Solving System of Inequalities with Examples

Using logical OR to combine inequalities.
I have a physical system that must satisfy one of two inequalities:
$x\le <!--\; \le \; -->y$ or $p\le <!--\; \le \; -->q$
But not necessarily both simultaneously.
Is there a way to combine this into a single inequality?

Find the minimum value of C subject to the given constraints
$C=2x+5y$
Constraints:
$x+y\ge <!--\; \ge \; -->2$
$2x-<!--\; -\; -->3y\le <!--\; \le \; -->-<!--\; -\; -->6$
$3x-<!--\; -\; -->2y\ge <!--\; \ge \; -->6$
A. $-<!--\; -\; -->42$
B. $-<!--\; -\; -->4$
C. $-<!--\; -\; -->49$
D. $-<!--\; -\; -->10$

Let $P_1$ and $P_2$ being two polytopes defined by their vertices.
Is there any efficient way to compute $P_3=P_1\cap <!--\; \cap \; -->P_2$?

Given a system of n linear equations. Prove that the system is inconsistent if and only if you can obtain $0=1$, by using linear combinations.
I do not want to apply theorems from linear algebra here. Instead of it I want to use Farkas' lemma.
Every equation can be rewrited in form of two inequalities with $\le <!--\; \le \; -->$ and $\ge <!--\; \ge \; -->$, now I got $2n$ inequalities instead of $n$ equations.
I think that's the moment for me to apply Farkas' lemma, but why I can always get $0\le <!--\; \le \; -->1$ and $0\ge <!--\; \ge \; -->1$?

Is it possible to solve a linear inequality system using SVD?
I have a large linear inequality system of the form $Ax\le 0$

trying to solve a systems of equations with one inequality
I am trying to create a website that would run off this mathematical formula. I have tried to solve it but I got that there was no answer. I am only in pre-algebra and want a second opinion on if I got the equation right.
$1.2=x-<!--\; -\; -->y$
$x>1$
$y>1$
$x>y+2.3$
Is there a solution and if so how do you get it?

Let $(a_1,a_2,...,a_n)\in <!--\; \in \; -->{\mathbb{R}}^{\mathbb{n}}$ and $b\in <!--\; \in \; -->\mathbb{R}$. Prove that the set:
$F_1:=\{(x_1,x_2,...,x_n)\in <!--\; \in \; -->{\mathbb{R}}^{\mathbb{n}}|\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{a}_i{x}_i\ge <!--\; \ge \; -->b\}$
is closed in ${\mathbb{R}}^{\mathbb{n}}$

Consider a linear optimization program like the following
$max\frac{\underset{i=1}{\overset{T}{\sum <!--\; \sum \; -->}}{x}_i}{T}$ s.t. $a{x}_i+b{x}_{i+1}\le <!--\; \le \; -->1,\mathrm{\forall <!--\; \forall \; -->}i\ge <!--\; \ge \; -->1,x_i\ge <!--\; \ge \; -->0$
Assume that we take $T\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$. Say that $a>0,b>0$ and hence the constraints make sure we have some nice properties like $x_i$ have to be in some bounded region.
There is a lot of structure in this problem and its almost 'symmetric' if we shift every index by 1. I wonder if something can be said about the solution. For example $x_i=c\mathrm{\forall <!--\; \forall \; -->}i$. Does this hold for more general linear inequalities that are almost shift-invariant?
There are some similarities to this problem with MDPs and in particular it resembles an Average reward maximization problem. However nothing is random in this problem

Consider a Linear Programming problem in dictionary form,
$max\{f^{\mathrm{\pi <!--\; \pi \; -->}}+\underset{j\in <!--\; \in \; -->D(\mathrm{\pi <!--\; \pi \; -->})}{\sum <!--\; \sum \; -->}{d}_j^{\mathrm{\pi <!--\; \pi \; -->}}{x}_j|\mathrm{\forall <!--\; \forall \; -->}i\in <!--\; \in \; -->B(\mathrm{\pi <!--\; \pi \; -->})b_i^{\mathrm{\pi <!--\; \pi \; -->}}+\underset{j\in <!--\; \in \; -->D(\mathrm{\pi <!--\; \pi \; -->})}{\sum <!--\; \sum \; -->}{G}_{ij}^{\mathrm{\pi <!--\; \pi \; -->}}{x}_j\ge <!--\; \ge \; -->0,\mathrm{\forall <!--\; \forall \; -->}j\in <!--\; \in \; -->D(\mathrm{\pi <!--\; \pi \; -->})x_j\ge <!--\; \ge \; -->0\},$
where indices $\mathrm{\pi <!--\; \pi \; -->}$ uniquely define the coefficients of each basic inequality system.
Now assume that the problem is both primal and dual infeasible. Primal infeasibility implies that there always exists a "primal-infeasible" basis α with some $r\in <!--\; \in \; -->B(\mathrm{\alpha <!--\; \alpha \; -->})$ having $b_r^{\mathrm{\alpha <!--\; \alpha \; -->}}<0$ and $\mathrm{\forall <!--\; \forall \; -->}j\in <!--\; \in \; -->D(\mathrm{\alpha <!--\; \alpha \; -->})G_{rj}^{\mathrm{\alpha <!--\; \alpha \; -->}}\le <!--\; \le \; -->0$, while dual infeasibility implies that there exists a (possibly different) "dual-infeasible" basis $\mathrm{\beta <!--\; \beta \; -->}$ with some $s\in <!--\; \in \; -->D(\mathrm{\beta <!--\; \beta \; -->})$ having $d_s^{\mathrm{\beta <!--\; \beta \; -->}}>0$ and $\mathrm{\forall <!--\; \forall \; -->}i\in <!--\; \in \; -->B(\mathrm{\beta <!--\; \beta \; -->})G_{is}^{\mathrm{\beta <!--\; \beta \; -->}}\ge <!--\; \ge \; -->0$.
It is easy to construct doubly infeasible problems where both of the above properties appear together at some of its bases, that is, where $\mathrm{\alpha <!--\; \alpha \; -->}=\mathrm{\beta <!--\; \beta \; -->}$. But is that true for all doubly infeasible problems? By a case by case analysis, I was able to prove that a counterexample to the existence of such doubly infeasible bases would have at least $|D(\mathrm{\pi <!--\; \pi \; -->})|\ge <!--\; \ge \; -->3$ and $|B(\mathrm{\pi <!--\; \pi \; -->})|\ge <!--\; \ge \; -->3$.
Does a doubly infeasible problem always have bases that are simultaneously primal-infeasible and dual-infeasible? Alternatively, is there a doubly infeasible problem such that the above two properties do only appear at different bases?

How to determine bounds on one variable in a system of inequalities?
I am interested in the point of 'cross-over' between a generalised harmonic number where the denominator of the summand is raised to a power, and a non-exponential harmonic sum operating on some subset of the natural numbers.
For example, take the generalised harmonic number $H_x^{(k)}=\underset{n=1}{\overset{x}{\sum <!--\; \sum \; -->}}\frac{1}{n^k}$, and a harmonic number operating only on odd denominators $G_x=\underset{n=1}{\overset{x}{\sum <!--\; \sum \; -->}}\frac{1}{2n-<!--\; -\; -->1}$.
Clearly, there exist values of x,k such that $G_x<H_x^{(k)}$ and values such that $H_x^{(k)}<G_x$. Thus there exists a value $c=G_{x_0}$ such that
$G_{x_0}=c<H_{x_0}^{(k)}=\underset{n=1}{\overset{x}{\sum <!--\; \sum \; -->}}\frac{1}{n^k}$
and
$H_{x_0+2}^{(k)}<G_{x_0+2}=c+\frac{1}{2x_0+1}+\frac{1}{2x_0+3}$
or
$H_{x_0+2}^{(k)}-<!--\; -\; -->c<\frac{1}{2x_0+1}+\frac{1}{2x_0+3}$
The values of $c,x_0,k$ are obviously co-dependent. I am searching for a way to solve for $x_0$ or at least put bounds on it.
I am interested in how to approach this algebraically rather than numerically. This is a single simple example of $G$ and I want to be able to explore how to solve such problems generally, for whatever pattern of $G$ I choose (provided it's formulable!).
Algebraically, how do I put bounds on $x_0$ in terms of $c,k$?

How to prove that there is no positive integer solution to this two variables inequalities system?
$\{\begin{array}{lll}77n& -<!--\; -\; -->154m& \le <!--\; \le \; -->-<!--\; -\; -->1\\ 154m& -<!--\; -\; -->77n& \le <!--\; \le \; -->76\end{array}$

Differential equation for a spring mass system
$m{y}^{″}+\mathrm{\beta <!--\; \beta \; -->}{y}^{\prime }+ky=0$
I have to find the displacement y at any time $t$. $\mathrm{\beta <!--\; \beta \; -->}=2\sqrt{mk}$ and the initial conditions are $y(0)=h>0$ and $y^{\prime }(0)=v$. I initally found the following
$y=(a+bt)e^{\frac{-<!--\; -\; -->t\sqrt{mk}}{m}}$
I am currently stuck with the initial conditions as I end up with $b=\frac{-<!--\; -\; -->v\sqrt{mk}}{ak}$ from$y^{\prime }(0)=v$ and I don't know how to interpret $y(0)=h>0$
Now have
$y=(h+v+ht\sqrt{\frac{k}{m}})e^{\frac{-<!--\; -\; -->t\sqrt{mk}}{m}}$
and have to express y=0 as an inequality $v<-<!--\; -\; -->f(h,k,m)$ for the function $f$

How to solve the following system of inequalities rigorously
$\{\begin{array}{l}\mathrm{\Phi <!--\; \Phi \; -->}\ge <!--\; \ge \; -->\frac{1}{2}(x^2-<!--\; -\; -->y^2-<!--\; -\; -->z^2)+x+y+z-<!--\; -\; -->\frac{3}{8},\\ \mathrm{\Phi <!--\; \Phi \; -->}\le <!--\; \le \; -->\frac{1}{2}(x^2-<!--\; -\; -->y^2-<!--\; -\; -->z^2)+\frac{1}{2}x+y+z-<!--\; -\; -->\frac{9}{8}\\ \mathrm{\Phi <!--\; \Phi \; -->}\le <!--\; \le \; -->\frac{1}{2}(x^2-<!--\; -\; -->y^2-<!--\; -\; -->z^2)+\frac{1}{8}\\ y\ge <!--\; \ge \; -->\frac{1}{2},z\ge <!--\; \ge \; -->1\end{array}$
The task is find the range of $\mathrm{\Phi <!--\; \Phi \; -->}$ such that this system of inequalities in $x,y,z$ has solutions in $\mathbb{R}$.

need to build the optimal controller, i.e. one that maximizes:
$J={\int <!--\; \int \; -->}_0^{t_f}f(u)\mathrm{d}t$
For the following time-dependent system:
$\stackrel{\dot{}<!--\; \dot{}\; -->}{x}=g(x,u,t)$
where $x(t)\in <!--\; \in \; -->\mathbb{R}$ is the state, $u(t)\in <!--\; \in \; -->\mathbb{R}$ the input, $t\in <!--\; \in \; -->\mathbb{R}$ the time and $l(t)\in <!--\; \in \; -->\mathbb{R}$ the time-varying lower bound for the state. Using the Pontryagin's maximum principle, I have defined the Hamiltonian as: $\mathcal{H}=\mathrm{\psi <!--\; \psi \; -->}g(x,u,t)+f(u)$ and used the necessary condition $\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\psi <!--\; \psi \; -->}}=-<!--\; -\; -->\frac{\mathrm{\partial <!--\; \partial \; -->}\mathcal{H}}{\mathrm{\partial <!--\; \partial \; -->}x}$. As you can see I have completely ignored the time-dependent inequality, so the solution I get is correct, but of course it doesn't enforce the lower bound $l$ on the state value.
How can I rewrite the hamiltonian so that also the inequality is considered?

System of inequalities. Points of intersection?
$x^2+y^2<=81$
$y<x$

When does a given system of linear inequalities form a bounded convex polytope?
We know that a Closed Convex Polytope may be regarded as the set of solutions to the system of linear inequalities:
$\begin{array}{c}{a}_{11}{x}_1+a_{12}{x}_2+\cdots <!--\; \cdots \; -->+a_{1n}{x}_n\le <!--\; \le \; -->b_1\\ a_{21}{x}_1+a_{22}{x}_2+\cdots <!--\; \cdots \; -->+a_{2n}{x}_n\le <!--\; \le \; -->b_2\\ ⋮<!--\; ⋮\; -->\\ a_{m1}{x}_1+a_{m2}{x}_2+\cdots <!--\; \cdots \; -->+a_{mn}{x}_n\le <!--\; \le \; -->b_m\end{array}$
This can be concisely written as the matrix inequality $Ax\le b$, where $A$ is an $m\times n$ matrix, $x$ is an $n\times 1$ column vector of variables, and $b$ is an $m\times 1$ column vector of constants.

Variable intervals from system of inequalities
$\left(\begin{array}{cc}-<!--\; -\; -->1& 0\\ 1& 0\\ 0& -<!--\; -\; -->1\\ 1& -<!--\; -\; -->1\\ -<!--\; -\; -->2& 1\end{array}\right)\left(\begin{array}{c}i\\ j\end{array}\right)<\left(\begin{array}{c}20\\ 20\\ 10\\ 0\\ 20\end{array}\right)$
and I need to find possible intervals of i and j.

Let $X$ be an inner product space, let $v\in <!--\; \in \; -->X$, $a\in <!--\; \in \; -->\mathbb{R}$.
Let us consider the system of equations:
$\Vert <!--\; \Vert \; -->x\Vert <!--\; \Vert \; -->=1,⟨<!--\; ⟨\; -->x,v⟩<!--\; ⟩\; -->=a.$
If it has a solution, then by the Schwarz inequality $|a|\le <!--\; \le \; -->\Vert <!--\; \Vert \; -->v\Vert <!--\; \Vert \; -->$. Is it true the converse, if $|a|\le <!--\; \le \; -->\Vert <!--\; \Vert \; -->v\Vert <!--\; \Vert \; -->$, then that sysem has a solution?

What is the solution of the following system of inequalities?
$x+y\le <!--\; \le \; -->0$
$x^2+y^2\le <!--\; \le \; -->1$

Let $(A,\mathfrak{m}\mathfrak{)}$ be a local Noetherian ring and let $x_1,\dots <!--\; \dots \; -->,x_d$ be a system of parameters, i.e. $\mathfrak{m}=(x_1,\dots <!--\; \dots \; -->,x_d)$. Then
$\dim <!--\; \; -->A/(x_1,\dots <!--\; \dots \; -->,x_i)=d-<!--\; -\; -->i$
$i=1,\dots <!--\; \dots \; -->,d$
I know just a few basic facts about dimension theory. I think I can prove the inequality $\le <!--\; \le \; -->$ via Krull's Hauptidealsatz in this way: the maximal ideal of $A/(x_1,\dots <!--\; \dots \; -->,x_i)$ is
${\mathfrak{m}}_i:=(\stackrel{¯<!--\; ¯\; -->}{x_{i+1}},\dots <!--\; \dots \; -->,\stackrel{¯<!--\; ¯\; -->}{x_d}).$
So it must be $\mathrm{h}\mathrm{t}({\mathfrak{m}}_i)\le <!--\; \le \; -->d-<!--\; -\; -->i$.
But how to prove the other inequality? I think I should do it by induction, but I cannot understand how to begin. So, if what I said so far is right, my question is: how can I prove that
$\dim <!--\; \; -->A/(x_1)\ge <!--\; \ge \; -->d-<!--\; -\; -->1?$