# linear programming problem given initial solution

While dealing with a linear programming problem we usually try to start with the basic feasible solution corresponding to the identity Matrix in the coefficient matrix. I have no idea how to solve the following type of LPP:

Solve the LPP using simplex method starting with the basic feasible solution $x_1=4$ and $x_2=0$

Max $z=-x_1+2x_2$ subject to $$3x_1+4x_2=12$$ $$2x_1-x_2\leq 12$$ $$x_1\geq 0, x_2\geq 0$$

This is a case where you have to introduce an artificial variable. I don't know why an initial solution in terms of $$x_1$$ and $$x_2$$ is given since it is not with these variables that we begin.
I introduced an artificial variable $$x_4$$ to account for the first constraint being an equation. The second constraint doesn't need an artificial variable because when we put it in the standard form, the slack variable $$x_3$$ gives us a basis to begin the iterations. So our basic feasible solution here is $$x_3 = 12$$, $$x_4 = 12$$, but we now want to minimize a new objective function $$w = x_4$$.
After following the simplex algorithm, I get $$w = 0$$ which is surely its minimum, since $$w$$ is a non-negative function, and I end up with a basis in terms of variables $$\{x_2, x_3\}$$. This means the original problem is feasible and attains its optimal solution at the same point $$w$$ attain its minimum --- which is at $$x_2 = 3, x_3 = 15$$, when $$z = 6$$.
So the final answer is $$z = 6$$ at $$(0, 3)$$ because the original problem is only involved with variables $$x_1$$, $$x_2$$, so we must not mention $$x_3$$ and $$x_4$$ here.