Solve the system using elimination and back substitution: [ \begincases x + 2y + z = 2 \ 2x + 5y + 3z = 7 \ x + 3y + 2z = 5 \endcases ]
The last row says (0=0) (consistent system). Pivot variables: (x, y). Free variable: (z).
[ \left[\beginarrayc 1 & 2 & 1 & 2 \ 2 & 5 & 3 & 7 \ 1 & 3 & 2 & 5 \endarray\right] ]