Question 20089
{{{2x+3y-z=14}}}....(i)
{{{x-y-z=0}}}...(ii)
{{{3x+2y-3z=15}}}....(iii)


subtracting 3 times (ii) from (iii), we get
{{{5y=15}}} which gives {{{y=3}}}

subtracting (ii) from (i), we get
{{{x+4y=14}}}
and using y=3 we get 
{{{x=2}}}

using the values of x and y in (i) we get {{{z=-1}}}