Determine whether the orderred pair is a solution to the system of inequlities x+y<4 y<=2x+1 a. (0,1) b. (3,1) c. (2,0) d. (1,4)

I have a problem on the form: find $x\in {\mathbb{R}}^n,y\in {\mathbb{R}}^m$ such that:
$Ax\ge a$
$By\ge b$
$f(x)\ge 0$
$f_i(x)\cdot y_j=0$
$x_i,y_j\ge 0$
In the later A and B are matrices, $a\in {\mathbb{R}}^n,b\in {\mathbb{R}}^m$, $f=(f_1\dots f_n)$ is a linear vectorial function of $x$ and $x=(x_1\dots x_n)$, $y=(y_1\dots y_m)$
There is a solver or a method that could solve that kind of inequalities?

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

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: $2x1+x2-3x3-1\le 10$ and $(2x1+x2-3x3-1)\le 10$?

What should $a$ be for the system of inequalities to have one solution?
$(1)\frac{3}{x-a}\ge 1(2)|x-2a-2|\le 1$
For them to have one solution, the solution must be an endpoint of an interval. (1) gives me
$x\le a+3$
but I don't know how to proceed from here. Any hint would be appreciated.

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

Find the smallest possible value of $\omega$ such that the system of inequalities
$\begin{array}{r}(1009\cdot 1010-2018)x+2018\le 2018\omega \\ (1008\cdot 1009-2018)x+2018\le 2016\omega \\ (1007\cdot 1008-2018)x+2018\le 2014\omega \\ ....................................\\ (1\cdot 2-2018)x+2018\le 2\omega \end{array}$
has a real solution in $x$.
Subtracting the first inequality from the second inequality, we get that $x=\frac{1009}{2017}$. This is also correct when we subtract the second inequality from the third one. When I plugged in $x$, I got an unruly fractional value for $\omega$. Can someone guide me through the solution? Help is much appreciated.

Given a system of inequalities with ordinal numbers:
$\begin{array}{rl}{a}_1& >b_1\\ & ⋮\\ a_n& >b_n\end{array}$
Let $\varphi$ be a permutation on $\{1,\dots ,n\}$. Is it true that $a_{\varphi (1)}+\cdots +a_{\varphi (n)}>b_1+\cdots +b_n$?