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Question Number 115035 by bobhans last updated on 23/Sep/20

solve ∣ 4−(3/x) ∣ < 8

$${solve}\:\mid\:\mathrm{4}−\frac{\mathrm{3}}{{x}}\:\mid\:<\:\mathrm{8} \\ $$

Commented bybemath last updated on 23/Sep/20

⇔ −8 < 4−(3/x) < 8 case(1) → −8 < 4−(3/x) (3/x) < 12 ; (1/x) < 4 ; ((1−4x)/x) < 0 ((4x−1)/x) > 0 ; x < 0 ∪ x > (1/4) case(2) 4−(3/x) < 8 ; −(3/x) < 4 (3/x) > −4 ; ((3+4x)/x) > 0 x < −(3/4) ∪ x > 0 solution set is (−∞, −(3/4)) ∪((1/4), ∞)

$$\Leftrightarrow\:−\mathrm{8}\:<\:\mathrm{4}−\frac{\mathrm{3}}{{x}}\:<\:\mathrm{8} \\ $$ $${case}\left(\mathrm{1}\right)\:\rightarrow\:−\mathrm{8}\:<\:\mathrm{4}−\frac{\mathrm{3}}{{x}} \\ $$ $$\:\:\:\:\:\frac{\mathrm{3}}{{x}}\:<\:\mathrm{12}\:;\:\frac{\mathrm{1}}{{x}}\:<\:\mathrm{4}\:;\:\frac{\mathrm{1}−\mathrm{4}{x}}{{x}}\:<\:\mathrm{0} \\ $$ $$\:\:\:\:\:\:\frac{\mathrm{4}{x}−\mathrm{1}}{{x}}\:>\:\mathrm{0}\:;\:{x}\:<\:\mathrm{0}\:\cup\:{x}\:>\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$ $${case}\left(\mathrm{2}\right)\:\mathrm{4}−\frac{\mathrm{3}}{{x}}\:<\:\mathrm{8}\:;\:−\frac{\mathrm{3}}{{x}}\:<\:\mathrm{4} \\ $$ $$\:\:\frac{\mathrm{3}}{{x}}\:>\:−\mathrm{4}\:;\:\frac{\mathrm{3}+\mathrm{4}{x}}{{x}}\:>\:\mathrm{0} \\ $$ $$\:\:{x}\:<\:−\frac{\mathrm{3}}{\mathrm{4}}\:\cup\:{x}\:>\:\mathrm{0} \\ $$ $${solution}\:{set}\:{is}\:\left(−\infty,\:−\frac{\mathrm{3}}{\mathrm{4}}\right)\:\cup\left(\frac{\mathrm{1}}{\mathrm{4}},\:\infty\right) \\ $$

Commented bybobhans last updated on 23/Sep/20

santuyy

$${santuyy} \\ $$

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