Let's multiply the first equation by 3 so that we can eliminate the y terms. What if we multiplied both sides by 3? Then we'd have $6x-3y=3a$. What if we add the second equation to that? We'd have $6x-3y+3y-6x=3a+b$. Which simplifies to $3a+b=0$. And we know that $b$ is not equal to $0$, so we can divide both sides by $b$. We get $\frac{3a}{b}+1=0$, so $\frac{a}{b}=-\frac{1}{3}$.


If the system of equations

\begin{align*}
2x-y&=a,\\
3y-6x &=b.
\end{align*}has a solution, find $\frac{a}{b},$ assuming $b \neq 0.$