Inequality with modulus I would be glad if someone will help me to understand how to solve inequalities as the following one:
$$\vert 6-3x\vert+x \leq \vert x+2\vert$$
I remember that I need to see where the modulus is zeroing, which is $2$ and $-2$. It's kinda pathetic, but I don't how to move on from this. 
 A: Hint: $$|6-3x|=\begin{cases}\phantom{-}6-3x, &x\le2 \\ -6+3x, &x\ge2\end{cases}$$ and similarly 
$$|x+2|=\begin{cases}\phantom{-}x+2, &x\ge-2 \\ -x-2, &x\le-2\end{cases}$$ Now take $3$ cases, $x\le-2$, $-2\le x \le 2$ and $x\ge2$ and solve. At the end of each case, compare the result with the condition you started with in order to determine the answer.

For example, for $x\ge2$ you have that $$|6-3x|+x\le|2+x| \iff 6-3x+x\le x+2 \iff x\ge \frac{4}{3}$$ Now, $x\ge 2$ and $x\ge 4/3$ are both true when $x\ge 2$. Thus the first interval that satisfies the inequality is $[2, +\infty)$. Now repeat for the other 2 cases to determine all solutions.
A: As you've pointed out, one of the moduli zeroes out at each of $x=\pm 2.$
One approach you can take, then, is to consider three cases: $x<-2,$ $-2\le x<2,$ and $x\ge 2.$ See what happens to the left- and right-hand side in each case, then solve, bearing the restriction in mind (since there may appear to be solutions that are then ruled out by the restriction in each case).
