Question 902201
From eq. 3,
{{{x-z=6}}}
{{{z=x-6}}}
Substitute into eq. 1,
{{{x+y+(x-6)=57}}}
{{{2x+y=63}}}
Then from eq. 2,
{{{y=3+2x}}}
Substituting,
{{{2x+(3+2x)=63}}}
{{{4x=60}}}
{{{x=15}}}
Now working backwards,
{{{y=3+2(15)}}}
{{{y=3+30}}}
{{{y=33}}}
and finally,
{{{z=15-6}}}
{{{z=9}}}