Question Number 179073 by yaslm last updated on 24/Oct/22
Answered by MJS_new last updated on 24/Oct/22
$$\mathrm{the}\:\mathrm{absolute}\:\mathrm{value}\:\mathrm{of}\:\mathrm{any}\:\mathrm{number}\:\mathrm{is}\:\mathrm{greater} \\ $$$$\mathrm{than}\:\mathrm{zero}\:\Rightarrow\:\mathrm{it}'\mathrm{s}\:\mathrm{greater}\:\mathrm{than}\:−\mathrm{2}\:\Rightarrow\:\mathrm{answer} \\ $$$$\mathrm{is} \\ $$$${x}\in\mathbb{C} \\ $$
Answered by thenxtkvng last updated on 24/Oct/22
$${x}−\mathrm{4}\geqslant−\mathrm{2}\Rightarrow\:{eqn}\mathrm{1} \\ $$$$−\left({x}−\mathrm{4}\right)\geqslant−\mathrm{2}\:\mathrm{eq}{n}\mathrm{2} \\ $$$${Eqn}\mathrm{1} \\ $$$${x}−\mathrm{4}\geqslant−\mathrm{2} \\ $$$${x}\geqslant−\mathrm{2}+\mathrm{4} \\ $$$${x}\geqslant\mathrm{2} \\ $$$${Eqn}\mathrm{2} \\ $$$$−\left({x}−\mathrm{4}\right)\geqslant−\mathrm{2} \\ $$$$−{x}+\mathrm{4}\geqslant−\mathrm{2} \\ $$$$−{x}\geqslant−\mathrm{2}−\mathrm{4} \\ $$$$\frac{−{x}}{−\mathrm{1}}\geqslant\frac{−\mathrm{6}}{−\mathrm{1}} \\ $$$${x}\geqslant\mathrm{6} \\ $$$$\therefore{x}\geqslant\mathrm{2},{x}\geqslant\mathrm{6} \\ $$
Commented by MJS_new last updated on 24/Oct/22
$$\mathrm{mistake}: \\ $$$$−{x}\geqslant−\mathrm{6} \\ $$$$\frac{−{x}}{−\mathrm{1}}\leqslant\frac{−\mathrm{6}}{−\mathrm{1}} \\ $$$${x}\leqslant\mathrm{6} \\ $$$$\Rightarrow \\ $$$${x}\geqslant\mathrm{2}\:\vee\:{x}\leqslant\mathrm{6}\:\Rightarrow\:{x}\in\mathbb{R} \\ $$
Answered by Acem last updated on 24/Oct/22
$$\:{x}\in\:\mathbb{R} \\ $$