3. For the system of equations $\left\{\begin{array}{l}x-y=2, \\ m x+y=3\end{array}\right.$, if the solution corresponds to a point in the first quadrant of the Cartesian plane, then the range of values for $m$ is
3. $-10$, solving gives $m>-1$.
Thus, $y=x-2=\frac{3-2 m}{m+1}$. But $m+1>0$,
so, $3-2 m>0, m<\frac{3}{2}$. Therefore, $-1<m<\frac{3}{2}$.