Question 966802
From eq. 2,
{{{x=1+2z}}}
Substitute into eq. 1,
{{{2(1+2z)-y-z=0}}}
{{{2+4z-y-z=0}}}
4.{{{-y+3z=-2}}}
Substitute into eq. 3,
{{{1+2z+y-5z=3}}}
5.{{{y-3z=2}}}
Equation 4 and 5 are the same equation.
There are infinitely many solutions.