why sometimes when solving inequalities, variables of both sides cancel? Case in point:

$|2x|\ge 2x+1$ can equivalent to $2x\ge 2x+1$ and $-2x\ge 2x+1$. The second resulting inequality yields you $x\le -1/4$ and is fine. But the first resulting inequality gives you $0\ge 1$, which doesn't make sense.

Why does this kind of things happen?
 A: In your case,
the second result
just means that
there are no solutions.
In general,
when a deduction
leads to a contradiction,
that means that
there are no solutions.
A: You have to look a little more closely at when the two cases apply. $|2x|\ge 2x+1$ is equivalent to $2x\ge 2x+1$ only when $2x\ge 0$, i.e., when $x\ge 0$; the fact that this inequality has no solutions simply tells you that $|2x|\ge 2x+1$ has no solutions with $x\ge 0$. The original inequality is equivalent to $-2x|ge 2x+1$ when $2x\le 0$, i.e., when $x\le 0$, and indeed, those are precisely the solutions that you get from this case.
Suppose that the inequality had been $|2x|\ge 2x-1$. Now when $x\ge 0$ this reduces to $2x\ge 2x-1$, which is true for every $x\in\Bbb R$. However, this case applies only when $x\ge 0$, so the actual solutions that you get from this branch of the analysis are the non-negative real numbers: $x\ge 0$.
When $x\le 0$, the inequality reduces to $-2x\ge 2x-1$, or $x\le\frac14$. However, it’s only when $x\le 0$ that this reduction is correct, so the solutions that you get in this case are only the $x$ such that $x\le 0$. Of course when you put the two branches of the analysis together, you find that the solution set is $x\le 0$ or $x\ge 0$, so every real number satisfies the inequality — but you couldn’t conclude that from just the first case, and you couldn’t conclude just from the second case that $\frac14$ satisfies it. You must always remember that each case comes with a built-in additional inequality: the one that puts you in that case in the first place, by making what’s inside the absolute value signs negative or non-negative.
A: Formally, $|2x|\ge 2x+1 \Longleftrightarrow x \in \{t:2t \geq 2t+1 \ \land t \geq 0\} \ \lor x \in \{t:-2t \geq 2t+1 \ \land t < 0\}$
In our situation, it just means that the set to the left of "$\lor$" is $\emptyset$. The solution set is therefore the set to the right, which is $x\leq-\frac 14$
