Solving System of Inequalities with Examples

How to prove if there are exist positive integer solutions to these two variables inequalities system?
$\{\begin{array}{lll}141n& -<!--\; -\; -->143m& \le <!--\; \le \; -->-<!--\; -\; -->60\\ 143m& -<!--\; -\; -->141n& \le <!--\; \le \; -->138\end{array}$

Cauchy inequalities in systems of equations
For positive real numbers solve the system of equations
$\{\begin{array}{l}\begin{array}{r}{x}_1+x_2+···+x_n=\frac{1}{4}\\ \frac{1}{x_1}+\frac{4}{x_2}+···+\frac{n^2}{x_n}=n^2(n+1{)}^2\end{array}\end{array}$

How to determine whether a system of linear inequalities has a positive solution or not?

Solve in positive real numbers the system of equations:
$\begin{array}{c}(x^3+y^3)(y^3+z^3)(z^3+x^3)=8,\\ \frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac{3}{2}.\end{array}$

How to solve this system of linear inequalities?
$\begin{array}{rl}{\mathrm{\theta <!--\; \theta \; -->}}_1+{\mathrm{\theta <!--\; \theta \; -->}}_2& \ge <!--\; \ge \; -->0\\ 2{\mathrm{\theta <!--\; \theta \; -->}}_1+k{\mathrm{\theta <!--\; \theta \; -->}}_2& \ge <!--\; \ge \; -->0\\ {\mathrm{\theta <!--\; \theta \; -->}}_1+3{\mathrm{\theta <!--\; \theta \; -->}}_2& \le <!--\; \le \; -->0\end{array}$

Solve the system:
$|3x+2|\ge <!--\; \ge \; -->4|x-<!--\; -\; -->1|$
$\frac{x^2+x-<!--\; -\; -->2}{2+3x-<!--\; -\; -->2x^2}\le <!--\; \le \; -->0$

Proving the image of a convex polyhedron under a linear map is a polyhedron
prove for $A\in <!--\; \in \; -->{\mathbb{R}}^{m\times <!--\; \times \; -->n}$ and a convex polyhedron $Q\subseteq <!--\; \subseteq \; -->{\mathbb{R}}^n$ that the set
is also a convex polyhedron. However, I am asked to do so using the following statement about convex polyhedra (which is easy to prove):
$P\subseteq <!--\; \subseteq \; -->{\mathbb{R}}^{m+n}\text{ is a polyhedron }⟹<!--\; ⟹\; -->\{x\in <!--\; \in \; -->{\mathbb{R}}^n\mid <!--\; \mid \; -->(x,y)\in <!--\; \in \; -->P\text{ for some }y\in <!--\; \in \; -->{\mathbb{R}}^m\}.(1)$
This seems like it should be easy but I'm having trouble.
One approach is to write $Q$ as a set of linear inequalities
$Q=\{x\in <!--\; \in \; -->{\mathbb{R}}^n\mid <!--\; \mid \; -->Bx\le <!--\; \le \; -->b\}$
and then try to write $A(Q)$ as as system of linear inequalities
$A(Q)=\{y\in <!--\; \in \; -->{\mathbb{R}}^m\mid <!--\; \mid \; -->B{A}^{-<!--\; -\; -->1}y\le <!--\; \le \; -->b\}.$
This doesn't quite work since A may not be invertible. More importantly, it does not use the statement (#1) given above.

The question reads: For which real numbers k is the zero state a stable equilibrium of the dynamic system $x_{t+1}=A{x}_t$?
$A=\left[\begin{array}{cc}0.1& k\\ 0.3& 0.3\end{array}\right]$
So, my thought is I need to find the eigenvalues. In order to do this I calculated the characteristic polynomial as
$x^2-<!--\; -\; -->0.4x+0.03-<!--\; -\; -->0.3k=0$, with $x$ representing eigenvalues.
Using the quadratic formula I found that the (real) eigenvalues are
$x=\frac{2\pm <!--\; \pm \; -->\sqrt{1+30k}}{10}$
and for the zero state to be in stable equilibrium $\sqrt{1+30k}<8$. Hence, $k<21/10$ (for stable equilibrium).
My question is how do I figure out the values for $k$ if the eigenvalues are complex?
Do I solve the inequality $\sqrt{-<!--\; -\; -->1-<!--\; -\; -->30k}<8$?

$x_i+y_i\cos <!--\; \; -->(p_{ik}-<!--\; -\; -->a_i)-<!--\; -\; -->z_i{q}_{ik}>x_j+y_j\cos <!--\; \; -->(p_{jk}-<!--\; -\; -->a_j)-<!--\; -\; -->z_j{q}_{jk}$
$x_i+y_i\cos <!--\; \; -->(p_{in}-<!--\; -\; -->a_i)-<!--\; -\; -->z_i{q}_{in}<x_m+y_m\cos <!--\; \; -->(p_{mn}-<!--\; -\; -->a_m)-<!--\; -\; -->z_m{q}_{mn}$
All $p$ and $q$ are parameters, I need to solve for $x_i$, $y_i$, $z_i$, $a_i$ for all $i$.

SAT inequality problem
$y\le 3x+1$
$x-<!--\; -\; -->y>1$
Which of the following ordered pairs $(x,y)$ satisfies the system of inequalities above?
$A)(-2,-1)$
$B)(-1,3)$
$C)(1,5)$
$D)(2,-<!--\; -\; -->1)$

Systems of Absolute Inequalities
$|3-<!--\; -\; -->2x|\ge <!--\; \ge \; -->1$
$x^2/(x+2)\le <!--\; \le \; -->0$
Please advise on this with the answer to this.

For $A\in <!--\; \in \; -->B(H)$ where $H$ is a separable Hilbert space (so a countable orthonormal basis exists or also known as a maximal orthonormal system) where $A$ is also a Hilbert-Schmidt operator and $B\in <!--\; \in \; -->B(H)$ (so $B$ does not need to be a Hilbert-Schmidt operator. It is just linear and bounded.) How do I show that $||AB|{|}_{HS}\le <!--\; \le \; -->||A|{|}_{HS}||B|{|}_H$ and $||BA|{|}_{HS}\le <!--\; \le \; -->||A|{|}_{HS}||B|{|}_H$ where $HS$ is the Hilbert Schmidt norm and $H$ is just the standard operator norm. I have never been introduced to the trace operator so I can't use that. I can only use the standard Hilbert space inequalities like Cauchy-Schwarz, Parseval's and Bessel's inequality as well as the fact that $\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}_H\le <!--\; \le \; -->\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}_{HS}$ I have not been able to turn the left hand side anywhere close to the right hand side so I'm not sure if there is some trick I'm missing or some inequality I haven't thought of.

Sign of a Linear Expression in Mixed Integer Linear Programming
I have a question about a system in mixed integer programming I am trying to solve using GLPK though the question mathematical and is software agnostic.
I have converted the whole problem to linear equations and inequalities except for one part. Suppose I have a real number $x$ and a binary variable $b$ (either 0 or 1). How can I "tie them together" such that $x>0$ if and only if $b=1$?

mixed system of equalities and inequalities, for example:
$$\begin{gathered} x+2y+3z=10 \\ 2x+4y+10z=20 \\ 4x+y+z<10 \end{gathered}$$
where unknown variables $x,y,z$ are all real-valued.
How can get a solution (or the range of all feasible solutions) for this system?

What is an equation to determine if a 3d point (x,y,z) is within a rectangular prism on a cartesian coordinate system? I want to check if a point is inside a rectangular prism.
e.g. the equation for a 2d rectangle is |x+y|+|x-y|<=2.

Let $P\subseteq <!--\; \subseteq \; -->{\mathbb{R}}^2$ be the convex hull of the points $(0,0)$, $(3,1)$ and $(-<!--\; -\; -->1,2)$.
Exercise: Give a TDI system $Ax\le <!--\; \le \; -->b$ describing $P$ with $A$ and $b$ integral.
I know the following:
Many combinatorial min-max relations can be understood as a result of LP-duality
$max\{w^{T}x:Ax\le <!--\; \le \; -->b\}=min\{y^{T}b:y^{T}A=w^T,y\ge <!--\; \ge \; -->0\}$
combined with integrality of optimal solutions on the primal and dual side.
Def: Let $Ax\le <!--\; \le \; -->b$ be a rational system of linear inequalities and let $P:=\{x:Ax\le <!--\; \le \; -->b\}$ be the associated polyhedron. The system $Ax\le <!--\; \le \; -->b$ is Totally Dual Integral (TDI) if for every integral objective vector w, the minimum in the dual is attained by an integral vector $y$ (if the minimum is finite).
What I should do: I need to find a TDI system $Ax\le <!--\; \le \; -->x$ such that $P=\{x:Ax\le <!--\; \le \; -->b\}$ is the convex hull of the points $(0,0),(3,1)$ and $(-<!--\; -\; -->1,2)$ and that $A$ and $b$ are integral. So I think I probably should identify $P$ first.

I was trying to solve this inequality with two absolute values:
$|2x-<!--\; -\; -->3|+7\le <!--\; \le \; -->3x-<!--\; -\; -->3|x-<!--\; -\; -->7|$
I've got an empty set of solutions, but it's not correct.

The problem at hand can be summarized as follows: we have to allocate a ressource to $n$ production units. The allocation to production unit $i$ is $x_i$. Each of the production unit will produce at different rate. Let us define $\mathbf{r}=(r_1,...,r_n)$ where the production rate of unit $i$ is $r_i$.
We have also a method to come up with an initial allocation ($(x_1,...,x_n)$ for a total desired production $R$ (a given constant). We do not give any details about this method because it is irrelevant to the question.
So the method finds a solution $\mathbf{x}=(x_1,...,x_n)$ to the system
$R=\mathbf{r}{\mathbf{x}}^T$
where $\mathbf{x}\ge <!--\; \ge \; -->\mathbf{0}$.
We also regroup the production units into categories. We would like to obtain a minimum production from certain categories and a maximum production in other categories. This can be expressed by a system of linear inequalities $B\mathbf{x}\ge <!--\; \ge \; -->\mathbf{b}$.
The initial solution found by the method does not necessarily meets all the constraints.
My Questions:
1) What would be the best approach to shift the production from one category of production units to another when the constraints are not met (and still have $R=\mathbf{r}{\mathbf{x}}^T$)? We would like the final solution not too far from the initial one and the all the constraints are respected.
2) Can you provide some reference (papers, website, etc.) for this type of problems?

For what a does this system of inequalities have only one solution?
$\{\begin{array}{l}y\ge <!--\; \ge \; -->x^2+4a^2\\ x\ge <!--\; \ge \; -->y^2+4a^2\end{array}$
So, I have tried getting $y$ from the first equation then putting it into second, but then I have two quadratic and $x$, which of these quadratics are the first term? I know that I have to put it into $\mathrm{\Delta <!--\; \Delta \; -->}=0$. But don't really get which of them are first, second etc. Even tried subtracting second to first.

How can I prove that for any real $(x,y)$ root this system of equations
$x^4+y^2=(a+\frac{1}{a}{)}^3$
$x^4-<!--\; -\; -->y^2=(a-<!--\; -\; -->\frac{1}{a}{)}^3$
the inequation $x^2+|y|\ge <!--\; \ge \; -->4$ is always true?
Also, when will $x^2+|y|=4$ be true for this system?