Let's determine all the triples of numbers $(x, y, m)$ for which

$$
-2 x + 3 y = 2 m, \quad x - 5 y = -11
$$

and $x$ is a negative integer, $y$ is a positive integer, and $m$ is a real number.
From the second equation, $x=5 y-11$. Since $x$ is negative according to our condition, the equality can only hold if $y<\frac{11}{5}$. From the constraint on $y$, we get $y=2$ or $y=1$, and therefore $x=-1$ or $x=-6$.

Now we can determine the values of $m$ from the first equation:

\[
\begin{aligned}
\text { if } y=2 \text { and } x=-1, \quad \text { then } m=4 \\
\text { and if } y=1 \text { and } x=-6, \quad \text { then } m=7.5
\end{aligned}
\]

Remark. During the solution, we only assumed that $x$ is negative, no constraint is needed for $m$, and for any real number pair $(x ; y)$, $-2 x+3 y$ is always real.