Solve for the value of x in the system of equations:
$\begin{array}{l}6 x+y-z=4 \\ 2 x+3 y+5 z=20 \\ -x+2 y-4 z=6\end{array}$

Step 1 ：We are given the system of equations: \(6x + y - z = 4\), \(2x + 3y + 5z = 20\), and \(-x + 2y - 4z = 6\).

Step 2 ：We can represent this system in matrix form as follows: \(\begin{bmatrix}6 & 1 & -1 \\ 2 & 3 & 5 \\ -1 & 2 & -4\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}4 \\ 20 \\ 6\end{bmatrix}\).

Step 3 ：Solving this system of equations gives us the values of x, y, and z. However, the value of x is extremely close to zero, which is likely due to the precision of the floating point arithmetic. We can consider this value as zero for practical purposes.

Step 4 ：Final Answer: The solution to the system of equations is \(x = 0\), \(y = 5\), and \(z = 1\). So, the value of \(x\) is \(\boxed{0}\).