Solving System of Inequalities with Examples

The system of inequalities is as follows:
$\{\begin{array}{l}a-<!--\; -\; -->\lfloor \frac{a}{b}\rfloor \cdot <!--\; \cdot \; -->b\le <!--\; \le \; -->M\\ a-<!--\; -\; -->\lfloor \frac{a}{b}\rfloor \cdot <!--\; \cdot \; -->b\ge <!--\; \ge \; -->m\end{array}$
I need to solve for $a$ and $b$. $M$ and $m$ are known constants.

Let $D=x\in R^3:|2x1-x2+3x3+1|+|x2+2x3-2|+|5x2-3x3|\le 10$. Express D as the feasible solution set of a linear system of inequalities (meaning, a system of the form $Ax\le b$).
How is the feasible solution set represented? Is this problem just a matter of removing the absolute signs and setting up two linear equations as such: $2x1+x2-<!--\; -\; -->3x3-<!--\; -\; -->1\le <!--\; \le \; -->10$ and $(2x1+x2-<!--\; -\; -->3x3-<!--\; -\; -->1)\le <!--\; \le \; -->10$?

The Fourier-Motzkin elimination (FME) is a procedure that reduces an $n$-variable problem to an equivalent $(n-1)$-variable problem. It is analogous to Gaussian elimination but for a system of inequalities. Does Fourier-Motzkin elimination also not change the solution set? Why or why not?

Consider the following system of inequalities:
$\{\begin{array}{l}{x}_{ab}+x_{ac}+x_{ad}+x_{abc}+x_{abd}+x_{acd}+x_{abcd}\ge <!--\; \ge \; -->4+x_{bc}+x_{bd}+x_{cd}+x_{bcd}\\ x_{ab}+x_{bc}+x_{bd}+x_{abc}+x_{abd}+x_{bcd}+x_{abcd}\ge <!--\; \ge \; -->4+x_{ac}+x_{cd}+x_{ad}+x_{acd}\\ x_{ac}+x_{bc}+x_{cd}+x_{abc}+x_{acd}+x_{bcd}+x_{abcd}\ge <!--\; \ge \; -->4+x_{ab}+x_{ad}+x_{bd}+x_{abd}\\ x_{ad}+x_{bd}+x_{cd}+x_{abd}+x_{acd}+x_{bcd}+x_{abcd}\ge <!--\; \ge \; -->4+x_{ab}+x_{ac}+x_{bc}+x_{abc}\end{array}$
where each $x\in <!--\; \in \; -->\{0,1\}$. Does this system of inequalities have a solution?

Suppose that we are given a subroutine which, given a system of linear inequalities $Ax\le <!--\; \le \; -->b$, $x\in <!--\; \in \; -->{\mathbb{R}}^n$, either produces a solution or decides that no solution exists.
Construct an algorithm that uses a single call of this subroutine and which returns an optimal solution (if it exists) to the following linear programming problem:
$\begin{array}{cccc}\text{minimize }& c^{T}x& & \\ \text{subject to: }& Ax& =& b\\ & x& \ge <!--\; \ge \; -->& 0\end{array}$
Well, I think I should take a look at the dual problem.
The dual is:
$\begin{array}{cccc}\text{maximize }& p^{T}b& & \\ \text{subject to: }& A^{T}p& \le <!--\; \le \; -->& c\end{array}$
Then I would call the subroutine for the matrix $A^T$ and to the vector $c$ to find a dual feasible point $y$.
I tried to use complementary slackness to construct a primal solution and to decide if $y$ is optimal but it didn't work .

Compute volume of $E=\{5x^8\le <!--\; \le \; -->y\le <!--\; \le \; -->7x^8;2y^5\le <!--\; \le \; -->z\le <!--\; \le \; -->3y^5;z^7\le <!--\; \le \; -->x\le <!--\; \le \; -->6z^7\}$
My progress:
It is easy to see that $x,y,z\ge <!--\; \ge \; -->0$. Therefore I can add all inequalities and after simple algebra get the following:
$5x^8+2y^5+z^7\le <!--\; \le \; -->x+y+z\le <!--\; \le \; -->7x^8+3y^5+6x^7$

Prove that the system
$a,b,d,e,f,g,h,i>0$
$ae+ai-bd+ei-fh=0$
$aei-hfa-<!--\; -\; -->bdi-gbf=0$
is inconsistent.

How would one go about solving the system of five equations:
$p^2=p+q-<!--\; -\; -->2r+2s+t-<!--\; -\; -->8$
$q^2=-<!--\; -\; -->p-<!--\; -\; -->2q-<!--\; -\; -->r+2s+2t-<!--\; -\; -->6$
$r^2=3p+2q+r+2s+2t-<!--\; -\; -->31$
$s^2=2p+q+r+2s+2t-<!--\; -\; -->2$
$t^2=p+2q+3r+2s+t-<!--\; -\; -->8$
over the integers?

Which quadrant contains no solutions to the system
If the system of inequalities
$y\ge <!--\; \ge \; -->2x+1$ and $y>0.5x-<!--\; -\; -->1$

Solving Linear System with inequalities.
$\begin{array}{r}b-<!--\; -\; -->x=0\\ a-<!--\; -\; -->0.33b-<!--\; -\; -->0.5x=0\\ d-<!--\; -\; -->0.33b=0\\ a-<!--\; -\; -->0.33b+c=0\\ a+b+c+d+2x=1\\ a+b+c+d-<!--\; -\; -->8.8x\le <!--\; \le \; -->0\\ a+b+c+d-<!--\; -\; -->7.27x\ge <!--\; \ge \; -->0\end{array}$
where a,b,c,d and x are unknown variables.
Is there any solution to this system and how to solve this system?

The given question is :
Consider the sets defined by the real solutions of the inequalities
$A=\{(x,y):x^2+y^4⩽<!--\; ⩽\; -->1\}B=\{(x,y):x^4+y^6⩽<!--\; ⩽\; -->1\}$
A. $B\subseteq <!--\; \subseteq \; -->A$
B. $A\subseteq <!--\; \subseteq \; -->B$
C. Each of the sets $A–B$,$B–A$ and $A\cup <!--\; \cup \; -->B$ is non-empty
D. None of the above
how to solve this without using any such plotter?

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
$\begin{array}{lcl}x& \le <!--\; \le \; -->& y\\ x& \ge <!--\; \ge \; -->& y\\ x+y& \le <!--\; \le \; -->& 1\\ x+y& \ge <!--\; \ge \; -->& 1\end{array}$
has 1 unique solution, namely $x=y=\frac{1}{2}$, while the system
$\begin{array}{lcl}x& \ge <!--\; \ge \; -->& y\\ x+y& \le <!--\; \le \; -->& 1\\ x+y& \ge <!--\; \ge \; -->& 1\end{array}$
has infinitely many solutions. Is there a away to find out how many solutions a system of linear inequalities has, if any?

How can we use Fourier-Motzkin elimination on system of inequalities with positive coefficients preceding each variable $x_1$ to $x_2$. Obviously, in this case we will only have an upper bound as a solution but how do we find it.
For example:
$5x_1+3x_2\le <!--\; \le \; -->8$
$2x_1+5x_2\le <!--\; \le \; -->15$

Inversion of a matrix in a system of linear inequalities
I would like to know if someone knows sufficient conditions on $A\in <!--\; \in \; -->{\mathbb{R}}^{n\times <!--\; \times \; -->n}$ and $b\in <!--\; \in \; -->{\mathbb{R}}^n$ such that for all $x\in <!--\; \in \; -->{\mathbb{R}}^n$:
$Ax\le <!--\; \le \; -->b\Rightarrow <!--\; \Rightarrow \; -->x\le <!--\; \le \; -->A^{-<!--\; -\; -->1}b\text{ or }x\ge <!--\; \ge \; -->A^{-<!--\; -\; -->1}b.$
$A$ is assumed to be positive definite.

Suppose, for an unknown quantity $x\in <!--\; \in \; -->(0,1)$ you observe the signs (positive, negative or zero) of all possible quantities of the form $\sum <!--\; \sum \; -->a_n{x}^n$ where $a_n\in <!--\; \in \; -->\{-<!--\; -\; -->1,1,0\}$ for $n\in <!--\; \in \; -->\{0,1,2,\mathrm{3...}\}$
Of course, if you encounter something like $x=0$ or $x^2+x=0$ among the system of (in)equalities, you can find a unique solution for $x$.
But, can you always find a unique solution for $x$ from there given the infinite set of (in)equalities at your disposal? Or is there a counter-example?

An airline company wants to install an air connection between two cities. Their goal is to transport $16000$ people and $96$ tons of cargo in a specific interval of time. At the moment, two types of airplanes are available: $11$ airplanes of type A and $8$ airplanes of type B. Type A costs $4000$ € and can carry $200$ people, and $6$ tons cargo per flight. Type B costs $1000$ € and can carry $100$ people, and $15$ tons cargo per flight.
We are supposed to determine the system of linear inequalities (and then continue calculating the best possible solution). But I would like to know only if the inequalities I found are correct, and maybe some other important things that need to be taken from the question?
So we have as constraints:
$I:200x+100y\ge <!--\; \ge \; -->1600$
$II:6x+15y\ge <!--\; \ge \; -->96$

Find a solution to the following system or prove that none exists:
$\begin{array}{rl}{x}_1-<!--\; -\; -->x_2& \le <!--\; \le \; -->4\\ x_1-<!--\; -\; -->x_5& \le <!--\; \le \; -->2\\ x_2-<!--\; -\; -->x_4& \le <!--\; \le \; -->-<!--\; -\; -->6\\ x_3-<!--\; -\; -->x_2& \le <!--\; \le \; -->1\\ x_4-<!--\; -\; -->x_1& \le <!--\; \le \; -->3\\ x_4-<!--\; -\; -->x_3& \le <!--\; \le \; -->5\\ x_4-<!--\; -\; -->x_5& \le <!--\; \le \; -->10\\ x_4-<!--\; -\; -->x_3& \le <!--\; \le \; -->-<!--\; -\; -->4\\ x_5-<!--\; -\; -->x_4& \le <!--\; \le \; -->-<!--\; -\; -->8\end{array}$

Find the maximums and minimums of $z=15x+14y$ with constraints $0\le <!--\; \le \; -->x\le <!--\; \le \; -->10,0\le <!--\; \le \; -->y\le <!--\; \le \; -->5,3x+2y\ge <!--\; \ge \; -->6$
I obviously can't take the partial derivatives of inequalities, so I'm at a loss here as my brain simply can't analyze/visualize 3 simultaneous constraints on the graph at once. What is the easiest/most efficient way of analyzing this system?

Find the number of integer numbers of $a$ such that the matrix
$\left(\begin{array}{ccc}-<!--\; -\; -->1& 0& 5\\ 0& a-<!--\; -\; -->50& 0\\ 5& 0& 6-<!--\; -\; -->a\end{array}\right)$
is negative definite.
For that, the eigenvalues must be negative, right?
So all these have to be negative and we get a system of inequalities:
$$\begin{gathered} a-<!--\; -\; -->50<0 \\ \frac{1}{2}(-<!--\; -\; -->\sqrt{a^2-<!--\; -\; -->14a+149}-<!--\; -\; -->a+5)<0 \\ \frac{1}{2}(\sqrt{a^2-<!--\; -\; -->14a+149}-<!--\; -\; -->a+5)<0 \end{gathered}$$
So we get from the first $a<50$, from the second $a\in <!--\; \in \; -->\mathbb{R}$ and from the last one $a>31$.
So the integer values for a are from $32$ to $49$, so there are $49-<!--\; -\; -->32+1=18$ values, right?

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