Solving System of Inequalities with Examples

Having its center on the line x=2 and passing through the point of intersection between the circles $x^2+y^2-<!--\; -\; -->3x=0\text{and}{x}^2+y^2=3y$

Center on the y-axis and touching the line x+y+2=0 at (2, -4)

Through (2,1) and (3.5) and having its center on the line 8x+5y=8

Graph the inequality y< 3x - 2 on a plane

Concentric with thw circle$x^2+y^2+8x-<!--\; -\; -->3y+16=0$ and touching the line 4x+3y=12

Graph the rational function: f(x) = $\frac{2x^2}{x^2-<!--\; -\; -->1}$ andgive the following information:
Domain:
y intercept:
Horizontal asymptote:
Vertical Asymptote(s):
x intercept(s):
Complete the Table:
x f(x)
-3
-2
-0.5
0.5
2
3
(b) using the answer from (A), solve the rationalequality:
$\frac{2x^2}{x^2-<!--\; -\; -->1}<0$

Find the intersection of two or more polygons in terms of linear inequalities
Given two or more closed polygons each defined by a system of linear inequalities, is there any method by which their intersection polygon may be determined, also in terms of a system of linear inequalities?

Maximization problem with inequalities restriction
Function $g(x,y,z)$, and $x+y+z=1,x\ge <!--\; \ge \; -->0,y\ge <!--\; \ge \; -->0,z\ge <!--\; \ge \; -->0$. Now I want to maximize $g$.
If I ignore the inequalities, then I can use lagrangian and can solve this thing for maximum. But I am not sure how to incorporate the inequalities in the system of equations.

Let's consider the following linear inequalities:
$$\begin{gathered} a-<!--\; -\; -->10\le <!--\; \le \; -->b\le <!--\; \le \; -->a-<!--\; -\; -->7 \\ b+3\le <!--\; \le \; -->c\le <!--\; \le \; -->b+6 \\ c+3\le <!--\; \le \; -->d\le <!--\; \le \; -->c+6 \\ d+3\le <!--\; \le \; -->e\le <!--\; \le \; -->d+6 \end{gathered}$$
Is there a way to find a solution to this system where $max(|a|,|b|,|c|,|d|,|e|)$ is minimized?

Consider the following linear system
$y=A_1{x}_1+A_2{x}_2$
subject to the linear constrains
$C_1{x}_1+C_2{x}_2\le <!--\; \le \; -->d$
I am looking for a solution for the above linear system that gives more priority to the coordinates corresponding to $x_1$ compared to those of $x_2$. This is what I precisely mean by priority
If $x_1$ alone can geenerated the given y while $x_2$ is kept at its minimum feasible value.
If there are infinite number of solutions for $x_1$ in step 1 then we take the minimum norm solution.
If $x_1$ alone cannot generate y then we allow $x_2$ to participate in generating y with the optimal $x_1$ obtained in step 1 and 2.
If there are infinite solutions for $x_2$, then we take the minimum norm solution.
How to formuate the above described problem as an optimization problem for instance QP?

Convex polyhedron $P$ is a subset of ${\mathbb{R}}^n$ that satisfies system of linear inequalities
$\begin{array}{rl}{a}_{11}{x}_1+\cdots <!--\; \cdots \; -->+a_{1n}{x}_n& {\sim <!--\; \sim \; -->}_1{c}_1\\ & ⋮<!--\; ⋮\; -->\\ a_{p1}{x}_1+\cdots <!--\; \cdots \; -->+a_{pn}{x}_n& {\sim <!--\; \sim \; -->}_p{c}_p,\end{array}$
where ${\sim <!--\; \sim \; -->}_i\in <!--\; \in \; -->\{\le <!--\; \le \; -->,\ge <!--\; \ge \; -->\}$. It can be alternatively represented by two finite sets of generators $V,W\subseteq <!--\; \subseteq \; -->{\mathbb{R}}^n$:
$P=\text{conv}(V)+\text{cone}(W),$
where conv(V) denotes all convex combinations of points in V and cone(W) all nonnegative linear combinations of points in W.
Now, what if we allow ${\sim <!--\; \sim \; -->}_i$ to be from $\{\ge <!--\; \ge \; -->,>,\le <!--\; \le \; -->,<\}$. Is there some similar representation in terms of generating points for such sets?

Determining existence of a solution for a system of linear inequalities
I have a set of vectors ${\mathbf{v}}_1,{\mathbf{v}}_2,...,{\mathbf{v}}_m\in <!--\; \in \; -->{\mathbb{R}}^n$ and I want to know if there exists a nonzero vector $\mathbf{x}$ such that $\mathbf{x}\cdot <!--\; \cdot \; -->{\mathbf{v}}_i\le <!--\; \le \; -->0$ for any $i$. This is the same as saying the equation ${\mathbf{x}}^{T}V\leqq <!--\; \leqq \; -->\mathbf{0}$ has a nonzero solution, where $V$ is an $n\times <!--\; \times \; -->m$ matrix whose columns are ${\mathbf{v}}_i$.

Given the following constraints
$\begin{array}{rlllllll}{x}_1& +& x_2& +& x_3& +& x_4& \le <!--\; \le \; -->10\\ x_1& -<!--\; -\; -->& x_2& & & & & \le <!--\; \le \; -->0\\ x_1& & & -<!--\; -\; -->& x_3& & & \le <!--\; \le \; -->2\\ x_1& +& x_2& & & -<!--\; -\; -->& x_4& \le <!--\; \le \; -->3\\ x_i& & & & & & & \in <!--\; \in \; -->{\mathbb{R}}_{\ge <!--\; \ge \; -->0}\end{array}$
I want to find all basic feasible solutions. They are the extreme points of the convex polyhedra induced by these constraints. However, to solve these system we introduce as many slack variables as we have inequalities. This leads us to
$\begin{array}{rllllllllllllllll}{x}_1& +& x_2& +& x_3& +& x_4& +& s_1& & & & & & & & =10\\ x_1& -<!--\; -\; -->& x_2& & & & & & & & +& s_2& & & & & =0\\ x_1& & & -<!--\; -\; -->& x_3& & & & & & & & +& s_3& & & =2\\ x_1& +& x_2& & & -<!--\; -\; -->& x_4& & & & & & & & +& s_4& =3\\ {\stackrel{^<!--\; ^\; -->}{x}}_i& & & & & & & & & & & & & & & & \in <!--\; \in \; -->{\mathbb{R}}_{\ge <!--\; \ge \; -->0}\end{array}$
Now, a basic feasible solution would be
$\stackrel{^<!--\; ^\; -->}{x}\equiv <!--\; \equiv \; -->(x_1,x_2,x_3,x_4|s_1,s_2{s}_3,s_4)=(0,0,0,0|10,0,2,3{)}^T$
However,
1. How do I find all basic feasible solutions from this starting basic feasible solution?
2. These basic feasible solutions are basic feasible solutions for the modified system. How do I get basic feasible solutions for the original problem?

A nonlinear system
$\left[\begin{array}{c}\stackrel{\dot{}<!--\; \dot{}\; -->}{x_1}\\ \stackrel{\dot{}<!--\; \dot{}\; -->}{x_2}\end{array}\right]=\left[\begin{array}{c}{x_2}^3+u\\ u\end{array}\right]$
$y=x_1$
as the control objective to make $y$ track $y_d$, and the tracking error is $e=y-<!--\; -\; -->y_d$.
It is, $\stackrel{\dot{}<!--\; \dot{}\; -->}{y}=\stackrel{\dot{}<!--\; \dot{}\; -->}{x_1}$ and choosing the control law
$u=-<!--\; -\; -->{x_2}^3-<!--\; -\; -->e+\stackrel{\dot{}<!--\; \dot{}\; -->}{y_d}$
gives
$\stackrel{\dot{}<!--\; \dot{}\; -->}{e}+e=0$
which is stable and converging to zero as the time $t\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$ (or in other words, the error equation has one pole in -1).
The same control input $u$ applies also to the second equation of the nonlinear system, representing the internal dynamics. With the choice of u as above, $\stackrel{\dot{}<!--\; \dot{}\; -->}{x_2}=u$ yields
$\stackrel{\dot{}<!--\; \dot{}\; -->}{x_2}+{x_2}^3=\stackrel{\dot{}<!--\; \dot{}\; -->}{y_d}-<!--\; -\; -->e.$
With the choice of a bounded $\stackrel{\dot{}<!--\; \dot{}\; -->}{y_d}$ and e (bounded since $\stackrel{\dot{}<!--\; \dot{}\; -->}{e}+e=0$), then it is
$|\stackrel{\dot{}<!--\; \dot{}\; -->}{y_d}-<!--\; -\; -->e|\le <!--\; \le \; -->D,$
being $D$ a positive constant.
From now on my question starts, as I would like to know what are the math steps to derive the following conclusion:
the example concludes that $|x_2|\le <!--\; \le \; -->D^{1/3}$(i.e. $x_2$ is bounded too!), since $\stackrel{\dot{}<!--\; \dot{}\; -->}{x_2}<0$ when $x_2>D^{1/3}$, and $\stackrel{\dot{}<!--\; \dot{}\; -->}{x_2}>0$ when $x_2<D^{1/3}$. Can someone explain how to derive these inequalities?

Consider $U=[u_1...u_N]$ to be a non-singular square matrix. I am interested in range space of $U$ but with one caveat, I want to find all $a$ such that each element of $Ua$ is non-negative. In other words for all natural basis vectors $e_i$
$0\le <!--\; \le \; -->e_i^{T}Ua,\mathrm{\forall <!--\; \forall \; -->}i$
Is there a non-graphical, non-numerical approach that I can take to obtain all possible $a$?

Strategies to work with system of trigonometric inequality
$[\exp <!--\; \; -->(-<!--\; -\; -->q_1\ast <!--\; \ast \; -->i){\cos }^3<!--\; \; -->(p_3)\sin <!--\; \; -->(p_1)\sin <!--\; \; -->(p_2)\sin <!--\; \; -->(p_3)]\cdot <!--\; \cdot \; -->a-<!--\; -\; -->[\exp <!--\; \; -->(-<!--\; -\; -->q_{2}i)\exp <!--\; \; -->(-<!--\; -\; -->q_{3}i)\cos <!--\; \; -->(p_1)\cos <!--\; \; -->(p_2){\cos }^2<!--\; \; -->(p_3)\sin <!--\; \; -->(p_2){\sin }^2<!--\; \; -->(p_3)]\cdot <!--\; \cdot \; -->b\ne <!--\; \ne \; -->0,$
where $a$ and $b$ are complex variables and $q_i$ and $p_i$ are real variables. The real system (the least of them) have 18 inequalities and 8 variables, I need know if there is a set of values (8 real values) that makes true at least one of the inequalities. Maybe a good path is know how determine the minimum value of a expression of type
$[\exp <!--\; \; -->(a_{1}i){\cos }^3<!--\; \; -->(a_2)\sin <!--\; \; -->(a_3)\sin <!--\; \; -->(a_4){\sin }^2<!--\; \; -->(a_5)]$
or a bit more complex like
$[\exp <!--\; \; -->(a_{1}i){\cos }^3<!--\; \; -->(a_2)\sin <!--\; \; -->(a_3)\sin <!--\; \; -->(a_4){\sin }^2<!--\; \; -->(a_5)]\cdot <!--\; \cdot \; -->[\exp <!--\; \; -->(-<!--\; -\; -->b_{1}i){\cos }^2<!--\; \; -->(b_2)\sin <!--\; \; -->(b_3)\sin <!--\; \; -->(b_4){\sin }^3<!--\; \; -->(b_5)].$
Then, some idea?

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

Solve the system of equations:
$\{\begin{array}{l}(x-<!--\; -\; -->1)\sqrt{x-<!--\; -\; -->y^2}=y(x-<!--\; -\; -->2y+1)\\ y\sqrt{x-<!--\; -\; -->1}+3\sqrt{x-<!--\; -\; -->y^2}=2x+y-<!--\; -\; -->1\end{array}$
This is my try
Condition: $x\ge <!--\; \ge \; -->1;x\ge <!--\; \ge \; -->y^2$
We have:
$x-<!--\; -\; -->2y+1\ge <!--\; \ge \; -->x-<!--\; -\; -->2|y|+1\ge <!--\; \ge \; -->x-<!--\; -\; -->(y^2+1)+1=x-<!--\; -\; -->y^2\ge <!--\; \ge \; -->0\Rightarrow <!--\; \Rightarrow \; -->y\ge <!--\; \ge \; -->0$
Applying AM-GM inequality, we have:
$2(2x+y-<!--\; -\; -->1)=2y\sqrt{x-<!--\; -\; -->1}+6\sqrt{x-<!--\; -\; -->y^2}\le <!--\; \le \; -->(y^2+x-<!--\; -\; -->1)+3(1+x-<!--\; -\; -->y^2)=4x-<!--\; -\; -->2y^2+2$
$\Rightarrow <!--\; \Rightarrow \; -->y^2+y-<!--\; -\; -->2\le <!--\; \le \; -->0\Rightarrow <!--\; \Rightarrow \; -->0\le <!--\; \le \; -->y\le <!--\; \le \; -->1$
Squaring both side of first equation, we get:
$(x-<!--\; -\; -->2y^2+2y-<!--\; -\; -->1)(x^2-<!--\; -\; -->x+2y^2-<!--\; -\; -->2xy)=0$
If $x-<!--\; -\; -->2y^2+2y-<!--\; -\; -->1=0$, we get ${\displaystyle \frac{1}{2}}\le <!--\; \le \; -->x\le <!--\; \le \; -->1$. Thus, we get $x=1$ and $y=0$ or $y=1$. Two pairs of $(x,y)$ don't satisfy the system of equations.
If $x^2-<!--\; -\; -->x+2y^2-<!--\; -\; -->2xy=0$...

System of 3 equations and 1 inequality in 2 variables
As part of optimization problem I am trying to solve the following system of equations in $x,y\in <!--\; \in \; -->\mathbb{R}$
$\{\begin{array}{}2x+1+y=\frac{a}{2}x\\ 2y+1+x=\frac{a}{8}y\\ x^2+y^2-<!--\; -\; -->9<0\\ \frac{x^2}{4}+\frac{y^2}{16}=1\end{array}$
where $a\in <!--\; \in \; -->\mathbb{R}$.

Consider the system
$\{\begin{array}{l}1=A+B=C+D\\ B\ge <!--\; \ge \; -->C\end{array}$
with $A,B,C,D$ positive.
Does the system imply that $A\le <!--\; \le \; -->D$?