解:
$ (1)\begin {cases}2x+y=5a,①\\x -3y=-a+7,②\end {cases}$
$ ①×3+②,$得$7x=14a+7,$解得$x=2a+1,$
$ $将$x=2a+1$代入$①,$得$2(2a+1)+y=5a,$
$ $解得$y=a-2.$所以原方程组的解为$\begin {cases}x=2a+1\\y =a-2\end {cases}.$
$ (2)$由题意,得$\begin {cases}2a+1≥0\\a -2<0\end {cases},$
解得$-\frac {1}{2}≤ a<2.$
$ $所以字母$a$的取值范围是$-\frac {1}{2}≤ a<2.$