Question Number 178600 by Acem last updated on 18/Oct/22
$${Solve}\:\mathrm{1}{st}:\:\mid{x}−\mathrm{9}\mid\leqslant\:−\mathrm{1}\:,\:\mathrm{2}{nd}:\:\mid\mathrm{10}{x}+\mathrm{1}\mid>\:−\mathrm{4} \\ $$
Answered by Acem last updated on 19/Oct/22
$$ \\ $$ $$\mathrm{1}{st}:\:{Impossible} \\ $$ $$ \\ $$ $$\mathrm{2}{nd}:\:{x}\in\:\mathbb{R} \\ $$ $$ \\ $$
Answered by Rasheed.Sindhi last updated on 19/Oct/22
$$\left.\mathrm{1}\right) \\ $$ $$\mid{x}−\mathrm{9}\mid\leqslant\:−\mathrm{1}\:{Contradict}\:{to}\:{the}\:{definition}. \\ $$ $$\left.\mathrm{2}\right) \\ $$ $$\mid\mathrm{10}{x}+\mathrm{1}\mid>\:−\mathrm{4} \\ $$ $$\Rightarrow\:\mid\mathrm{10}{x}+\mathrm{1}\mid\geqslant\:\mathrm{0} \\ $$ $$\Rightarrow{x}\geqslant−\frac{\mathrm{1}}{\mathrm{10}}\:\mathbb{A}\mathrm{ns} \\ $$
Commented bymr W last updated on 19/Oct/22
$$\mid\mathrm{10}{x}+\mathrm{1}\mid\:{is}\:{always}\:\geqslant\mathrm{0}>−\mathrm{4},\:{so}\:{x}\:{can} \\ $$ $${be}\:{any}\:{value}\:\in{R}. \\ $$
Commented byRasheed.Sindhi last updated on 19/Oct/22
$${Sorry}\:{for}\:{my}\:{mistake}\:\boldsymbol{{sir}}!\: \\ $$
Commented byAcem last updated on 19/Oct/22
$${My}\:{friend},\:{even}\:{if}\:{the}\:{qusetion}\:{was}\:{solving} \\ $$ $$\:\mid\mathrm{10}{x}+\mathrm{1}\mid\geqslant\:\mathrm{0}\:,\:{it}\:{has}\:{two}\:{separeted}\:{solutions}\:{not}\:{only}\:{one} \\ $$ $$\:{Either}\:{x}\geqslant\:\frac{−\mathrm{1}}{\mathrm{10}}\:\:\:\:\:\:{or}\:\:\:\:{x}\leqslant\:\frac{−\mathrm{1}}{\mathrm{10}} \\ $$