解下列方程组：
(1)$\{\begin{array}{l} x+y+z=6,\\ 3x-y+2z=12,\\ x-y-3z=-4;\end{array} $
(2)$\{\begin{array}{l} 2x-y=4,\\ 2x+y+z=1,\\ x-z=5;\end{array} $
(3)$\{\begin{array}{l} x-2y=-9,\\ y-z=2,\\ 2z+x=47;\end{array} $
(4)$\{\begin{array}{l} \dfrac {x}{2}=\dfrac {y}{3}=\dfrac {z}{5},\\ x-2y+3z=33.\end{array} $
答案：(1)$\begin{cases}x=3,\\y=1,\\z=2\end{cases}$
(2)$\begin{cases}x=2,\\y=0,\\z=-3\end{cases}$
(3)$\begin{cases}x=21,\\y=15,\\z=13\end{cases}$
(4)$\begin{cases}x=6,\\y=9,\\z=15\end{cases}$
