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Question Number 207502 by hardmath last updated on 17/May/24

$$\frac{6}{\mid \mathbf{x}-4\mid -3}⩾1$$

Answered by efronzo1 last updated on 17/May/24

$$\mathrm{Let}\mid \mathrm{x}-4\mid =\mathrm{y}⩾0$$$$\Rightarrow \frac{6}{\mathrm{y}-3}⩾1$$$$\Rightarrow \frac{6-(\mathrm{y}-3)}{\mathrm{y}-3}⩾0$$$$\Rightarrow \frac{9-\mathrm{y}}{\mathrm{y}-3}⩾0$$$$\Rightarrow \frac{\mathrm{y}-9}{\mathrm{y}-3}⩽0$$$$\Rightarrow 3<\mathrm{y}⩽9$$$$\{\begin{array}{l}\mathrm{y}>3\Rightarrow \mid \mathrm{x}-4\mid >3\\ \mathrm{y}⩽9\Rightarrow \mid \mathrm{x}-4\mid ⩽9\end{array}$$$$\{\begin{array}{l}\mathrm{x}<1\mathrm{or}\mathrm{x}>7\\ -5⩽\mathrm{x}⩽13\end{array}$$$$\therefore \{-5⩽\mathrm{x}<1\cup 7<\mathrm{x}⩽13\}$$

Answered by A5T last updated on 17/May/24

$$when\mid x-4\mid >3[x>7orx<1]\Rightarrow 6⩾\mid x-4\mid -3$$$$\Rightarrow \mid x-4\mid ⩽9$$$$-9⩽x-4⩽9\Rightarrow -5⩽x⩽13$$$$\Rightarrow -5⩽x<1or7<x⩽13...\left(i\right)$$$$when\mid x-4\mid <3[1<x<7]\Rightarrow 6⩽\mid x-4\mid -3$$$$\Rightarrow \mid x-4\mid ⩾9\Rightarrow x⩾13orx⩽-5$$$$\Rightarrow \mathrm{\varnothing }..\left(ii\right)$$$$\left(i\right)\mathrm{\&}\left(ii\right)\Rightarrow -5⩽x<1or7<x⩽13$$

Commented by hardmath last updated on 17/May/24

$$\mathrm{thank}\mathrm{you}\mathrm{professors}$$

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