Solving System of Inequalities with Examples

Inequality solution verification(positive solution)
I have the following system of equation where all the free variables live in the integers that is are allowed to be integers only.
$x=l$
$y=5n-<!--\; -\; -->2l+25$
$z=10-<!--\; -\; -->2n$
For $l,n\in <!--\; \in \; -->\mathbb{N}$
I need to compute how many positive solutions are there so I computed the following inequalities:
$x>0\to <!--\; \to \; -->l>0$
$y>0\to <!--\; \to \; -->n>-<!--\; -\; -->5+2l/5$
$z>0\to <!--\; \to \; -->n<-<!--\; -\; -->5$
To get how many solution we keep a fixed l and see how many n we get and do that for each n until we get values that pop us out of the inequality for example for $l=1$ we have $n>-<!--\; -\; -->4.6$ hence we have $-<!--\; -\; -->4\le <!--\; \le \; -->n<5$ so in this case their is $10$ solutions and we proceed like this I just want to verify my answer I got $105$ possible positive solutions is that correct?

Show that the system of equations
$\{\begin{array}{l}{x}^4+y^4+\phantom{2}{z}^4=1\\ x^2+y^2+2z^2=\sqrt{7}\end{array}$
has no real solutions.

Elementary system of inequalities - set of solutions
$3x-<!--\; -\; -->1\le <!--\; \le \; -->1+4x$
$\frac{5}{6}x+\frac{2}{3}\ge <!--\; \ge \; -->1+\frac{1}{2}x$
$\frac{x-<!--\; -\; -->1}{6}<<!--\; <\; -->\frac{1}{4}$
$\Rightarrow <!--\; \Rightarrow \; -->$
$x\ge <!--\; \ge \; -->-<!--\; -\; -->2$
$x\ge <!--\; \ge \; -->1$
$x<<!--\; <\; -->\frac{5}{2}$
I compared the above reduced inequalities which gave me as answer that the solutions are from [$\frac{5}{2}$]. I think it is correct but I'm not sure. Can someone affirm this or debunk this please and what's the reason?

Suppose that $x$, $y$, $z$, $\mathrm{\nu <!--\; \nu \; -->}$, $N$, $g$ are some postitive parameters and $A$, $B$ and $C$ are variables that belong to $(0,+\mathrm{\infty <!--\; \infty \; -->})$. I want to identify one or all of the feasible points taht solve the following system of inequalities with respect to $A$, $B$ and $C$
$$\begin{gathered} xA-<!--\; -\; -->\mathrm{\nu <!--\; \nu \; -->}gB-<!--\; -\; -->NgC>0 \\ -<!--\; -\; -->\mathrm{\nu <!--\; \nu \; -->}gA+yB-<!--\; -\; -->NgC>0 \\ -<!--\; -\; -->\mathrm{\nu <!--\; \nu \; -->}gA-<!--\; -\; -->\mathrm{\nu <!--\; \nu \; -->}gB+zC>0 \end{gathered}$$

Suppose that the polytope $P=\{x\mid <!--\; \mid \; -->Ax\le <!--\; \le \; -->b\}$ is totally dual integral (TDI), and that the inequality ${\stackrel{~<!--\; ~\; -->}{a}}^{T}x\le <!--\; \le \; -->\stackrel{~<!--\; ~\; -->}{b}$ is satisfied at every point of $P$.
Does this imply that the system $\{Ax\le <!--\; \le \; -->b,{\stackrel{~<!--\; ~\; -->}{a}}^{T}x\le <!--\; \le \; -->\stackrel{~<!--\; ~\; -->}{b}\}$ is TDI?

Given a set of n inequalities each of the form ax+by+cz≤d for some a,b,c,d in Q, determine if there exists x, y and z in Q that satisfy all the inequalities.
Here is an O(n4) algorithm for solving this: for each triple of inequalities, intersect their corresponding planes to get a point (x,y,z) iff possible. If no such intersecting point exists continue on to the next triple of inequalities. Test each of these intersection points against all the inequalities. If a particular point satisfies all the inequalities the solution has been found. If none of these points satisfy all the inequalities then there is no point satisfying the system of inequalities. There are O(n3) such intersection points and there are n inequalities thus the algorithm is O(n4).
I would like a faster algorithm for solving this (eg: O(n3), O(n2), O(n*logn), O(n)). If you can provide such an algorithm in an answer that would be great. You may notice this problem is a subset of the more general k-dimensional problem where there are points in k-dimensions instead of 3 dimensions as in this problem or 2 dimensions as in my previous problem mentioned above. The time complexity of my algorithm generalized to k dimensions is O(nk+1). Ideally I would like something that is a polynomial time algorithm however any improvements over my naive algorithm would be great. Thanks

Let $\mathrm{\mu <!--\; \mu \; -->}$, $\mathrm{\nu <!--\; \nu \; -->}$ be finite measures on the non-degenerate compact interval $[a,b]\subseteq <!--\; \subseteq \; -->\mathbb{R}$ provided with the Borel $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra. It is well-known that if $\mathrm{\mu <!--\; \mu \; -->}(B)=\mathrm{\nu <!--\; \nu \; -->}(B)$ for every Borel set $B$ belonging to a $\mathrm{\pi <!--\; \pi \; -->}$-system that generates $\mathcal{B}([a,b])$ and that includes $[a,b]$, then $\mathrm{\mu <!--\; \mu \; -->}=\mathrm{\nu <!--\; \nu \; -->}$. Is the same true if the equality is replaced by an inequality? In other words, suppose $\mathrm{\mu <!--\; \mu \; -->}(B)\le <!--\; \le \; -->\mathrm{\nu <!--\; \nu \; -->}(B)$ for every Borel set $B$ belonging to a $\mathrm{\pi <!--\; \pi \; -->}$-system that generates $\mathcal{B}([a,b])$ and that includes $[a,b]$. Is it the case that $\mathrm{\mu <!--\; \mu \; -->}\le <!--\; \le \; -->\mathrm{\nu <!--\; \nu \; -->}$?

Mixed systems of linear equations and inequalities,
$a_{1,1}{x}_1+\cdots <!--\; \cdots \; -->+a_{1,n}{x}_n=b_1$
$a_{2,1}{x}_1+\cdots <!--\; \cdots \; -->+a_{2,n}{x}_n=b_2$
$\dots <!--\; \dots \; -->$
$a_{m-<!--\; -\; -->1,1}{x}_1+\cdots <!--\; \cdots \; -->+a_{m-<!--\; -\; -->1,n}{x}_n\ge <!--\; \ge \; -->b_{m-<!--\; -\; -->1}$
$a_{m,1}{x}_1+\cdots <!--\; \cdots \; -->+a_{m,n}{x}_n\ge <!--\; \ge \; -->b_m$
The number of equations $m$ might be less than, equal to or greater than $n$ (the number of unknowns).
Example1,
$\{\begin{array}{l}2x-<!--\; -\; -->y\ge <!--\; \ge \; -->-<!--\; -\; -->3\\ -<!--\; -\; -->4x-<!--\; -\; -->y\ge <!--\; \ge \; -->-<!--\; -\; -->5\\ -<!--\; -\; -->x+y=4\end{array}$
There is no solution for above system. I'm looking for a systematic (algorithmic) way to determine if such systems of equations have any solution or not.

Given a quadratic equation $y(x)$ and three points through which it passes, determine the value of a constant so that $y(x)$ has a minimum
$y=a{x}^2+bx+c$
passes through $(1,1)$, $(2,m)$ and $(m,2)$ where $m\ne <!--\; \ne \; -->2$. Determine an interval for $m$ such that $y(x)$ has a minimum.
Progress:
If it has a minimum, then $a>0$.
Also, by the points given:
$a+b+c=1$
$m=4a+2b+c$
$a{m}^2+bm+c=2$
Using the fact that $a$ is positive, I got the expressions:
$a+b+c=1⟹<!--\; ⟹\; -->b+c<1$
$a{m}^2+bm+c=2⟹<!--\; ⟹\; -->bm+c<2$
$m=4a+2b+c⟹<!--\; ⟹\; -->2b+c<m$
But I don't know how to solve it.

Given: For some $X$, $Var(X)=9$, $E(X)=2$, $E(X^2)=13$
Problem: $Pr[X=2]>0$
The solution in my book says to construct a r.v. to satisfy the above conditions and confirm or deny the statement from there. For simplicity, assume $X$ can take on two values $Pr[a]=\frac{1}{2}$ and $Pr[b]=\frac{1}{2}$. Finding that $a$ and $b$ are not $2$ is enough to disprove the statement. We can apply the constraints:
$\frac{1}{2}a+\frac{1}{2}b=2$
$\frac{1}{2}{a}^2+\frac{1}{2}{b}^2=13$
So I did this:
$a=4-<!--\; -\; -->b$ from the first equation
$b^2-<!--\; -\; -->4b-<!--\; -\; -->5=0$ by substituting $a$
The above quadratic has two solutions, $b=-<!--\; -\; -->1$ and $b=5$ which happen to be the same solutions the book got for $a$ and $b$ respectively.
I'm not sure if this is the right way to solve this problem since I never explicitly solved for $a$ and the two solutions for $b$ may coincidentally be the same.

I have a task to calculate a double integral on a region$D=A\setminus <!--\; \setminus \; -->B$, where $A=\{(x,y):\text{some system of inequalities}\},B=\{(x,y):\text{another system of inequalities}\}$. What does operator "$\setminus <!--\; \setminus \; -->$" mean?

given the question:
$y-<!--\; -\; -->2>-<!--\; -\; -->3x$
$x\le <!--\; \le \; -->-<!--\; -\; -->1-<!--\; -\; -->y$
In the $xy$ plane, if the point with co-ordinates $(a,b)$ lies in the solution set of the system of inequalities above, which of the following relationships between $a$ and $b$ must be true ?
1. $a<0$
2. $b>0$
3. $ab>0$
4. $ab<0$
5. $a>b$

Solve absolute value inequality with x and y
I have difficulties to solve this system of inequalities. Normally you would solve it like a normal system of equations just writing $\le <!--\; \le \; -->$ or $\ge <!--\; \ge \; -->$ instead of $=$. But these contain an absolute value which confuses me. The system I want to solve is this:
$\{\begin{array}{c}|x-<!--\; -\; -->y|\le <!--\; \le \; -->1\\ |x+y|\le <!--\; \le \; -->1\end{array}$
I would have written it like that
$\{\begin{array}{c}-<!--\; -\; -->1\le <!--\; \le \; -->x-<!--\; -\; -->y\le <!--\; \le \; -->1\\ -<!--\; -\; -->1\le <!--\; \le \; -->x+y\le <!--\; \le \; -->1\end{array}$
And by adding the inequalities I would obtain $-<!--\; -\; -->1\le <!--\; \le \; -->x\le <!--\; \le \; -->1$. I can’t get y though.
I don’t understand how my textbook got a third inequality out of the two initial ones
$\{\begin{array}{c}-<!--\; -\; -->1\le <!--\; \le \; -->x-<!--\; -\; -->y\le <!--\; \le \; -->1\\ -<!--\; -\; -->1\le <!--\; \le \; -->y-<!--\; -\; -->x\le <!--\; \le \; -->1\\ -<!--\; -\; -->1\le <!--\; \le \; -->x+y\le <!--\; \le \; -->1\end{array}$
But with that they can solve the system like so: $(1)+(3):-<!--\; -\; -->1\le <!--\; \le \; -->x\le <!--\; \le \; -->1$ and $(2)+(3):-<!--\; -\; -->1\le <!--\; \le \; -->y\le <!--\; \le \; -->1$.

In the euclidean space, the points $(x,y,z)$ belonging to a regular octahedron are those that satisfy the inequalities
$\pm <!--\; \pm \; -->x\pm <!--\; \pm \; -->y\pm <!--\; \pm \; -->z\le <!--\; \le \; -->a$
where $a\ge <!--\; \ge \; -->0$. These eight inequalities can be divided into two groups of four according to the number (even or odd) of negative signs they contain. For example, the inequalities
$\begin{array}{r}x-<!--\; -\; -->y+z\le <!--\; \le \; -->a\\ -<!--\; -\; -->x+y+z\le <!--\; \le \; -->a\\ x+y-<!--\; -\; -->z\le <!--\; \le \; -->a\\ -<!--\; -\; -->x-<!--\; -\; -->y-<!--\; -\; -->z\le <!--\; \le \; -->a\end{array}$
all have one or three negative signs and the points satisfying these form a tetrahedron. The other four inequalities correspond to the dual tetrahedron of the first, which shows that the intersection of two regular dual tetrahedra form a regular octahedron. Moreover, the vertices of the two tetrahedra can be seen as the eight vertices of a cube.
I am wondering if there exists a similar relationship between regular polytopes in four dimensions. As it is another case of a regular cross-polytope, the hexadecachoron (or 16-cell) is defined by the sixteen inequalities
$\pm <!--\; \pm \; -->x\pm <!--\; \pm \; -->y\pm <!--\; \pm \; -->z\pm <!--\; \pm \; -->w\le <!--\; \le \; -->a.$
If one were to take the eight inequalities containing an odd number of negative signs, say
$\begin{array}{r}x-<!--\; -\; -->y-<!--\; -\; -->z-<!--\; -\; -->w\le <!--\; \le \; -->a\\ -<!--\; -\; -->x+y-<!--\; -\; -->z-<!--\; -\; -->w\le <!--\; \le \; -->a\\ -<!--\; -\; -->x-<!--\; -\; -->y+z-<!--\; -\; -->w\le <!--\; \le \; -->a\\ -<!--\; -\; -->x-<!--\; -\; -->y-<!--\; -\; -->z+w\le <!--\; \le \; -->a\\ x+y+z-<!--\; -\; -->w\le <!--\; \le \; -->a\\ x+y-<!--\; -\; -->z+w\le <!--\; \le \; -->a\\ x-<!--\; -\; -->y+z+w\le <!--\; \le \; -->a\\ -<!--\; -\; -->x+y+z+w\le <!--\; \le \; -->a\end{array}$
which 4-polytope would be obtained ? I doubt it would be a regular 5-cell, since (obviously) the number of cells and the number of hyperplanes don't add up. Besides, the intersection of the two 4-polytopes corresponding to the two sets of eight inequalities should technically correspond to the 16-cell.
The tesseract, having eight cells, could be a candidate, but I have been unable to show that these eight inequalities define one (or any other 4-polytope). Any ideas?

Can any subset of ${\mathbb{R}}^2$ can be expressed by equations /inqualities?
For example the set of all points on a circle in ${\mathbb{R}}^2$ can be expressed by an equation. Similarly square, rectangle, parabola, interior of a circle, triangular regions, etc.
Likewise, can any subset of ${\mathbb{R}}^2$ can be expressed by a system (finite or infinite number) of equations or by inequalities?

Problem: To find all real solutions of the system:
$3a=(b+c+d{)}^3$
$3b=(c+d+e{)}^3$
$3c=(d+e+a{)}^3$
$3d=(e+a+b{)}^3$
$3e=(a+b+c{)}^3$

Suppose we have a $1$-dimensional differential inequality
$\frac{dx}{dt}\le <!--\; \le \; -->x-<!--\; -\; -->x^3$
We can apply the Comparison principle to claim that if $y(t)$ is the solution to $\frac{dy}{dt}=y-<!--\; -\; -->y^3$, then $x(t)\le <!--\; \le \; -->y(t)$ (assuming $x(0)\le <!--\; \le \; -->y(0)$. Can we extend similar argument to a $2$-dimensional system? For example, let us consider the following system of equations
$\frac{d{x}_1}{dt}=x_1-<!--\; -\; -->x_2,\frac{d{x}_2}{dt}\le <!--\; \le \; -->x_1+x_2-<!--\; -\; -->\frac{x_2^4}{x_1^2}$
Is the solution to following
$\frac{d{y}_1}{dt}=y_1-<!--\; -\; -->y_2,\frac{d{y}_2}{dt}=y_1+y_2-<!--\; -\; -->\frac{y_2^4}{y_1^2}$
related with the solution of the original problem. Specifically, can we apply the Comparison principle to first say that $x_2(t)\le <!--\; \le \; -->y_2(t)$ and then subsequently use it to claim $x_1(t)\ge <!--\; \ge \; -->y_1(t)$?

Computing bounding box of polytope (system of linear inequalities)
Given a N real valued variables and a set of linear inequality constraints, I would like to find a minimal bounding box which encapsulates the convex polytope defined by these constraints.

With $\mathbf{x}$ and $\mathbf{y}$ two vectors, the triangle inequality can be written as:
$\begin{array}{}\text{(A)}& |\mathbf{x}|-<!--\; -\; -->|\mathbf{y}|\le <!--\; \le \; -->|\mathbf{x}+\mathbf{y}|\le <!--\; \le \; -->|\mathbf{x}|+|\mathbf{y}|\end{array}$
which can be broken down into a system of inequalities (labelled as (1) and (2)):
$\{\begin{array}{l}|\mathbf{x}|-<!--\; -\; -->|\mathbf{y}|\le <!--\; \le \; -->|\mathbf{x}+\mathbf{y}|\\ |\mathbf{x}+\mathbf{y}|\le <!--\; \le \; -->|\mathbf{x}|+|\mathbf{y}|\end{array}$
then, with the assumption of $|\mathbf{y}|>0$, (1)-(2) yields:
$\begin{array}{}\text{(B)}& |\mathbf{x}|\le <!--\; \le \; -->|\mathbf{x}+\mathbf{y}|\end{array}$
But is this in fact true? Is (B) a direct consequence of (A) when $|\mathbf{y}|>0$? or is (B) is invalid because (1)-(2) is not allowed as it is not obvious if the subtraction preserves the inequality?

Are there an infinite number of prime quadruples of the form $10n+1$, $10n+3$, $10n+7$, $10n+9$?
In base $10$, any prime number greater than 5 must end with the digits 1$1$, $3$, $7$, or $9$. For some $n$, $10n+1$, $10n+3$, $10n+7$, $10n+9$ are all prime: for example, when $n=1$, we have that $11$, $13$, $17$, and $19$ are all prime. My question is, can anyone disprove the claim that there are an infinite number of such primes.
The only progress I've been able to make is to show that $n$ must be of the form $3k+1$ by considering the system of modular inequalities
$p\not\equiv 0\mathrm{mod}2$
$p\not\equiv 0\mathrm{mod}3$
$p\not\equiv 0\mathrm{mod}5$