#### Answer

Solution from Gaussian elimination procedure is $ x=10,y=6\text{ and }z=4$

#### Work Step by Step

A dependent solution obtained was $\left[ \begin{matrix}
k+4 \\
k \\
k-2 \\
\end{matrix} \right]$
And here, $ z $ has a construction limit of 4 cars per min so, $\begin{align}
& z=4 \\
& 4=k-2 \\
& \text{ }k=6
\end{align}$
And, $\begin{align}
& x=6+4 \\
& y=6 \\
& z=6-2
\end{align}$
So, $ x=10,y=6\text{ and }z=4$
So to keep the traffic moving: 10 cars should be at $ X $, 6 cars at $ Y $ and 4 cars at $ Z $.