## Precalculus (6th Edition) Blitzer

The solution of the system of equations is $\left( 2,3,3 \right)$.
Let us consider the given equation: $x+y+z=11$ …… (I) $x+y+3z=14$ …… (II) $x+2y-z=5$ …… (III) Putting the values of equations (I) and (II): \begin{align} & x+y+2z=11 \\ & x+y+3z=14 \\ & \overline{\begin{align} & -z=-3 \\ & \text{ }z=3 \\ \end{align}} \\ \end{align} Now, we plug in z: \begin{align} & x+y=5 \\ & x+2y=8 \\ & \overline{\begin{align} & -y=-3 \\ & \text{ }y=3 \\ \end{align}} \\ \end{align} Putting in the value of y and z in equation (I): \begin{align} & x+y+2z=11 \\ & x+3+2\left( 3 \right)=11 \\ & x+3+6=11 \\ & x=11-9 \end{align} Further simplify, $x=2$ Thus, the order triple $\left( 2,3,3 \right)$ satisfies the three equations.