Graph the solution set of the inequality or system of inequalities. {(x>=0 and y>=0),(3x+y<=9),(2x+3y>=6):}

Let's say that we have a system of linear inequalities:
$\left[\begin{array}{cccc}{c}_{1,1}& c_{1,2}& \dots & c_{1,n}\\ c_{2,1}& c_{2,2}& \dots & c_{2,n}\\ ⋮& ⋮& \ddots & ⋮\\ c_{m,1}& c_{m,2}& \dots & c_{m,n}\end{array}\right]\times \left[\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right]\ge \left[\begin{array}{c}{b}_1\\ b_2\\ ⋮\\ b_m\end{array}\right]$
It can be represented in a matrix form:
$\mathbf{C}\mathbf{x}\ge \mathbf{b}$
Does it hold that:
$\mathbf{x}\ge {\mathbf{C}}^{-1}\mathbf{b}$
and that $\mathbf{C}$ is invertible if and only if the whole system is solvable?
P.S. All the numbers $x_i,b_i,c_{i,j}$ are real. Would restricting them to be integers change the answer?
EDIT 1: for matrices $\mathbf{x}$ and $\mathbf{y}$ it holds that $\mathbf{x}\ge \mathbf{y}$ if and only if every element of $\mathbf{x}$ is $\ge$ to corresponding element in $\mathbf{y}$.
EDIT 2: the $x_i$ are bounded to $[-2,2]$.

Suppose that $x$, $y$, $z$, $\nu$, $N$, $g$ are some postitive parameters and $A$, $B$ and $C$ are variables that belong to $(0,+\mathrm{\infty })$. I want to identify one or all of the feasible points taht solve the following system of inequalities with respect to $A$, $B$ and $C$
$$\begin{gathered} xA-\nu gB-NgC>0 \\ -\nu gA+yB-NgC>0 \\ -\nu gA-\nu gB+zC>0 \end{gathered}$$

Let $P$ be a polyhedron in $[0,1{]}^n$ defined by the constraints$Ax\le b$ for $A\in {\mathbb{R}}^{m\times n}$, $x\in {\mathbb{R}}^n$, and $b\in {\mathbb{R}}^m$.
In the solutions of an exercise, the following is mentioned:
"Since the first $n$ constraints are linearly independent, they correspond to a basic solution of the system which, a priori, may be feasible or infeasible. This solution is obtained by replacing inequalities with equalities and computing the unique solution of this linear system."
So I am quite confused about this:
1) Why does "linear independent constraints" imply that "basic solution"?
2) Is a basic solution not always feasible?

detecting hidden equality constraints in optimization
I have a system of linear inequalities $Ax\le b$. It is possible that some inequality constraints actually create a linear equality constraint. for e.g. $ax\le b$ and $ax\ge b$ implies $ax=b$. A more involved linear combination of inequalities can also lead to equality constraints. I want to detect the final equality constraint given that $Ax\le b$. Can it be done?

How to solve this system with equation and inequality?
$\{\begin{array}{r}{2}^{x+2}=\frac{49}{4}{x}^2+4,\\ 2^{x+2}-4\le x^2(14-2^{x+2})\ast 2^x\end{array}$

Describe the solution set to the system of inequalities. ${\displaystyle x\ge 0,y\ge 0,x\le 1,y\le 1}$

Integer programming, system of linear inequalities.
$t_{01}+t_{11}+t_{21}\ge 4$
$t_{02}+t_{12}+t_{22}\ge 4$
$t_{10}+t_{11}+t_{12}\ge 4$
$t_{10}+t_{01}+t_{22}\ge 4$
$t_{10}+t_{02}+t_{21}\ge 4$
$t_{20}+t_{21}+t_{22}\ge 4$
$t_{20}+t_{01}+t_{12}\ge 4$
$t_{20}+t_{02}+t_{11}\ge 4$
There are 8 integer unknowns, $t_{ij}\ge 0,0\le i\le 2,0\le j\le 2,t_{00}$ is missing. Every unknown is used in exacly 3 inequalitites. The problem is to find solution that minimizes sum of all $t_{ij}$
$3\times \sum \ge 8\times 4$
$\sum \ge 11$
I have got this solution $t_{ij}=\left[\begin{array}{ccc}& 2& 2\\ 2& 2& 0\\ 2& 0& 2\end{array}\right]$ and I need hint to prove that it optimal. Sum of all unknowns is 12 here. Maybe linear programming method can be used?