#### Answer

The solution to this system is:
$$x=-\frac{34}{25},\quad y=\frac{51}{25}.$$

#### Work Step by Step

As instructed in the problem we will set $A=1/x$ and $B=1/y$. Then we get a system of LINEAR equations in $A$ and $B$:
\begin{align*}
2A+3B=&0\\
3A-4B=&-\frac{25}{6}.
\end{align*}
Now let's solve this system:
Step 1: Express $B$ in terms of $A$ from the first equation:
$$2A+3B=0\Rightarrow 3B=-2A$$
and that gives
$$B=-\frac{2}{3}A.$$
Step 2: Put this into the second equation and solve for $A$:
$$3A-4\times(-\frac{2}{3}A) = -\frac{25}{6}.$$
This further gives
$$3A+\frac{8}{3}A=-\frac{25}{6}\Rightarrow \frac{9}{3}A+\frac{8}{3}A=-\frac{25}{6}\Rightarrow \frac{17}{3}A=-\frac{25}{6}.$$
This now yields
$$A=-\frac{25}{6}\times\frac{3}{17} = -\frac{25}{34}.$$
Step 3: Use this calculated value to solve for $B$:
$$B=-\frac{2}{3}A=-\frac{2}{3}\times \left(-\frac{25}{34}\right) = \frac{25}{51}.$$
Now we have to solve for $x$ and $y$ and to do this we have to remember that:
$$A=\frac{1}{x}$$ and this means that
$$x=\frac{1}{A}=-\frac{34}{25}.$$
Also the fact that $$B=\frac{1}{y}$$ means that $$y=\frac{1}{B} = \frac{51}{25}.$$