[tex]\bf \begin{cases}
5x+3y+z=-14\\
2x-2y-z=-13\\
3x+y+3z=-2
\end{cases}[/tex]
so.. let's us eliminate, say "z" first, so for that, let's pick the 1s and 2nd equations there [tex]\bf \begin{array}{llll}
5x+3y+z=-14\\
2x-2y-z=-13\\
----------\\
\boxed{7x+y+0\ =-27}
\end{array}[/tex]
so, that's our first two-variables resultant
let's pick other two, and again, eliminate the same "z" variable, this time, let's use the 2nd and 3rd equations [tex]\bf \begin{array}{llll}
2x-2y-z=-13&\times 3\implies &6x-6y-3z=-39\\
3x+y+3z=-2\implies &&3x+y+3z\ =-2\\
&&----------\\
&&\boxed{9x-5y+0\ =-41}
\end{array}[/tex]
now, we have two two-variables equations, let's use them then
and say, we'll eliminate the "y" variable from them
[tex]\bf \begin{array}{llll}
7x+y=-27& \times 5\implies &35x+5y=-135\\
9x-5y=-41\implies &&9x-5y=-41\\
&&-------\\
&&44x+0=-176
\end{array}
\\\\\\
x=\cfrac{-176}{44}\implies \boxed{x=-4}
\\\\\\
\textit{now, that we know x = -4, let's plug that in }7x+y=-27
\\\\\\
7(-4)+y=-27\implies y=-27+28\implies \boxed{y=1}[/tex]
[tex]\bf \textit{now, let's plug those two at }5x+3y+z=-14
\\\\\\
5(-4)+3(1)+z=-14\implies -20+3+z=-14\implies z=-14+17
\\\\\\
\boxed{z=3}[/tex]
