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Question Number 175706 by daus last updated on 05/Sep/22

Commented by daus last updated on 05/Sep/22

$$justhelpmeinj,k,l$$

Answered by mahdipoor last updated on 05/Sep/22

$$j:\mid \frac{2x+1}{x-3}\mid =1\Rightarrow \frac{2x+1}{x-3}=\pm 1\Rightarrow$$$$x=\frac{\mp 3-1}{2\mp 1}=-4and\frac{2}{3}$$$$(-\mathrm{\infty },-4)\Rightarrow \mid \frac{2x+1}{x-3}\mid >1$$$$(-4,\frac{2}{3})\Rightarrow \mid \frac{2x+1}{x-3}\mid <1$$$$(\frac{2}{3},+\mathrm{\infty })\Rightarrow \mid \frac{2x+1}{x-3}\mid >1$$$$youcananswerk\mathrm{\&}llikethis...$$

Commented by Tawa11 last updated on 05/Sep/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by cortano1 last updated on 05/Sep/22

$$\left(\mathrm{l}\right)\mid \frac{\mathrm{x}-3}{\mathrm{x}+1}\mid <2,\mathrm{x}\ne -1$$$$\mid \mathrm{x}-3\mid <\mid 2\mathrm{x}+2\mid$$$$(3\mathrm{x}-1)(-\mathrm{x}-5)<0$$$$(3\mathrm{x}-1)(\mathrm{x}+5)>0$$$$\mathrm{x}<-5\vee \mathrm{x}>\frac{1}{3}$$$$(-\mathrm{\infty },-5)\cup (\frac{1}{3},\mathrm{\infty })$$

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