Graph the solution set for the of linear inequalities x+y<=2-x+y<=2y<=-2

With $\mathbf{x}$ and $\mathbf{y}$ two vectors, the triangle inequality can be written as:
$\begin{array}{}\text{(A)}& |\mathbf{x}|-|\mathbf{y}|\le |\mathbf{x}+\mathbf{y}|\le |\mathbf{x}|+|\mathbf{y}|\end{array}$
which can be broken down into a system of inequalities (labelled as (1) and (2)):
$\{\begin{array}{l}|\mathbf{x}|-|\mathbf{y}|\le |\mathbf{x}+\mathbf{y}|\\ |\mathbf{x}+\mathbf{y}|\le |\mathbf{x}|+|\mathbf{y}|\end{array}$
then, with the assumption of $|\mathbf{y}|>0$, (1)-(2) yields:
$\begin{array}{}\text{(B)}& |\mathbf{x}|\le |\mathbf{x}+\mathbf{y}|\end{array}$
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?

Graph the solution set of the inequality or system of inequalities.
${\displaystyle \{\begin{array}{c}x\ge 0\text{and}y\ge 0\\ 3x+y\le 9\\ 2x+3y\ge 6\end{array}}$

Find the slope and y-intercept of the line. Graph the line ${\displaystyle 5x+y=5}$

Graph the solution set of each system if inequalities.
X is greater than or = 0
X+Y is less than or equal to 4
2x+y is less than or equal to 5

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

Property of a system of two inequalities
$\{\begin{array}{l}a+b>1\\ a-b>1\end{array}$
can I sum the second inequality to the first getting $a>1$? Or this property can be used only equations?

Use the given graph off to find a number ${\displaystyle \delta }$ such that if ${\displaystyle |x-1|<\delta }$ then ${\displaystyle |f\left(x\right)-1|<0.2}$