Sketch the graph of the solution of the system of linear inequlities -3x+2y<6 x+4y>-2 2x+y<3

How to solve this complex inequality system
$1\le z\overline{z}\le 4$ and $|\mathrm{\Im }(z)|<\mathrm{\Re }(z)$

Solve a system of inequalities
$\{\begin{array}{ll}{\log }_2^2(-{\log }_{2}x)+{\log }_2{\log }_2^{2}x\le 3& \\ -4|x^2-1|-3\ge \frac{1}{x^2-1}& \end{array}$
What I've tried:Make substitution $t=x^2-1$ and solve second inequality:
$-4|t|-3\ge \frac{1}{t}<=>-4|t|-3-\frac{1}{t}\ge 0$
Where I got interval for $t$: $-\frac{1}{4}\le t<0$
And then for $x$: $x\in (-1;-\frac{\sqrt{3}}{2}]\cup [\frac{\sqrt{3}}{2};1)$
But I have no thoughts how to solve first inequality.

Need to find a solution to the following system of linear inequalities:
$\begin{array}{rl}{x}_1-x_2& \le 1\\ x_1-x_4& \le -4\\ x_2-x_3& \le 2\\ x_2-x_5& \le 7\\ x_2-x_6& \le 5\\ x_3-x_6& \le 10\\ x_4-x_2& \le 2\\ x_5-x_1& \le -1\\ x_5-x_4& \le 3\\ x_6-x_3& \le -8\end{array}$
Is there any systematic way to find a solution?

How can we express union of two or more constraints, for example $x\ge 0\vee y\ge 0$, in an inequality system as below:
$$\begin{gathered} f_1(x,y)\ge 0 \\ {f}_2(x,y)\ge 0 \\ {f}_3(x,y)\ge 0 \\ ⋮ \end{gathered}$$
where $f_i\in {\mathbb{C}}^1$. This feasible set is not necessarily a convex set, hence there is not any linear inequality system for this problem. Is there any non-linear one?

Quadratic inequalities - why does the solution seem different for $(x-a)(x-b)<0$ and $(x-a)(x-b)>0$
$3^{2x}-10\cdot 3^x+9>0$
$(3^x-9)(3^x-1)>0$
$3^x\in (-\mathrm{\infty };1)\cup (9;+\mathrm{\infty })$
So to find x we need to solve:
$3^x>9$ and $3^x>1$ (1)
and we get:
$x\in (-\mathrm{\infty };0)\cup (2;+\mathrm{\infty })$
Second example:
$5^{2x}-6\cdot 5^x+5<0$
$(5^x-5)(5^x-1)<0$
$5^x\in (1;5)$
So to find x we need to solve:
$\{\begin{array}{l}{5}^x<5\\ 5^x>1\end{array}$ (2)
and we get:
$x\in (0;1)$
My question is why in first example I get answer by solving 2 equations (1) seperately but in second example I get answer by solving system of equations (2)

Basic proof showing a system of equations is inconsistent
I have 8 variables, all between 0 and 1. I have a system of 4 inequalities as follows
$\mid a_1-x_1\mid -\mid b_1-x_1\mid +b_2-a_2<0$
$\mid b_1-y_1\mid -\mid a_1-y_1\mid +a_2-b_2<0$
$\mid a_1-y_1\mid -\mid a_1-x_1\mid +x_2-y_2<0$
$\mid b_1-x_1\mid -\mid b_1-y_1\mid +y_2-x_2<0$
Is there a way to prve it without expanding the absolute values?

How would one go about solving the system of five equations:
$p^2=p+q-2r+2s+t-8$
$q^2=-p-2q-r+2s+2t-6$
$r^2=3p+2q+r+2s+2t-31$
$s^2=2p+q+r+2s+2t-2$
$t^2=p+2q+3r+2s+t-8$
over the integers?