Answer
The solution of the system of equations is $\left( 2,3,3 \right)$.
Work Step by Step
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.
