Question Number 175618 by daus last updated on 04/Sep/22
Commented by mr W last updated on 04/Sep/22
$${x}\geqslant\mathrm{0} \\ $$
Commented by mr W last updated on 05/Sep/22
Answered by greougoury555 last updated on 04/Sep/22
$$\:\mid\mathrm{2}{x}+\mathrm{1}\mid\:\geqslant\:\mid\mathrm{2}{x}−\mathrm{1}\mid\: \\ $$$$\:\mathrm{2}{x}+\mathrm{1}\:\geqslant\mathrm{2}{x}−\mathrm{1}\:\cup\:\mathrm{2}{x}+\mathrm{1}\:\leqslant\mathrm{1}−\mathrm{2}{x} \\ $$$$\:\forall{x}\epsilon\:\mathbb{R}\:\cup\:{x}\leqslant\:\mathrm{0}\Rightarrow\:{x}\leqslant\mathrm{0} \\ $$