Solving System of Inequalities with Examples

Determining all $(a,b)$ on the unit circle such that $2x+3y+1\le <!--\; \le \; -->a(x+2)+b(y+3)$ for all $(x,y)$ in the unit disk
In the middle of another problem, I came up with the following inequality which needed to be solve for $(a,b)$ :
$2x+3y+1\le <!--\; \le \; -->a(x+2)+b(y+3)$
for all $(x,y)\in <!--\; \in \; -->{\mathbb{R}}^2$ with $x^2+y^2\le <!--\; \le \; -->1.$.
Here the solution for $(a,b)$ must be a subset of the unit circle, and I believe that it is a singleton.

Prove that following systems of inequalities are equivalent.
There are given two systems of inequalities: $b,d,c,e>0$ (first system) and $b+d>0,ec>0,(b+d)(e+c+bd)>be+cd,(be+cd)((b+d)(e+c+bd)-<!--\; -\; -->(be+cd))>(b+d{)}^{2}ec$ (second system). I need to show that they are equivalent. It is easy to see that first system implies second, because last two inequalities (from the second one) can be transformed to form $bc+b^{2}d+de+b{d}^2>0$ and $bd((e-<!--\; -\; -->c{)}^2+b^{2}e+bde+bdc+c{d}^2)>0$ and now it is clear that all inequalities from second system must be true, if $b,c,d,e$ are positive.

Using Banach's Fixed Point Theorem, show that the following system has at least one solution:
$x=0.000001x^2+10\sin <!--\; \; -->y+1$
$y=0.000001y^3-<!--\; -\; -->0.01\cos <!--\; \; -->x-<!--\; -\; -->1$
Here is what I have tried:
Consider the function defined by
$f(x,y)=(0.000001x^2+10\sin <!--\; \; -->y+1,0.000001y^3-<!--\; -\; -->0.01\cos <!--\; \; -->x-<!--\; -\; -->1)$
and show that it has a fixed point. If it does, then that fixed point is clearly a solution to our system.
I used the standard Euclidean distance between two points and attempted to show that
$d(f(x_1,y_1),f(x_2,y_2))\le d((x_1,y_1),(x_2,y_2))$
In doing so, I greatly struggle to manipulate the left-hand side so that it is clearly less than or equal to the right-hand side of the inequality above. I tremendously appreciate any advice, and if there is a smarter way to approach this, I appreciate any hints you are willing to offer.

Let $RLS(k)$ be the problem where we try to determine whether a Linear System of $n$ unknown variables and $m$ equations with Rational coefficients, has a $k$-solution which is basically a vector $x=(x_1,x_2,\dots <!--\; \dots \; -->,x_n)$ such that at most $k$ of the $x_i$'s are non-zero.
Prove that $RLS(k)$ is NP-hard

While doing some training in optimisation I was hit by the system of polynomial equations
$\begin{array}{}\text{(1)}& t{a}^{6}b+b{d}^2& =& 2a^3\text{(2)}& t{b}^{6}c+c{a}^2& =& 2b^3\text{(3)}& t{c}^{6}d+d{b}^2& =& 2c^3\text{(4)}& t{d}^{6}a+a{c}^2& =& 2d^3\text{(5)}& a^5+b^5+c^5+d^5& =& 4\end{array}$
where $a,b,c,d,t$ are real variables being strictly positive.
- The system is cyclic in $a,b,c,d$
- The remaining variable $t$ gets easily isolated: Multiply $(1)$ through $(4)$ with the variables not present in the t-term, then sum up and use $(5)$ which yields
$\begin{array}{}\text{(6)}& t\cdot <!--\; \cdot \; -->4abcd=a^{3}cd+b^{3}da+c^{3}ab+d^{3}bc.\end{array}$
- There's the solution with all variables $=1$.
Are there more solutions?

Why in module inequality $|f(x)|<A$ with a positive A we do not explicitly check that $f(x)>=0$ and $f(x)<0$?
$|f(x)|<A$ when $А>0$ is equivalent to the following system:
$\{\begin{array}{l}f(x)<A\\ f(x)>-<!--\; -\; -->A\end{array}$
I do not completely understand, why we do not exclude those $x$, for which $f(x)<0$ in the first equation, and $x$ for which $f(x)<0$ in the second equation - I mean why we do not write explicitly two systems of inequalities, uniting their solutions like this:
$\{\begin{array}{l}f(x)<A\\ f(x)>=0\end{array}$
$\{\begin{array}{l}f(x)>-<!--\; -\; -->A\\ f(x)<0\end{array}$

System of linear inequalities - mixture
We have 4 types bottles of water. Bottle 1 - with capacity 750ml - we have 2 for use with price 0,25 Bottle 2 - with capacity 500ml - we have 3 for use with price 0,5 Bottle 3 - with capacity 250ml - we have 5 for use with price 0,75 Bottle 4 - with capacity 100ml - we have 10 for use with price 1
Using water from these 4 types of bottles we have to create "aqueous mixture" containing 3050ml of water in total.
How we should pour water from those bottles that cost as low as possible and to have the least possible loss of water.
How to solve this?

Properties of solutions to system of linear equations with diagnonally dominant, positive (definite) matrix
Consider a system of linear equations $Ax=b$, where
$\begin{array}{rcl}A& =& \left[\begin{array}{ccccc}2& 1& & & \\ 1& 4& 1& & \\ & \ddots <!--\; \ddots \; -->& \ddots <!--\; \ddots \; -->& \ddots <!--\; \ddots \; -->& \\ & & 1& 4& 1\\ & & & 1& 2\end{array}\right]\in <!--\; \in \; -->{\mathbb{R}}^{n\times <!--\; \times \; -->n}\end{array}$
and $n\ge <!--\; \ge \; -->3$
I would like to prove the following two statements, where the inequalities are element-wise:
$\begin{array}{rcl}b\ge <!--\; \ge \; -->0& ⟹<!--\; ⟹\; -->& x\ge <!--\; \ge \; -->0\\ b>0& ⟹<!--\; ⟹\; -->& x>0.\end{array}$

If
$\begin{array}{r}x=cy+bz\\ y=az+cx\\ z=bx+ay\end{array}$
where $x,y,z$ are not all zero, prove that $a^2+b^2+c^2+2abc=1$
Further if at least one of $a,b,c$ is a proper fraction, prove that:
(i) $a^2+b^2+c^2<<!--\; <\; -->3$
(ii) $abc><!--\; >\; -->-<!--\; -\; -->1$

Find all the real values of $k$ for which the following system of inequalities
$-<!--\; -\; -->3<\frac{x^2+kx-<!--\; -\; -->2}{x^2-<!--\; -\; -->x+1}<2$
is fullfilled for all real values of $x$
I have attempted this question but my teacher told me that we have to take $\mathrm{\Delta <!--\; \Delta \; -->}<0$ in $-<!--\; -\; -->3(x^2-<!--\; -\; -->x+1)<x^2+kx-<!--\; -\; -->2$
The answer is to be provided in intervals

For each $m,n\in <!--\; \in \; -->\mathbb{N}$ think ${\mathbb{R}}^m$ as the space of real column vectors of size $m$ and ${\mathbb{R}}^{m\times <!--\; \times \; -->n}$ as the space of matrices of size $m\times <!--\; \times \; -->n$.
Let $d\in <!--\; \in \; -->\mathbb{N}$
Let $a:\{1,\dots <!--\; \dots \; -->,2^d\}\to <!--\; \to \; -->\{1\}\times <!--\; \times \; -->\{0,1{\}}^d$ be an enumeration (injective and surjective map).
Let $A\in <!--\; \in \; -->{\mathbb{R}}^{(d+1)\times <!--\; \times \; -->2^d}$ be the matrix whose columns are $a(1),\dots <!--\; \dots \; -->,a(2^d)$
Is it true that for each $b\in <!--\; \in \; -->\{1\}\times <!--\; \times \; -->[0,1{]}^d$ there exists $x\in <!--\; \in \; -->[0,+\mathrm{\infty <!--\; \infty \; -->}{)}^{2^d}$ such that $Ax=b?$?

When does a system of inequalities have only the solution $x=0$ ?
Let $A$ be a $n\times <!--\; \times \; -->n$ real valued matrix, $n\ge <!--\; \ge \; -->2$. For every column n-vector $x$ from $A\cdot <!--\; \cdot \; -->x=0$ we get $n$ equations
$$\begin{gathered} a_1^1{x}_1+a_1^2{x}_2+\cdots <!--\; \cdots \; -->+a_1^n{x}_n=0 \\ {a}_2^1{x}_1+a_2^2{x}_2+\cdots <!--\; \cdots \; -->+a_2^n{x}_n=0 \\ \dots <!--\; \dots \; --> \\ {a}_n^1{x}_1+a_n^2{x}_2+\cdots <!--\; \cdots \; -->+a_n^n{x}_n=0 \end{gathered}$$
I know that if $det(A)\ne <!--\; \ne \; -->0$then the only solution to $A\cdot <!--\; \cdot \; -->x=0$ is $x=0$.
Now replace all the previous equalities with inequalities in order to obtain the following system:
$$\begin{gathered} a_1^1{x}_1+a_1^2{x}_2+\cdots <!--\; \cdots \; -->+a_1^n{x}_n\ge <!--\; \ge \; -->0 \\ {a}_2^1{x}_1+a_2^2{x}_2+\cdots <!--\; \cdots \; -->+a_2^n{x}_n\ge <!--\; \ge \; -->0 \\ \dots <!--\; \dots \; --> \\ {a}_n^1{x}_1+a_n^2{x}_2+\cdots <!--\; \cdots \; -->+a_n^n{x}_n\ge <!--\; \ge \; -->0 \end{gathered}$$
Which conditions on the matrix A will guarantee that the only solution of this system of inequalities is $x=0$?

I'm currently learning something about Sperner's Lemma and then the LYM Inequality. In trying to prove the LYM Inequality, the proof uses the concept of a shadow but I can't seem to get a proper grip of what it actually means. I've tried 'drawing it out' using Hasse diagrams but really have no clue whether I'm right or not.
For reference, this is the definition that I'm working with:
Let $F\subset <!--\; \subset \; -->X^{(r)}$ be a set system. The shadow $\mathrm{\delta <!--\; \delta \; -->}F$ of $F$ is $\mathrm{\delta <!--\; \delta \; -->}F=\{A\in <!--\; \in \; -->X^{(r-<!--\; -\; -->1)}:\mathrm{\exists <!--\; \exists \; -->}B\in <!--\; \in \; -->Fs.t.A\subset <!--\; \subset \; -->B\}$.

Test if a system of linear inequalities has a solution.
system of linear inequalities $Ax>0$ where $A$ is a $m\times <!--\; \times \; -->n$ matrix with $m\ge <!--\; \ge \; -->n$. Is there a simple way to check whether it has a solution?
What if one of the columns of $A$ is parametrized? In a more general case, I have one of the columns of $A(t)$ as $a(t)=tv+w$, where $t\in <!--\; \in \; -->[0,1]$. Is there a way to check whether all of the systems
$A(t)x>0,t\in <!--\; \in \; -->[0,1]$
have a solution?

Question: Use the definition of absolute value and systems of inequalities to prove that for any real numbers x and c, and any positive real number δ, the given statement is true:
$|x-c|<\delta \iff x\in (c-\delta ,c+\delta )$

Question on application of Chinese Remainder Theorem
$x\equiv 3\left(\text{mod}30\right)$
$x\equiv 5\left(\text{mod}56\right)$
I have a system of modular equation that I want to solve. However, I thought that this system has no solution because the modulos are not coprime. Further, attempting to solve using chinese remainder theorem:
$x\equiv 56p+30q$
where $p$ is such that
$56p\equiv 3\left(\text{mod}30\right)$
and $q$ is such that
$30q\equiv 5\left(\text{mod}56\right)$
However, again the modular inverses of these do not exist.
Yet, one solution to this system of modular inequalities is 1293. How come the Chinese remainder theorem gives that no solution exists?

In an oil mill, they decide to make a mixture from two types of oil: the extra virgin whose price is $4 per liter and the virgin $3 per liter. They have 200 L of extra virgin olive oil and 500 L of virgin olive oil and they also want the cost of the mixed oil to be at least 1500 dollars.
The extra virgin olive oil will be $e$, while the virgin olive oil will be $v$.
a) Set up a system of inequalities that contains the restrictions set out in the problem.
The total number of mixed oil is:
$e+v=m$
$200+500=700$
Considering $4e+3v$, the maximum total price of the mixed oil is $2300$:
$4(200)+3(500)=2300$
Then, I've formed a system of inequalities
$\{\begin{array}{}e+v\le <!--\; \le \; -->700\\ 4e+3v\ge <!--\; \ge \; -->1500\end{array}$
I'm not sure if I've transformed the problem into a system correctly.
b) Is it possible to meet the conditions if the mixture contains the same amount of the two oils?
Since I wasn't sure about my system of equations, I've done the last two questions using
$4e+3v$
I'm taking the max amount of extra virgin oil, $200$L, and this is what we get:
$4(200)+3(200)=800+600=\mathrm{\$<!--\; \$\; -->}1400$
So, the answer would be
$1400<<!--\; <\; -->1500$
So, no, it would meet the conditions.
c) If they decide to use all the extra virgin oil they have in the mixture, what will be the amount of virgin oil that they should use if they want the cost of the mixture to be $$1700$?
$1700=4(200)+3v$
$\frac{1700-<!--\; -\; -->800}{3}=v$
$300L=v$
If you could help me identify my errors I would greatly appreciate it.

Here is a system of inequalities I've been trying to solve, and nothing so far;
$y>x,$, $1>y,$, $1>x,$, $x>y/2,x>(2-<!--\; -\; -->y)/3,y>0,x>0,1>x+y$

Find the values of $a,b$ so that the polyhedral set defined by the system of inequalities
$$\begin{gathered} ax+(b+1)y\le <!--\; \le \; -->120 \\ x+(a+b)y\le <!--\; \le \; -->160 \\ (a-<!--\; -\; -->b)x+y\le <!--\; \le \; -->30 \\ x\ge <!--\; \ge \; -->0 \\ y\ge <!--\; \ge \; -->0 \end{gathered}$$
has extreme direction.

The first part of the problem was to solve for $a=3$ implies $x\in <!--\; \in \; -->[0,2]$. This makes me think that I will somehow use this. I try to find the solution for the second part of the problem by squaring the inequality. I get $2\sqrt{(-<!--\; -\; -->x^2+2x)}\ge <!--\; \ge \; -->a-<!--\; -\; -->2$.
This is equal to two systems:
$\{\begin{array}{l}a<2\\ x\in <!--\; \in \; -->[0,2]\end{array}\text{and}\{\begin{array}{l}a\ge <!--\; \ge \; -->2\\ 4x^2-<!--\; -\; -->8x+a^2-<!--\; -\; -->4a+2\ge <!--\; \ge \; -->0\end{array}$
What do I do know and is this actually correct?