Question Number 115035 by bobhans last updated on 23/Sep/20
$${solve}\:\mid\:\mathrm{4}−\frac{\mathrm{3}}{{x}}\:\mid\:<\:\mathrm{8} \\ $$
Commented by bemath last updated on 23/Sep/20
$$\Leftrightarrow\:−\mathrm{8}\:<\:\mathrm{4}−\frac{\mathrm{3}}{{x}}\:<\:\mathrm{8} \\ $$$${case}\left(\mathrm{1}\right)\:\rightarrow\:−\mathrm{8}\:<\:\mathrm{4}−\frac{\mathrm{3}}{{x}} \\ $$$$\:\:\:\:\:\frac{\mathrm{3}}{{x}}\:<\:\mathrm{12}\:;\:\frac{\mathrm{1}}{{x}}\:<\:\mathrm{4}\:;\:\frac{\mathrm{1}−\mathrm{4}{x}}{{x}}\:<\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\frac{\mathrm{4}{x}−\mathrm{1}}{{x}}\:>\:\mathrm{0}\:;\:{x}\:<\:\mathrm{0}\:\cup\:{x}\:>\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${case}\left(\mathrm{2}\right)\:\mathrm{4}−\frac{\mathrm{3}}{{x}}\:<\:\mathrm{8}\:;\:−\frac{\mathrm{3}}{{x}}\:<\:\mathrm{4} \\ $$$$\:\:\frac{\mathrm{3}}{{x}}\:>\:−\mathrm{4}\:;\:\frac{\mathrm{3}+\mathrm{4}{x}}{{x}}\:>\:\mathrm{0} \\ $$$$\:\:{x}\:<\:−\frac{\mathrm{3}}{\mathrm{4}}\:\cup\:{x}\:>\:\mathrm{0} \\ $$$${solution}\:{set}\:{is}\:\left(−\infty,\:−\frac{\mathrm{3}}{\mathrm{4}}\right)\:\cup\left(\frac{\mathrm{1}}{\mathrm{4}},\:\infty\right) \\ $$
Commented by bobhans last updated on 23/Sep/20
$${santuyy} \\ $$