#### Answer

The solution to this system is $$x=3,\quad y=-4.$$

#### Work Step by Step

As instructed in the problem here we will let $A=1/x$ and $B=1/y$. Applying this to our system we get
\begin{align*}
12A-12B =&7\\
3A+4B=&0
\end{align*}
Let's solve this system for $A$ and $B$:
Step 1: Express $B$ in terms of $A$ from the second equation:
$$4B=-3A\Rightarrow B=-\frac{3}{4}A.$$
Step 2: Put this into the first equation and solve for $A$
$$12A-12\times(-\frac{3}{4}A) = 7.$$
This gives
$$12A+9A=7\Rightarrow 21A = 7$$
so we have
$$A=\frac{7}{21} = \frac{1}{3}.$$
Step 3: Now we use this calculated value to solve for $B$:
$$B=-\frac{3}{4}A=-\frac{3}{4}\times\frac{1}{3} = -\frac{1}{4}.$$
Now we have to find $x$ and $y$.
Since $$A=\frac{1}{x} = \frac{1}{3}$$
we have that $$x=3.$$
Since $$B=\frac{1}{y} = -\frac{1}{4}$$
we have that
$$y=-4.$$