Determine graphically the solution set for the system of inequalities: x+y <= 4 2x+y <= 6 2x−y >= −1 x >= 0, y >= 0

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?

Inequality solution verification(positive solution)
I have the following system of equation where all the free variables live in the integers that is are allowed to be integers only.
$x=l$
$y=5n-2l+25$
$z=10-2n$
For $l,n\in \mathbb{N}$
I need to compute how many positive solutions are there so I computed the following inequalities:
$x>0\to l>0$
$y>0\to n>-5+2l/5$
$z>0\to n<-5$
To get how many solution we keep a fixed l and see how many n we get and do that for each n until we get values that pop us out of the inequality for example for $l=1$ we have $n>-4.6$ hence we have $-4\le n<5$ so in this case their is $10$ solutions and we proceed like this I just want to verify my answer I got $105$ possible positive solutions is that correct?

Graph the feasible region for the system of inequalities
$y>4x-1$
$y\leftarrow 2x+3$

Solve a system of linear inequalities without graphing, i.e. using pure algebra?
$y\ge 2x+1$
$y>(x/2)-1$

the linear system of inequalities
$(S):Ax\ge 0,x\ge 0$
has a solution $x\in {\mathbb{R}}^n$ that is not zero $x\ne 0$. Here $A$ is a $n\times n$ square matrix with real entries.
Of course I could combine the two inequalities in (S) into one and write it in the form of $\tilde{A}x\ge 0$ where $\tilde{A}$ is $2n\times n$.

Describe the graphs of the equations and inequalities. ${\displaystyle x^2+y^2=0}$

What is the domain of $1/\sqrt{x^2-4}$ using the set builder notation?
I'm confused when it comes to getting the domain of a function using set builder notation. Please explain also how to do the system of inequalities when getting the domain of the given function
$x\mapsto \frac{1}{\sqrt{x^2-4}}$