Feasible point of a system of linear inequalities 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?

Clever guessing is ok, random guessing isn't advised. Since $x\geq 1$, $y\geq 2$ and $z\geq 3$. You should try some values that are close to the boundary, since then they are less likely to break the other conditions. So an $x$ value slightly greater than $1$, a $y$ value that is slightly greater than $2$, and a $z$ value slightly greater than $3$. 
So an interior point would be $(1.5, 2.5, 3.5)$ since  
$$-2(1.5)+(2.5)+(3.5)=3< 4$$ $$1.5 > 1$$ $$2.5>2$$ $$ 3.5 > 3 $$ $$(1.5)-2(2.5)+(3.5)=0 < 1$$ $$ 2(1.5)+2(2.5)-(3.5)=4.5 < 5$$ 
