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Question Number 84394 by jagoll last updated on 12/Mar/20

$$\mathrm{find}\mathrm{the}\mathrm{solution}$$$$\frac{2\mathrm{x}}{\mathrm{x}-2}⩽\mid \mathrm{x}-3\mid$$

Commented by john santu last updated on 12/Mar/20

$$\left(\mathrm{i}\right)\mathrm{x}⩾3\Rightarrow \frac{2\mathrm{x}}{\mathrm{x}-2}⩽\mathrm{x}-3$$$$\Rightarrow 2\mathrm{x}⩽(\mathrm{x}-3)(\mathrm{x}-2)$$$$\Rightarrow 2\mathrm{x}⩽{\mathrm{x}}^2-5\mathrm{x}+6$$$$\Rightarrow {\mathrm{x}}^2-7\mathrm{x}+6⩾0,(\mathrm{x}-1)(\mathrm{x}-6)⩾0$$$$\{\mathrm{x}⩽1\vee \mathrm{x}⩾6\}\wedge \mathrm{x}⩾3$$$$\bullet )\mathrm{x}⩾6$$$$\left(\mathrm{ii}\right)\mathrm{x}<3\Rightarrow \frac{2\mathrm{x}}{\mathrm{x}-2}⩽3-\mathrm{x}$$$$\Rightarrow \frac{2\mathrm{x}-(3-\mathrm{x})(\mathrm{x}-2)}{\mathrm{x}-2}⩽0$$$$\frac{2\mathrm{x}-(-{\mathrm{x}}^2+5\mathrm{x}-6)}{\mathrm{x}-2}⩽0$$$$\frac{{\mathrm{x}}^2-3\mathrm{x}+6}{\mathrm{x}-2}⩽0\Rightarrow \bullet )\mathrm{x}<2$$$$\mathrm{the}\mathrm{solution}(-\mathrm{\infty },2)\vee [6,\mathrm{\infty })$$

Commented by jagoll last updated on 12/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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