TITLE: Feasible point of a system of linear inequalities
QUESTION [5 upvotes]: Let $P$ denote $(x,y,z)\in \mathbb R^3$, which satisfies the inequalities:
$$-2x+y+z\leq 4$$ $$x \geq 1$$ $$y\geq2$$ $$ z \geq 3 $$ $$x-2y+z \leq 1$$ $$ 2x+2y-z \leq 5$$
How do I find an interior point in $P$? 
Is there a specific method, or should I just try some random combinations and then logically find an interior point?

REPLY [2 votes]: You could also use the simplex method to solve the following problem:
$$
\min\limits_{x,y,z,e,a}\quad Z= a_1+a_2+a_3
$$
subject to
$$-2x+y+z+e_1 =4$$ 
$$x -e_2+a_1= 1$$ 
$$y-e_3+a_2=2$$ 
$$ z-e_4+a_3 = 3 $$ 
$$x-2y+z +e_5= 1$$ 
$$ 2x+2y-z +e_6= 5$$
$$
x,y,z,e,a\ge 0
$$
If $\min \{a_1+a_2+a_3\} = 0$, then you have a feasible point (with $a_1=a_2=a_3=0$).