Answer
The solution set is $\left\{ \left( 6,3,5 \right) \right\}$
Work Step by Step
We have the system given as:
$\begin{align}
& x+y+2z=19\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& y+2z=13\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \\
& z=5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{III} \right)
\end{align}$
Put the value $z=5$ into equation (II).
Then,
$\begin{align}
& y+2z=13 \\
& y+2\left( 5 \right)=13 \\
& y+10=13 \\
& y=3
\end{align}$
Now, put the value of $y=3$ and $z=5$ into equation (I) and then find the value of x:
$\begin{align}
& x+y+2z=19 \\
& x+3+2\left( 5 \right)=19 \\
& x+3+10=19 \\
& x+13=19
\end{align}$
Finally,
$\begin{align}
& x=19-13 \\
& x=6
\end{align}$
Hence, the solution set is $\left\{ \left( 6,3,5 \right) \right\}$.
