Smallest value that a certain variable can take in a system of equations. 
Consider the solutions $(x,y,z,u)$ of the system of equations:
  $$\begin{cases}
x+y=3(z+u)\\
x+z=4(y+u)\\
x+u=5(y+z)\\
\end{cases}$$
  where $x,y,z \text{ and } u$ are positive integers.
What is the smallest value that $x$ can have?


What I have done:
$$\begin{pmatrix}
1&1&-3\\
1&-4&1\\
1&-5&-5\\
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
z\\
\end{pmatrix}
=\begin{pmatrix}
3\\
4\\
-1\\
\end{pmatrix}\ u
$$
$$ \left(      \begin{array}{lcr|c}
        1&1&-3&3\\
        1&-4&1&4\\
1&-5&-5&-1\\
      \end{array}
    \right)
\equiv
 \left(      \begin{array}{lcr|c}
        17&0&0&83\\
        0&17&0&7\\
0&0&17&13\\
      \end{array}
    \right)
$$
Hence
$$\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
z\\
\end{pmatrix}
=\frac{1}{17}\begin{pmatrix}
83\\
7\\
13\\
\end{pmatrix}\ u
$$
$$x=\frac{83}{17}u,\; y=\frac{7}{17}u,\; z=\frac{13}{17}u$$
Since $x,y,z,u \in \mathbb{Z^+},x_{min}=83 \quad $ Is that right?

Given that the variables have to be positive integers, $83$ is correct for the minimum value of $x$.  You have done well.
