How to find all basic feasible solutions of a linear system? I'm trying to solve this problem but need some help getting started. The problem asks to find all the basic feasible solutions of the following system:
$$ \begin{aligned} -4x_2+x_3 &= 6\\ 2x_1-2x_2-x_4 &= 1\\ x_1, x_2, x_3, x_4 &\geq 0 \end{aligned} $$
How might one go about solving this problem? Thanks!
 A: Given the constraints of a model,
$$-4x_2 + x_3 = 6$$
$$2x_1-2x_2-x_4=1$$
$$x_1, x_2, x_3, x_4 \ge 0$$
Let's put it in standard form:
$$-4x_2 + x_3 +a_1= 6$$
$$2x_1-2x_2-x_4 + a_2=1$$
$$x_1, x_2, x_3, x_4, a_1, a_2\ge 0$$
To start, let's put these values into a tableau as such:

From here, notice that the $x_3$ column is like the columns of the basic, artificial variables, and as such, we can swap $a_1$ and $x_3$ to achieve a new tableau as shown here:

In addition, we can divide the second row by 2 to make $x_1$ a basic variable. By doing all of this, we should find two more tableaus as depicted below,


The last tableau will be the only feasible solution we are able to reach with these constraints, as modifying the $x_2$ and $x_4$ columns to make them basic variables will turn the RHS negative, which will then invalidate the $x_4\ge0$ and $x_2\ge0$ constraints and make the solution infeasible.
Since we can only have a max of two basic variables with the two constraints we have, and the coefficients of $x_2$ and $x_4$ are all negative, there will exist no combination of points that will allow $x_2$ and $x_4$ to become basic variables in the constraints. Thus, there is only one possible solution.
