-1)& =4(2x)
\\\Rightarrow \qquad 27x-9& =8x
\\ \Rightarrow \qquad19x&=9
\\ \Rightarrow \qquad x&=\boxed{\frac9{19}}.
\end{align*}Checking, we see that this value of $x$ does indeed work, so it is not an extraneous solution.