## Precalculus (6th Edition) Blitzer

The solution of the system is $\left( 1,0,-3 \right)$.
Let us consider the given equation: $2x+y=2$ …… (I) $x+y-z=4$ …… (II) $3x+2y+z=0$ …… (III) Now, adding equation (II) and equation(III): \begin{align} & x+y-z=4 \\ & 3x+2y+z=0 \\ & \overline{\,\,\,\,\,4x+3y=4} \\ \end{align} $4x+3y=4$ …… (IV) Equation (I) is multiplied by 2 and we subtract from equation(IV): \begin{align} & 4x+3y=4 \\ & 4x+2y=4 \\ & \overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=0\,\,\,\,\,} \\ \end{align} Now, putting the value of $y=0$ in equation (I): \begin{align} & 2x+0=2 \\ & x=1\, \\ \end{align} $x=1,y=0$ Putting in equation (III): \begin{align} & 3\left( 1 \right)+2(0)+z=0 \\ & 3+0+z=0 \\ & z=-3 \end{align} Thus, the ordered triple $\left( 1,0,-3 \right)$ satisfies the three equations.