$\displaystyle 2x-z=1 \Rightarrow z = 2x-1 $
$\displaystyle 4y+az = -3 \Rightarrow 4y + a (2x-1) = -3 \Rightarrow 4y+2ax-a = -3 $
$\displaystyle 9x-5y-7z=2 \Rightarrow 9x-5y -7(2x-1) = 2 \Rightarrow 9x-5y-14x+7 = 2 $
so now you have
$\displaystyle 2ax+4y =a -3 $
$\displaystyle -5x-5y = -5 $
can you continue from here ?
P.S. you should get something like :
$\displaystyle \displaystyle x = \frac {a-7}{2(a-2)} $
$\displaystyle \displaystyle y= \frac {a+3}{2 (a-2)}$
$\displaystyle \displaystyle z=- \frac {5}{a-2} $
of course $\displaystyle a\neq 2$