Question 406673
{{{2x+5y+z=22}}}.................1.........solve for {{{z}}} and substitute in 2
{{{3x-3y-z=-22}}}..................2
{{{5x+y+2z=-3}}}..................3 



{{{2x+5y+z=22}}}.................1

{{{z= -2x-5y + 22}}}


{{{3x-3y-(-2x-5y + 22)=-22}}}..................2

{{{3x-3y+2x+5y - 22=-22}}}

{{{5x+2y =-22+22}}}

{{{5x+2y = 0}}}................solve for {{{y}}} and substitute in 3


{{{5x+2y = 0}}}

{{{2y = -5x}}}

{{{y = -(5/2)x}}}

substitute {{{y = -(5/2)x}}} in {{{z= -2x-5y + 22}}}

{{{z= -2x-5(-(5/2)x) + 22}}}

{{{z= -2x+(25/2)x + 22}}}

{{{z= -2x+(12.5)x + 22}}}

{{{z= 10.5x + 22}}}


{{{5x+y+2z=-3}}}..................3 ..substitute {{{z= 10.5x + 22}}} and {{{y = -(5/2)x}}}


{{{5x+(-(5/2)x)+2(10.5x + 22)=-3}}}..................3...calculate {{{x}}}


{{{5x-(5/2)x+21x +44=-3}}}

{{{26x-2.5x =-3-44}}}

{{{23.5x = -47}}}
{{{x = -47/23.5}}}

{{{x = -2}}}


now find {{{y}}} and {{{z}}}


{{{y = -(5/2)x}}}

{{{y = -(5/2)(-2)}}}

{{{y = -(5/cross(2))(-cross(2))}}}

{{{y = 5}}}



{{{z= 10.5(-2) + 22}}}

{{{z= -21 + 22}}}

{{{z= 1}}}

so, your answer is:

{{{x = -2}}}

{{{y = 5}}}

{{{z= 1}}}

check:

{{{2x+5y+z=22}}}.................1.

{{{2(-2)+5*5+1=22}}}

{{{-4+25+1=22}}}

{{{-4+26=22}}}

{{{22=22}}}