Solving System of Inequalities with Examples

Suppose, we have a system of diophantine equations and also restrictions to the variables such as $0\le <!--\; \le \; -->a\le <!--\; \le \; -->x$ that can also be inequalities.
Can we transform this system in a diophantine equation that has a solution if and only if the given system has a solution ? In other words, is the system equivalent to some diophantine equation ?

There are 10 decision variables, $x_1,x_2,...,x_{10}$, with the following goals:
Max $0.067x_1+...+0.0590x_{10}$
Subject to:
$x_1+...+x_{10}<100,000$ (Budget)
- $x_1\le <!--\; \le \; -->25,000,...,x_{10}\le <!--\; \le \; -->25,000$ (Divestment Constraint)
- $x_9+x_{10}\ge <!--\; \ge \; -->x_1+...+x_4$
- $x_1+...+x_4\le <!--\; \le \; -->50,000$
- $x_1+...+x_4\ge <!--\; \ge \; -->20,000$
To fully capture the problem, I have another constraint I'd like to add: each decision variable has to be 0, OR has to be at least 10,000. Is there a good way through a system of inequalities to create this, or should I go backwards and make my decision variables binary to accomplish this?

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

How can we solve this system of linear inequalities?
Let $c_i$ be a given non-negative integer for all $i\in <!--\; \in \; -->\{1,\dots <!--\; \dots \; -->,n\}$. would like to find the non-negative integers $a_i$ and $b_i$ for all $i\in <!--\; \in \; -->\{1,\dots <!--\; \dots \; -->,n\}$ such that:
$\begin{array}{r}\{\begin{array}{ll}{c}_i=a_i+b_i,& \text{ for all }i\in <!--\; \in \; -->\{1,\dots <!--\; \dots \; -->,n\}\\ a_i<a_{i+1},& \text{ for all }i\in <!--\; \in \; -->\{1,\dots <!--\; \dots \; -->,n-<!--\; -\; -->1\}\\ b_i>b_{i+1},& \text{ for all }i\in <!--\; \in \; -->\{1,\dots <!--\; \dots \; -->,n-<!--\; -\; -->1\}\\ b_i\ge <!--\; \ge \; -->0\text{ and integer },& \text{ for all }i\in <!--\; \in \; -->\{1,\dots <!--\; \dots \; -->,n\}\\ a_i\ge <!--\; \ge \; -->0\text{ and integer },& \text{ for all }i\in <!--\; \in \; -->\{1,\dots <!--\; \dots \; -->,n\}\end{array}\end{array}$

How to solve this system with equation and inequality?
$\{\begin{array}{r}{2}^{x+2}=\frac{49}{4}{x}^2+4,\\ 2^{x+2}-<!--\; -\; -->4\le <!--\; \le \; -->x^2(14-<!--\; -\; -->2^{x+2})\ast <!--\; \ast \; -->2^x\end{array}$

Prove vertices of a simplex are affinely independent
Given that the definition of a simplex $T$ is $x\in <!--\; \in \; -->{\mathbb{R}}^n$ such that $x$ satisfies $n+1$ linear inequalities: $(u_k,x)<<!--\; <\; -->c_k$ for $k=1,\dots <!--\; \dots \; -->,n+1$ (i.e. $T$ is the intersection of $n+1$ half spaces)
If we additionally impose the condition that $T$ must be bounded and nonempty, I was able to show that the $k^{\text{th}}$ vertex of $T$ can be defined $v_k$ as the unique solution to the system $(v_k,u_j)=c_j$ for all $j\ne <!--\; \ne \; -->k$The existence of such a solution comes from the fact that any size n subset of the $u_k$ are independent. Or else we could write $u_k$'s out as rows of a matrix and pick an arbitrarily large vector l in the nullspace. Adding that on to any existing $x\in <!--\; \in \; -->T$ would show $T$ is unbounded since $x+l\in <!--\; \in \; -->T$ since the inequalities $(u_k,x)<<!--\; <\; -->c_k⟺<!--\; ⟺\; -->(u_k,x+l)<<!--\; <\; -->c_k$.
It's intuitive to me that $v_k$'s must be affinely independent. It's also intuitive that the closure of $T$ should be the convex hull of all the $v_k$. In fact, I think either fact implies the other. But I'm unable to prove either of them. I was, however, able to prove that the convex hull of $v_k\subset <!--\; \subset \; -->T$.
Is there a simple proof that the vk are affinely independent?

A retired couple wishes to have an additional annual income of $6000 per year.As their financial consultant, you recommend that they invest some money in Treasury Bills ($t$) that yield 6%, some in corporate paper ($p$) yielding 3%, some in corporate bonds ($b$) that yield 4%, and some in junk bonds ($j$) that yield 10%. Suppose the couple have $120000 to invest. Set up and solve a system of equations for this situation. Note any natural constraints on the variables (at least four inequalities needed). Graph the domain on a labeled and scaled coordinate axis (that is graph the inequalities and shade the appropriate region). Scale the axes so that the graph is large and clear. Copy and fill in the table below to show various ways their goal can be achieved. Include one option where no category has zero invested.What I have so far: I put the system into a matrix through the equations$\begin{array}{rl}t+p+b+j& =120000\\ 0.06t+0.03p+0.04b+0.1j& =6000\end{array}$
From the ref, I got the following info:
$\begin{array}{rl}t+\frac{b}{3}+\frac{7}{3}j& =8000\\ p+\frac{2}{3}b-<!--\; -\; -->\frac{4}{3}j& =40000\end{array}$
The free variables are $b+j⩽<!--\; ⩽\; -->120000$ and $b+j⩾<!--\; ⩾\; -->0$.
Next, we have to determine the natural constraints. Obviously, two are $b,j⩾<!--\; ⩾\; -->0$. I put down the other two as $b+j⩽<!--\; ⩽\; -->120000$ and $b+j⩾<!--\; ⩾\; -->0$.
But I think I need more constraints, since after plotting the domain of $b$ and $j$, and identifying possible combinations, I was getting totals over $120,000$, which is the maximum that the couple can spend.
The question that I have is what "natural constraint" am I missing?

Omar needs to eat at least 800 calories before going to his team practice. All he wants is hamburgers and cookies, and he doesn’t want to spend more than $5. At the hamburger restaurant near his college, each hamburger has 240 calories and costs $1.40. Each cookie has 160 calories and costs $0.50.
a. Write a system of inequalities to model this situation.
b. Graph the system.
c. Could he eat 3 hamburgers and 1 cookie?
d. Could he eat 2 hamburgers and 4 cookies?
I think the answer for question a is:
$\{\begin{array}{l}1.4h+0.5c\le <!--\; \le \; -->5\\ 240h+160c\ge <!--\; \ge \; -->800\end{array}$
If I was to solve this by graphing, how will I know which would be the x and y variable?

How to prove this basic fact: $(a)(-<!--\; -\; -->1)=-<!--\; -\; -->a$. How to prove that an additive inverse can be expressed multiplicatively?

Let $u,v:[0,\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->[0,\mathrm{\infty <!--\; \infty \; -->})$ satsfying the following system of differential inequalities:
$$\begin{gathered} u^{\prime }(t)\le <!--\; \le \; -->a_{1}u(t)+a_{2}v(t)+a_0 \\ {v}^{\prime }(t)\le <!--\; \le \; -->b_{1}u(t)+b_{2}v(t)+b_0 \end{gathered}$$
for suitable coefficients $a_0,a_1,a_2,b_0,b_1,b_2\in <!--\; \in \; -->\mathbb{R}.$ In particular I have $a_1,b_2<0$ and $0<b_1<|b_2|$
Is there a Gronwall lemma for this system of linear differential inqualities? Namely an (optimal) inequality of type
$$\begin{gathered} u(t)\le <!--\; \le \; -->F(t) \\ v(t)\le <!--\; \le \; -->G(t) \end{gathered}$$
where the functions $F,G:[0,\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->[0,\mathrm{\infty <!--\; \infty \; -->})$ depend on $u,v$ only through their initial values $u(0),v(0)$?

Suppose I have an element-wise inequality: $Ax\ge <!--\; \ge \; -->b$, where $A$ is a rectangular matrix with full row rank, and $x$ and $b$ are appropriately sized column vectors. I need to check if the inequality holds to branch some code I'm writing.
If I (thin) LQ decompose $A$, I have:
where $L$ is a lower triangular square matrix of full rank and Q is an orthogonal matrix.
As it turns out, I need to calculate $L^{-<!--\; -\; -->1}b$ anyway, so my first thought was to calculate it and just check:
$Qx\ge <!--\; \ge \; -->L^{-<!--\; -\; -->1}b$
However, this doesn't necessarily work, as depending on exactly how the matrix $A$ was decomposed, it might change the $\ge <!--\; \ge \; -->$ in to a $\le <!--\; \le \; -->$. For example, if $A=1$, and we decompose it as $L=-<!--\; -\; -->1$, $Q=-<!--\; -\; -->1$, multiplying both sides by $L^{-<!--\; -\; -->1}$ switches the direction of the inequality.
Under what circumstances will multiplying both sides of an inquality system by a matrix not change the direction of the inequality? My intuitive guess is that as long as the eigenvalues of the matrix are positive (that is, the matrix is positive definite), the inequality direction won't change. But I'm not sure how to prove/disprove that.

$q_i^{\prime }(0)=0$, for $i=1,2,3,4.$
$q_2(t)q_3(t)>q_1(t)q_4(t)$ for $0<=t<T$. $T$ is finite. At $t=T$ the inequality breaks.
$q_4(0)\ast <!--\; \ast \; -->q_3^{″}(0)>q_4^{″}(0)\ast <!--\; \ast \; -->q_3(0)$
$(q_4^{″}(0)+q_2^{″}(0))(q_1(0)+q_3(0)>(q_4(0)+q_2(0))(q_1^{″}(0)+q_3^{″}(0))$
This also holds but I suspect it is superfluous:
$q_2(0)\ast <!--\; \ast \; -->q_1^{″}(0)<q_2^{″}(0)\ast <!--\; \ast \; -->q_1(0)$
One can also add this without loss of generality:
$q_1(0)=1,q_2(0)=-<!--\; -\; -->1$
Prove or disprove that:
$q_4(t)\ast <!--\; \ast \; -->q_3^{″}(t)>q_4^{″}(t)\ast <!--\; \ast \; -->q_3(t)$
$(q_4^{″}(t)+q_2^{″}(t))(q_1(t)+q_3(t)>(q_4(t)+q_2(t))(q_1^{″}(t)+q_3^{″}(t))$
for $0<=t<T$

Calculating volume of body enclosed by surfaces
$x^2+y^2=9$; (cylinder)
$x+y+z=10$
$z=0$

find the solution of the following two inequalities for variables p and q:
$p<\sqrt{p^2-<!--\; -\; -->4q}$
$p<-<!--\; -\; -->\sqrt{p^2-<!--\; -\; -->4q}$

How to algebraically (without graphing) find coordinates delimiting the solution region of a system of linear inequalities?
For example
$\{\begin{array}{c}x\ge <!--\; \ge \; -->2\\ y\ge <!--\; \ge \; -->3\\ y\le <!--\; \le \; -->-<!--\; -\; -->x+5\\ x\le <!--\; \le \; -->3\end{array}$
There are 5 intersections in total: (2,3), (2,1), (3,2), (3,1) and (4,1), but the last is not part of the solution region.

Need a confirmation Let's say I have a system of inequalities
$\{\begin{array}{l}Ax\le <!--\; \le \; -->b\\ x\ge <!--\; \ge \; -->0\end{array}$
$A\in <!--\; \in \; -->{\mathbb{R}}^{m\times <!--\; \times \; -->n}$, $x\in <!--\; \in \; -->{\mathbb{R}}^{n\times <!--\; \times \; -->1}$, $b\in <!--\; \in \; -->{\mathbb{R}}^{m\times <!--\; \times \; -->1}$. $n\ge <!--\; \ge \; -->m$ Assume also that $A$ has full rank. Is it possible to manipulates the inequalities involved such that for $i=1\dots <!--\; \dots \; -->m$I have that $x_i$ is bounded (above or below) by $x_{i-<!--\; -\; -->1},...,x_0$? Something like
$a_{ii}{x}_i\le <!--\; \le \; -->C_i+\underset{j=0}{\overset{i-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}{c}_{ij}{x}_j$

Solve the system of linear inequalities with parameters
$\begin{array}{}\text{(*)}& \{\begin{array}{l}0\le <!--\; \le \; -->\phantom{-<!--\; -\; -->2}x+2y-<!--\; -\; -->3b+3a\le <!--\; \le \; -->2\\ 0\le <!--\; \le \; -->-<!--\; -\; -->2x-<!--\; -\; -->3y+6b\phantom{+3a}\le <!--\; \le \; -->1\\ 0\le <!--\; \le \; -->x\le <!--\; \le \; -->1\\ 0\le <!--\; \le \; -->y\le <!--\; \le \; -->2\\ 0\le <!--\; \le \; -->a\le <!--\; \le \; -->1\\ 0\le <!--\; \le \; -->b\le <!--\; \le \; -->1\end{array}\end{array}$
Here x,y are unknown variables and a,b are parameters.
My attempt. By adding the inequalities with some coeficients I separated the variables and get the simple system
$\begin{array}{}\text{(**)}& \{\begin{array}{l}0\le <!--\; \le \; -->y+6a\le <!--\; \le \; -->5,\\ 0\le <!--\; \le \; -->-<!--\; -\; -->x+9a+3b\le <!--\; \le \; -->8.\end{array}\end{array}$
and I am able to solve it. But the solutions of the last system are not solution of the initial system!
Maple and wolframAlpha cant solve the system.
P.S.1 For $a=\frac{63}{100}$ and $b=\frac{59}{100}$ Maple gives the solutions
$\begin{array}{c}\{x=1,\frac{9}{50}\le <!--\; \le \; -->y,y\le <!--\; \le \; -->\frac{11}{25}\},\{x=-<!--\; -\; -->3/2y+\frac{127}{100},\frac{9}{50}<y,y<\frac{11}{25}\},\{\frac{9}{50}<y,x<1,y<\frac{11}{25},-<!--\; -\; -->3/2y+\frac{127}{100}<x\},\{y=\frac{11}{25},\frac{61}{100}\le <!--\; \le \; -->x,x<1\},\{x=-<!--\; -\; -->3/2y+\frac{127}{100},\frac{11}{25}<y,y<\frac{127}{150}\},\{\frac{11}{25}<y,x<-<!--\; -\; -->2y+\frac{47}{25},y<\frac{127}{150},-<!--\; -\; -->3/2y+\frac{127}{100}<x\},\{x=-<!--\; -\; -->2y+\frac{47}{25},\frac{11}{25}<y,y<\frac{127}{150}\},\{x=0,\frac{127}{150}\le <!--\; \le \; -->y,y\le <!--\; \le \; -->\frac{47}{50}\},\{y=\frac{127}{150},x\le <!--\; \le \; -->\frac{14}{75},0<x\},\{0<x,\frac{127}{150}<y,x<-<!--\; -\; -->2y+\frac{47}{25},y<\frac{47}{50}\},\\ \{x=-<!--\; -\; -->2y+\frac{47}{25},\frac{127}{150}<y,y<\frac{47}{50}\}\end{array}$

Integer programming, system of linear inequalities.
$t_{01}+t_{11}+t_{21}\ge <!--\; \ge \; -->4$
$t_{02}+t_{12}+t_{22}\ge <!--\; \ge \; -->4$
$t_{10}+t_{11}+t_{12}\ge <!--\; \ge \; -->4$
$t_{10}+t_{01}+t_{22}\ge <!--\; \ge \; -->4$
$t_{10}+t_{02}+t_{21}\ge <!--\; \ge \; -->4$
$t_{20}+t_{21}+t_{22}\ge <!--\; \ge \; -->4$
$t_{20}+t_{01}+t_{12}\ge <!--\; \ge \; -->4$
$t_{20}+t_{02}+t_{11}\ge <!--\; \ge \; -->4$
There are 8 integer unknowns, $t_{ij}\ge <!--\; \ge \; -->0,0\le <!--\; \le \; -->i\le <!--\; \le \; -->2,0\le <!--\; \le \; -->j\le <!--\; \le \; -->2,t_{00}$ is missing. Every unknown is used in exacly 3 inequalitites. The problem is to find solution that minimizes sum of all $t_{ij}$
$3\times <!--\; \times \; -->\sum <!--\; \sum \; -->\ge <!--\; \ge \; -->8\times <!--\; \times \; -->4$
$\sum <!--\; \sum \; -->\ge <!--\; \ge \; -->11$
I have got this solution $t_{ij}=\left[\begin{array}{ccc}& 2& 2\\ 2& 2& 0\\ 2& 0& 2\end{array}\right]$ and I need hint to prove that it optimal. Sum of all unknowns is 12 here. Maybe linear programming method can be used?

The sum of the prices of a pen, an eraser and a notebook is 100 rupees. The price of a notebook is greater than the price of two pens. The price of three pens is greater than the price of four erasers and the price of three erasers is greater than the price of a notebook. If all of the prices are in integers, find each of their prices.
Let the prices of a pen, an eraser and a notebook be $p$, $e$ and $n$ respectively. Then we have the following:
$$\begin{gathered} p+e+n=100 \\ n>2p \\ 3p>4e \\ 3e>n \end{gathered}$$
We have $p>\frac{4}{3}e$ and $n>2p>\frac{8}{3}e$.
So,
$$\begin{gathered} \frac{4}{3}e+e+\frac{8}{3}e<100 \\ ⟹<!--\; ⟹\; -->e<20 \end{gathered}$$
Similarly, we get
$$\begin{gathered} p\le <!--\; \le \; -->27\text{and} \\ n\le <!--\; \le \; -->56 \end{gathered}$$
Setting the values of $e$ and $p$ and with a bit brute forcing, I got $e=19$, $p=26$ and $n=55$ which I think is the only solution.
Is the solution correct?

Graphically solving system of inequalitiesJillian is going to college next year and has decided to start working this summer. She makes $20$ dollars an hour working at a farm and $16$ teaching gymnastics. Write $3$ inequalities.
1. Make at most $\mathrm{\$<!--\; \$\; -->}800$
2. Work more than $10$ hours per week
3. Work less hours as a gymnastics teacher.
If we set $Y$ = gymnastics, $X$ = farm.
Would the following answers fit?
1. $20x+16y<800$ (making at most $\mathrm{\$<!--\; \$\; -->}800$)
2. $x+y>10$
(working more than 10 hours a week).3. $x>y$ (working more at the farm than the gymastics class)