Graph the solution set for the system of linear inequalities x>=3 y>=-1

If u is an algebraic expression and c is a positive number,
1. The solutions of $|u|<c$ are the numbers that satisfy$-c<u<c$.
2. The solutions of $|u|>c$ are the numbers that satisfy$u<-coru<c$.
These rules are valid if $<$ is replaced be ${\displaystyle \le }$ and $>$ is replaced by ${\displaystyle \ge }$

Graph the solution set of each system if inequalities.
X is greater than or = 0
X+Y is less than or equal to 4
2x+y is less than or equal to 5

Given a system $Ax\le b$ of linear inequalities, describe a linear programming problem whose optimal solution immediately tells us which inequalities among $Ax\le b$ are always satisfied with equality.
I wonder if I could use the theorem: assume both optima of $max(c^{T}x|Ax\le b)=min(y^{T}b|y\le 0,y^{T}A=c^T)$ are finite, then $x_0$ and $y_0$ are optimum solutions iff if a component of $y_0$ is positive, the corresponding inequality in $Ax\le b$ is satisfied by $x_0$ with equality $y_0^T(b-A{x}_0)=0$.

Graph the solution set of the system of inqualities
x-2y<3
2x+y>3

Graph the following system of inequalities. Show (by shading in) the feasible region.
$x+2y\le 12$
$2x+y\le 12$
$x\ge 0,y\ge 0$

Given a set of n inequalities each of the form ax+by+cz≤d for some a,b,c,d in Q, determine if there exists x, y and z in Q that satisfy all the inequalities.
Here is an O(n4) algorithm for solving this: for each triple of inequalities, intersect their corresponding planes to get a point (x,y,z) iff possible. If no such intersecting point exists continue on to the next triple of inequalities. Test each of these intersection points against all the inequalities. If a particular point satisfies all the inequalities the solution has been found. If none of these points satisfy all the inequalities then there is no point satisfying the system of inequalities. There are O(n3) such intersection points and there are n inequalities thus the algorithm is O(n4).
I would like a faster algorithm for solving this (eg: O(n3), O(n2), O(n*logn), O(n)). If you can provide such an algorithm in an answer that would be great. You may notice this problem is a subset of the more general k-dimensional problem where there are points in k-dimensions instead of 3 dimensions as in this problem or 2 dimensions as in my previous problem mentioned above. The time complexity of my algorithm generalized to k dimensions is O(nk+1). Ideally I would like something that is a polynomial time algorithm however any improvements over my naive algorithm would be great. Thanks