Question 1077277
I'm assuming you mean,
{{{2x+2y+2z=0}}}
{{{-2x+5y+2z=1}}}
{{{8x+2y+4z=-1}}}
R1+R2,
{{{2x+2y+2z-2x+5y+2z=1}}}
R3:{{{7y+4z=1}}}
.
.
4R2+R3,
{{{-8x+20y+8z+8x+2y+4z=4-1}}}
R4:{{{22y+12z=-3}}}
So then,
R4-3R3,
{{{22y+12z-21y-12z=3-(3)}}}
{{{y=0}}}
Now that you now y work backwards to solve for x and z.