$解:记\begin{cases}{x-2y=t①}\\ {2x+3y=2t+4②}\end{cases}$
$由②-①,得x+5y=t+4$
$由①+②,得3x+y=3t+4$
$∵\begin{cases}{x+5y>0}\\ {3x+y\leqslant 0}\end{cases}$
$∴\begin{cases}{t+4>0}\\ {3t+4\leqslant 0}\end{cases}$
$解得-4<t\leqslant -\frac{4}{3}$
$∵t为整数$
$∴t=-3,-2$
$∴满足条件的t的整数值为-3,-2$