Question

I keep getting confused I don't understand what I keep doing wrong

Answer 1

$\bf \begin{cases} 5x+3y+z=-14\\ 2x-2y-z=-13\\ 3x+y+3z=-2 \end{cases}$

so.. let's us eliminate, say "z" first, so for that, let's pick the 1s and 2nd equations there  $\bf \begin{array}{llll} 5x+3y+z=-14\\ 2x-2y-z=-13\\ ----------\\ \boxed{7x+y+0\ =-27} \end{array}$

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   $\bf \begin{array}{llll} 2x-2y-z=-13&\times 3\implies &6x-6y-3z=-39\\ 3x+y+3z=-2\implies &&3x+y+3z\ =-2\\ &&----------\\ &&\boxed{9x-5y+0\ =-41} \end{array}$

now, we have two two-variables equations, let's use them then
and say, we'll eliminate the "y" variable from them

$\bf \begin{array}{llll} 7x+y=-27& \times 5\implies &35x+5y=-135\\ 9x-5y=-41\implies &&9x-5y=-41\\ &&-------\\ &&44x+0=-176 \end{array} \\\\\\ x=\cfrac{-176}{44}\implies \boxed{x=-4} \\\\\\ \textit{now, that we know x = -4, let's plug that in }7x+y=-27 \\\\\\ 7(-4)+y=-27\implies y=-27+28\implies \boxed{y=1}$

$\bf \textit{now, let's plug those two at }5x+3y+z=-14 \\\\\\ 5(-4)+3(1)+z=-14\implies -20+3+z=-14\implies z=-14+17 \\\\\\ \boxed{z=3}$