You can multiply both equations by constants before adding. Here if you multiply the first equation by $3$ and the second by $4$, you get
$\left\{\begin{align*}-12A-9B&=-6\\12A-16B&=0\;,\end{align*}\right.$
which you can add to get $-25B=-6$. Dividing both sides by $-25$ now gives you $B=\frac6{25}$, which you can substitute into either original equation to solve for $A$. I’ll use the second:
$3A-4\cdot\frac6{25}=0\;,$ so $3A=\frac{24}{25}\;,$ and $A=\frac8{25}$.
To check a potential answer, just substitute it into the original equations and see whether it ‘works’. Here we have $-4A-3B=-4\cdot\frac{8}{25}-3\cdot\frac6{25}=\frac{-32-18}{25}=-2\;,$ as we should, and you can easily check that the second equation is also satisfied by these numbers.