解:$(1)\begin {cases} 3x-2y=8, &① \\2x+y=3; &② \end {cases}$
$ $由$②$得$y=-2x+3 ③,$
把③代入①,得$3x-2(-2x+3)=8,$
$ $解得$x=2,$
$ $把$x=2$代入$②,$得$y=-1,$
$ $则原方程组的解是$\begin {cases} x=2 \\y=-1 \end {cases}。$
$ (2) \begin {cases} x+4y=14, &① \\\dfrac {x-3}{4}-\dfrac {y-3}{3}=\dfrac {1}{12}. &② \end {cases}$
$ $由$②$得$3x-4y=-2 ③,$
①+③,得$4x=12,$解得$x=3,$
$ $把$x=3$代入$③,$解得$y=\dfrac {11}{4},$
$ $则原方程组的解是$\begin {cases} x=3 \\y=\dfrac {11}{4} \end {cases}$