$解:\left\{\begin{array}{l}3 x+2 y=2 k(1) \\5 x+4 y=k+3(2)\end{array}\right.$
$(2) -2 (1)得5 x+4 y-6 x-4 y=k+3-4 k$
$解得\ x=3 k-3\ $
$将\ x=3 k-3\ 代入 (1)得$
$9 k-9+2 y=2 k$
$解得\ y=\frac{9-7 k}{2}\ $
$x+y=3 k-3+\frac{9-7 k}{2}\ $
$\because x+y=-5\ $
$\therefore 3 k-3+\frac{9-7 k}{2}=-5$
$解得\ k=13\ $
$ $