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.
If we multiply the first equation by $-\frac{3}{2}$, we obtain

$$6y-9x=-\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have

$$-\frac{3}{2}a=b\Rightarrow\frac{a}{b}=\boxed{-\frac{2}{3}}.$$
-\frac{2}{3}
Level 4
Algebra
Problem: 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.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the ratio \(\frac{a}{b}\), we start by manipulating the given equations. 

First, we can express \(a\) and \(b\) in terms of \(x\) and \(y\):
1. From the first equation, we have \(a = 6x - 4y\).
2. From the second equation, we can rearrange it to find \(b\): \(b = 6y - 9x\).

Next, we need to find a relationship between \(a\) and \(b\). We can express \(a\) in terms of \(b\) by substituting \(y\) from one equation into the other. 

To do this, we can solve for \(y\) in terms of \(x\) from the first equation:
\[
4y = 6x - a \implies y = \frac{6x - a}{4}.
\]

Now, substitute this expression for \(y\) into the equation for \(b\):
\[
b = 6\left(\frac{6x - a}{4}\right) - 9x.
\]
Simplifying this gives:
\[
b = \frac{36x - 6a}{4} - 9x = \frac{36x - 6a - 36x}{4} = -\frac{6a}{4} = -\frac{3a}{2}.
\]

Now we have a direct relationship between \(a\) and \(b\):
\[
b = -\frac{3a}{2}.
\]

To find \(\frac{a}{b}\), we can rearrange this equation:
\[
\frac{a}{b} = \frac{a}{-\frac{3a}{2}} = -\frac{2}{3}.
\]

This derivation shows that the ratio \(\frac{a}{b}\) is indeed \(-\frac{2}{3}\), confirming the final answer.<|END_OF_LATENT|> The final answer is $-\frac{2}{3}$.