If u is an algebraic expression and c is a positive number,1. The solutions of |u| < c are the numbers that satisfy

Instead of solving the equality Ax=b, I want to solve the inequality Ax>=b. In general, how can I solve this problem with n unknowns, or when A has n columns?
When A has 1 column (or, when each equation in this system has only 1 unknown x) I can solve it easily by isolating x in each inequality and combining the inequalities to yield 1 inequality that bounds x. When choosing a satisfactory x, I simply refer to the yielded inequality.
When A has 2 columns (or, when each equations has 2 unknowns x and y), I graph the constraints and find the shaded regions, and use the appropriate equations for different x intervals (i.e., for x = 0 to x = 10 use x + y <= 50 and for x = 11 to x = 20 use 10 x +10 y <= 30). When choosing a satisfactory pair (x , y), I plug in x to the equation appropriate for the x interval, and I can choose any y value within the bounds for y.

System of parametric inequalities
I'm doing some systems of parametric inequalities, but I can not understand how to proceed.This is the system:
$x-2a<1+a$
$\frac{x}{2}-2a<2x-(a-3)$
$\frac{1}{3}-x<a$ per $a>0$
These are the solutions I've found:
$x<3a+1$
$x>\frac{-2a-6}{3}$
$x>\frac{-3a+1}{3}$
But now I do not know how to proceed. I thought I could compare the solutions like:
$3a+1>\frac{-2a-6}{3}>\frac{-3a+1}{3}$

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)

Solve ${\displaystyle (x+1)(x-3)>0}$.

Let:
$f(x_1,\cdots ,x_n)=\prod _i{x}_i(1-x_i)\prod _{i<j}|x_i-x_j|$
Suppose all $x_i\in (0,1)$ are fixed and $\sum _i{x}_i<\frac{n}{2}$. Show that there is some $i$ and a sufficiently small $ϵ$ so that $x_i\mapsto x_i+ϵ$ doesn't decrease the value of $f$.
That is to say, at least one of the partial derivatives of $f$ is non-negative.
After taking the derivative in each $x_i$ one gets a system of inequalities. I was able to prove the statement for $n=2,3$ this way through basically brute force. This doesn't generalize well though.

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

Draw the graph of inequalities
${\displaystyle \{\begin{array}{c}|x-2|\le 5\\ |y-4|>2\end{array}}$