Giải các hệ phương trình:
$a) \begin{cases}\frac{x+3y}{5} +\frac{y+z}{6}  =z \\ \frac{2x+5}{7}+\frac{4z+5}{3}  =z+1\\ \frac{3y+7}{8}+\frac{2z+1}{3}=y-1  \end{cases}                     b)  \begin{cases}\frac{1}{x-1}+\frac{1}{y-2}+\frac{1}{z-3}   =1 \\ \frac{1}{x-1}+\frac{2}{y-2}+\frac{4}{z-3}   =8\\\frac{1}{x-1}+\frac{3}{y-2}+\frac{9}{z-3}   =27 \end{cases} $

$c) \begin{cases}\frac{1}{x}+\frac{1}{y}-\frac{1}{z}   =0\\\frac{1}{x} -\frac{1}{y}+\frac{1}{z}=10   \\ -\frac{1}{x}+\frac{1}{y}+\frac{1}{z} =-6 \end{cases}                         d) \begin{cases}\frac{x-3}{5}=\frac{y+9}{13}  =\frac{6z-1}{7}  \\3x+2 y-3z=12.\end{cases} $

$e) \begin{cases}\frac{x}{5} =\frac{y}{7} =\frac{z}{13}  \\ x+2y+3z=174 \end{cases}                          f) \begin{cases}\frac{2x-7}{3} = \frac{3y+1}{2} =\frac{6z-1}{7} \\ 3x+2y-z=61 \end{cases} $

$g) \begin{cases}x+y+z=-2 \\ y+z+t=4\\z+t+x=-3\\t+x+y=1 \end{cases} $