Graph the solution set of the system of inequalities or indicate that the system has no solution. {(y>=x^2-4),(x-y>=2):}

Solve the equations and inequalities. Write the solution sets to the inequalities in interval notation.
${\displaystyle 5\le 3+|2x-7|}$

I have a task to calculate a double integral on a region$D=A\setminus B$, where $A=\{(x,y):\text{some system of inequalities}\},B=\{(x,y):\text{another system of inequalities}\}$. What does operator "$\setminus$" mean?

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?

Solve a system of inequalities
$\{\begin{array}{ll}{\log }_2^2(-{\log }_{2}x)+{\log }_2{\log }_2^{2}x\le 3& \\ -4|x^2-1|-3\ge \frac{1}{x^2-1}& \end{array}$
What I've tried:Make substitution $t=x^2-1$ and solve second inequality:
$-4|t|-3\ge \frac{1}{t}<=>-4|t|-3-\frac{1}{t}\ge 0$
Where I got interval for $t$: $-\frac{1}{4}\le t<0$
And then for $x$: $x\in (-1;-\frac{\sqrt{3}}{2}]\cup [\frac{\sqrt{3}}{2};1)$
But I have no thoughts how to solve first inequality.

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

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

I want to determine the number of solutions of a system of linear inequalities, and I was wondering if there was a simple way to to that. I know that linear programming is often used to check whether there are a zero or non-zero number of solutions, i.e. if the system/bounds is/are feasible, but is it possible to distinguish between there being a finite amount of unique solutions or infinitely many solutions? For instance, the system
$\begin{array}{lcl}x& \le & y\\ x& \ge & y\\ x+y& \le & 1\\ x+y& \ge & 1\end{array}$
has 1 unique solution, namely $x=y=\frac{1}{2}$, while the system
$\begin{array}{lcl}x& \ge & y\\ x+y& \le & 1\\ x+y& \ge & 1\end{array}$
has infinitely many solutions. Is there a away to find out how many solutions a system of linear inequalities has, if any?