Answer :
Elimination Method
[tex]\begin{gathered} -2X+Y-2Z=-8 \\ 7X+Y+Z=-1 \\ 5X+2Y-Z=-9 \end{gathered}[/tex]If we multiply the equation 3 by (-1) we obtain this:
[tex]\begin{gathered} -2X+Y-2Z=-8 \\ 7X+Y+Z=-1 \\ -5X-2Y+Z=9 \end{gathered}[/tex]If we add them we obtain 0, therefore there are infinite solutions. So, let's write it in terms of Z
1. Using the 3rd equation we can obtain X(Y,Z)
[tex]\begin{gathered} 5X=-9-2Y+Z \\ X=\frac{-9-2Y+Z}{5} \\ \end{gathered}[/tex]2. We can replace this value of X in the 1st and 2nd equations
[tex]\begin{gathered} -2\cdot(\frac{-9-2Y+Z}{5})+Y-2Z=-8 \\ 7\cdot(\frac{-9-2Y+Z}{5})+Y+Z=-1 \end{gathered}[/tex]3. If we simplify:
[tex]\begin{gathered} \frac{-9Y+12Z-63}{5}=-1 \\ \frac{9Y-12Z+18}{5}=-8 \end{gathered}[/tex]4. We can obtain Y from this two equations:
[tex]\begin{gathered} Y=-\frac{-12Z+58}{9} \\ \end{gathered}[/tex]5. Now, we need to obtain X(Z). We can replace Y in X(Y,Z)
[tex]\begin{gathered} X=\frac{-9-2Y+Z}{5} \\ X=\frac{-9-2(-\frac{-12Z+58}{9})+Z}{5} \end{gathered}[/tex]6. If we simplify, we obtain:
[tex]X=\frac{-3Z+7}{9}[/tex]7. In conclusion, we obtain that
(X,Y,Z) =
[tex](\frac{-3Z+7}{9},-\frac{-12Z+58}{9},Z)[/tex]