### Instruction: 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.### Response: I notice that the coefficients of $x$ and $y$ in both equations are multiples of 3, so I can divide both equations by 3 to simplify them.
This gives me

\begin{align*}
2x-\frac{4}{3}y&=\frac{a}{3},\\
2y-3x &=\frac{b}{3}.
\end{align*}
Now I want to eliminate one of the variables by adding or subtracting the equations.
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*}
Then I can add the equations and get
$0=a+\frac{2b}{3},$ which simplifies to $a=-\frac{2b}{3}.$
This means that $\frac{a}{b}=-\frac{2}{3}.$