Solving System of Inequalities with Examples

Suppose we are given a rational $q$. Is it possible that there infinitely many integer solutions $(h,k)$ to following system of inequalities: $0<|q-<!--\; -\; -->h/k|<1/2k^2$? I think that it is easy to see that for $q=1/2$ (and probably for any rational $a/b$ with $2|b$ and $gcd(a,b)$$=1$) it is impossible, but what about other rationals? What about changing constant in the right inequality (making it stronger or weaker)?

I am looking for a way to express an "or" option in a system of linear inequalities for a linear program I am working on.
I will explain what I mean precisely: Lets say I have a set of inequalities $e{q}_1$ to $e{q}_n$, which must hold. But, additionally, I have several more inequalities $eq{a}_1,....,eq{a}_k$ and $eq{b}_1,...,eq{b}_k$, that can be partitioned to couples (of the form $⟨eq{a}_i,eq{b}_i⟩$ for all $1\le <!--\; \le \; -->i\le <!--\; \le \; -->k$), such that either $eq{a}_i$ holds or if it doesn't then $eq{b}_i$. Nonetheless, it is also possible for both of them to hold. What I am looking for is an option to encode such an "or" condition on a set of inequalities.
Ofcourse, if we say that $eq{a}_i$ is lets say $f(...)\le <!--\; \le \; -->K$ and $eq{b}_i$ is $g(...)\le <!--\; \le \; -->M$ I can write $f(...)>K+M$, but this demand is too strong, because in my original problem I may allow $g(...)\le <!--\; \le \; -->M$ (for example) as long as g(...)≤M, but then $f(...)+g(...)>K+M$. So requiring $f(...)+g(...)\le <!--\; \le \; -->K+M$ is too strong for me.
My question can be divided to 2 questions:
1. Is there any idea for a nice trick here or for some sort of a reduction to Linear Programming?
2. Perhaps there exists a software or an online site which allow this kind of "or"?

Could someone kindly suggest a method of solving the following constrained (equality and inequality) system of equations in the least squares fashion?
$\underset{x}{min}\frac{1}{2}\Vert <!--\; \Vert \; -->Ax-<!--\; -\; -->b{\Vert <!--\; \Vert \; -->}_2^2$
such that
$\begin{array}{c}{x}_i+x_j=1\text{for a certain }i,j\\ 0<x_k<{\mathrm{\tau <!--\; \tau \; -->}}_k\text{where }k\ne <!--\; \ne \; -->i,j\end{array}$
${\mathrm{\tau <!--\; \tau \; -->}}_k$ is the upper bound for $x_k$.

$x\ge 0$,
$y\ge 0$,
$2x+y<8$
$x+2y<10$
Does $x+y$ have a maximum value under the above conditions?

Definition [Monomial of max-degree 1]. Given $n$ variables $x_1,...,x_n$, a multivariate monomial of max-degree 1 is an expression of the form: $r(x_1^{e_1}\cdot <!--\; \cdot \; -->x_2^{e_2}\cdot <!--\; \cdot \; -->\cdots <!--\; \cdots \; -->\cdot <!--\; \cdot \; -->x_n^{e_n})$, where $r\in <!--\; \in \; -->\mathbb{Q}$ and all exponents $e_i$ are either 0 or 1.
For example $2(x_1{x}_2{x}_5)$ is a monomial of max-degree 1, but $3x_1^2$ is not.
Definition [Polynomial of max-degree 1]. A polynomial of max-degree 1 is a sum $f=m_1+\dots <!--\; \dots \; -->m_k$ of multivariate monomials of max-degree 1.
For example: $2x_1{x}_2+3x_1{x}_3$ is a Polynomial of max-degree 1.
Definition [System of Polynomial inequalities of max-degree 1]. A system of polynomial inequalities is a finite conjunction of inequalities of the form $f=0$ or $f=0$ or $f\le <!--\; \le \; -->0$.
I have came across this notion recently, and I am not at all an expert. I have the following question
QUESTION Can a system of polynomial inequalities of max-degree 1 have a solution in the reals R but none in rationals $\mathbb{Q}$? Any example?

A room can have up to 800 people. There are two types of tables: one, rectangular, seating 20 people; another, circular, that seats 9. If the organizers want to have approximately the same amount of rectangular and circular tables to seat the maximum of people possible, what combination of tables can they use?
Initially, I started off with a simple let statement, stating that $x$ will represent rectangular tables, and $y$ will represent circular tables. So my inequality so far would be something like $20x+9y\le <!--\; \le \; -->800$. I also know that $x\ge <!--\; \ge \; -->0$ and $y\ge <!--\; \ge \; -->0$. However, with this equation, I am not sure if I can actually find out an equal amount of rectangular and circular tables. I have also considered that this is a maximum/minimum problem that I have learned before, but the inequality I came up with is not a quadratic equation.
The question is multiple choice, and, substituting the correct values in, I can see that it works, but I have no idea how they arrived at the numbers. It seems to me that the question can be solved without this guess-and-check, so I'm interested in what you have to say about this problem.

How to solve the system of inequalities when using topological proof to show that there are exactly 5 platonic solids?
I have seen other answers explaining the topological proof up until the point of
$1/p+1/q>1/2$ and $p$, $q$ are greater than or equal to three
Then they proceed to say that the 5 platonic solids have the only values that satisfy these conditions. My question is how do I solve for these values?

Suppose I have a string of inequalities, i.e.
$f(x,y,z)<g(x,y,z)<h(x,y,z)$
Clearly, if I find when $g(x,y,z)<h(x,y,z)$ and $f<g$, then $f<h$ when both these conditions hold.
Is it ever useful to compare $f(x,y,z)<h(x,y,z)$ in addition to $f<g$ and $g<h$
Or is it useful to compare the conditions (assuming I can get them in a comparable form) from $g<h$ and $f<h$ to each other? Maybe comparing the two can give a single, simplified condition for $f<g<h$ to hold?

Comparison function for the Rossler system
$\begin{array}{rl}& {\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}_1=-<!--\; -\; -->x_2-<!--\; -\; -->x_3\\ & {\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}_2=x_1+\mathrm{\alpha <!--\; \alpha \; -->}{x}_2\\ & {\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}_3=\mathrm{\beta <!--\; \beta \; -->}+x_3(x_1-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->})\end{array}$
With $\mathrm{\alpha <!--\; \alpha \; -->}=\mathrm{\beta <!--\; \beta \; -->}=0.1$ and $\mathrm{\gamma <!--\; \gamma \; -->}=14$, and $x_3(0)>0$, such that $x_3>0$ $\mathrm{\forall <!--\; \forall \; -->}$ $t\ge <!--\; \ge \; -->0$.
we have found that for the comparison function $W=x_1^2+x_2^2+2x_3$ the following inequality holds:
$\stackrel{\dot{}<!--\; \dot{}\; -->}{W}\le <!--\; \le \; -->2\mathrm{\alpha <!--\; \alpha \; -->}W+2\mathrm{\beta <!--\; \beta \; -->}$
This implies that the solutions of this system are well defined on the infinite time interval $[0,\mathrm{\infty <!--\; \infty \; -->})$, and implies that solutions can not escape to infinity in finite time.

Consider the following system of inequalities.
$$\begin{gathered} x_1-x_2\le 3 \\ {x}_2-x_3\le -2, \\ {x}_3-x_4\le 10, \\ {x}_4-x_2\le \alpha , \\ {x}_4-x_3\le -4, \end{gathered}$$
where $\mathrm{\alpha <!--\; \alpha \; -->}$ is a real number. A value for α for which the system has a solution is
(a) −16,
(b)−12,
(c)−10,
(d) None.

How to prove this? The existence of solutions to linear inequalities
A system of real homogeneous linear inequalities ${\mathrm{\lambda <!--\; \lambda \; -->}}_i>0$, $i=1,2,\dots <!--\; \dots \; -->,m$, has a solution if and only if there is no nontrivial linear dependence with nonnegative coefficients among the ${\mathrm{\lambda <!--\; \lambda \; -->}}_i$. For example, ${\mathrm{\lambda <!--\; \lambda \; -->}}_i=\sum <!--\; \sum \; -->a_{ij}{x}_j$.

Determining for what values is a system of inequalities true.
$\begin{array}{}\text{(1)}& 8A^2\mathrm{\alpha <!--\; \alpha \; -->}+44A\mathrm{\alpha <!--\; \alpha \; -->}+60\mathrm{\alpha <!--\; \alpha \; -->}>3A^2{\mathrm{\alpha <!--\; \alpha \; -->}}^2+12A^2+18A{\mathrm{\alpha <!--\; \alpha \; -->}}^2+40A+27{\mathrm{\alpha <!--\; \alpha \; -->}}^2+40\end{array}$
$\begin{array}{}\text{(2)}& A>0\end{array}$
$\begin{array}{}\text{(3)}& \mathrm{\alpha <!--\; \alpha \; -->}>1\end{array}$
Is there a procedure that determines for what value range(s) of $\mathrm{\alpha <!--\; \alpha \; -->}$ inequalities (1) and (2) are satisifed?

Let $\mathrm{\Delta <!--\; \Delta \; -->}$ be an indecomposable root system in a real inner product space $E$, and suppose that $\mathrm{\Phi <!--\; \Phi \; -->}$ is a simple system of roots in $\mathrm{\Delta <!--\; \Delta \; -->}$, with respect to an ordering of $E$. If $\mathrm{\Phi <!--\; \Phi \; -->}=\{{\mathrm{\alpha <!--\; \alpha \; -->}}_1,\dots <!--\; \dots \; -->,{\mathrm{\alpha <!--\; \alpha \; -->}}_l\}$, prove that
${\mathrm{\alpha <!--\; \alpha \; -->}}_1+\cdots <!--\; \cdots \; -->+{\mathrm{\alpha <!--\; \alpha \; -->}}_l\in <!--\; \in \; -->\mathrm{\Delta <!--\; \Delta \; -->}$
I know that any positive root $\mathrm{\gamma <!--\; \gamma \; -->}$ may be written as a sum of simple roots, and furthermore that every partial sum is itself a root, but I am unsure if that will help me or not. Any hints to get me started?

Consider the folloing ODE system:
$x^{\prime }(t)=A(t)x(t)+b(t)$
$x(t_0)=x_0$
Show that the solutions of the system are globally defined.

When is $||Ax|{|}_2\le <!--\; \le \; -->||x|{|}_2$ true for all $x$?

One similiarly shows, with the help of Farkas' lemma (...): if a system Ax <= b of n variables has no solution, then Ax <= b has a subsystem A'x <= b' of at most n + 1 inequalities having no solution.

Help interpreting question about a system of equations
For fixed $a,b,m$, and $n$, solve the system of equations:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $mx+ny=k$
choosing $k$ so that the resulting quadratic equation has a double root. I know how to solve a system of equations, but I'm not entirely sure what they're asking for here. I'm not really sure how I'd even end up with a factorable quadratic. Maybe just add the two together and try to factor?

Solving a system of inequalities in modulo N
I have a problem that boils down to two unknowns, $X_1$ and $X_2$, where:
$X_1\cdot <!--\; \cdot \; -->M+AmodN=X_2$
And:
$X_1<<!--\; <\; -->L_{1}modN$
$X_2<<!--\; <\; -->L_{2}modN$

Say I have a system of linear equalities and inequalities with integer coefficients in n variables, and let $R^n$ be the space of all possible solutions. I know that $\stackrel{\to <!--\; \to \; -->}{0}$ is a solution.
Is there any efficient algorithm to check if there are any other solutions but zero? In other words, given a linear optimization problem, is there a way to check if the feasible region is a point?

Given a number of $3in$ squares and $2in$ squares, how many of each are needed to get a total area of 35 $i{n}^2$?
Through quick trial and error (the method they wanted I believe) you find that you need 3 $3in$ squares and 2 $2in$ squares, but I got to thinking on how to solve this exactly.
You have 2 unknowns and the following info:
$4x+9y=35$
$x\ge <!--\; \ge \; -->0,y\ge <!--\; \ge \; -->0,x\text{and}y$ are both integers.
It also follows then that $x\le <!--\; \le \; -->8\text{and}y\le <!--\; \le \; -->3$