If I multiply the first equation by 3 and the second equation by 2, I get

\begin{align*}
6x-4y&=a,\\
4y-6x &=\frac{2b}{3}.
\end{align*} $0=a+\frac{2b}{3},$ which simplifies to $a=-\frac{2b}{3}.$ This means that $\frac{a}{b}=-\frac{2}{3}.$ 
If the system of equations

\begin{align*}
6x-4y&=a,\\
6y-9x &=b.
\end{align*}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\frac{a}{b},$ assuming $b$ is nonzero.