Question Number 27977 by NECx last updated on 18/Jan/18
$$\mid\mathrm{2}{x}+\mathrm{1}\mid\leqslant\mathrm{2} \\ $$
Answered by Rasheed.Sindhi last updated on 18/Jan/18
$$\mid\mathrm{2x}+\mathrm{1}\mid\leqslant\mathrm{2} \\ $$$$\pm\left(\mathrm{2x}+\mathrm{1}\right)\leqslant\mathrm{2} \\ $$$$\mathrm{2x}+\mathrm{1}\leqslant\mathrm{2}\:\wedge\:−\mathrm{2x}−\mathrm{1}\leqslant\mathrm{2} \\ $$$$\mathrm{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\:\wedge\:\mathrm{x}\geqslant−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:\:\:\:−\frac{\mathrm{3}}{\mathrm{2}}\leqslant\mathrm{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$