Question Number 207502 by hardmath last updated on 17/May/24
$$\frac{\mathrm{6}}{\mid\boldsymbol{\mathrm{x}}\:−\:\mathrm{4}\mid\:−\:\mathrm{3}}\:\:\geqslant\:\:\mathrm{1} \\ $$
Answered by efronzo1 last updated on 17/May/24
$$\:\:\:\mathrm{Let}\:\mid\mathrm{x}−\mathrm{4}\mid\:=\:\mathrm{y}\:\geqslant\mathrm{0} \\ $$$$\:\Rightarrow\frac{\mathrm{6}}{\mathrm{y}−\mathrm{3}}\:\geqslant\mathrm{1}\: \\ $$$$\:\Rightarrow\:\frac{\mathrm{6}−\left(\mathrm{y}−\mathrm{3}\right)}{\mathrm{y}−\mathrm{3}}\:\geqslant\:\mathrm{0}\: \\ $$$$\:\:\Rightarrow\:\frac{\mathrm{9}−\mathrm{y}}{\mathrm{y}−\mathrm{3}}\:\geqslant\:\mathrm{0}\: \\ $$$$\:\Rightarrow\frac{\mathrm{y}−\mathrm{9}}{\mathrm{y}−\mathrm{3}}\:\leqslant\:\mathrm{0}\: \\ $$$$\:\Rightarrow\mathrm{3}\:<\:\mathrm{y}\:\leqslant\mathrm{9}\: \\ $$$$\:\:\begin{cases}{\mathrm{y}>\mathrm{3}\Rightarrow\mid\mathrm{x}−\mathrm{4}\mid>\mathrm{3}\:}\\{\mathrm{y}\leqslant\mathrm{9}\Rightarrow\mid\mathrm{x}−\mathrm{4}\mid\leqslant\mathrm{9}}\end{cases} \\ $$$$\:\:\:\begin{cases}{\mathrm{x}<\mathrm{1}\:\mathrm{or}\:\mathrm{x}>\mathrm{7}}\\{−\mathrm{5}\leqslant\mathrm{x}\leqslant\mathrm{13}}\end{cases} \\ $$$$\:\therefore\:\left\{\:−\mathrm{5}\leqslant\mathrm{x}<\mathrm{1}\:\cup\:\mathrm{7}<\mathrm{x}\leqslant\mathrm{13}\:\right\}\: \\ $$
Answered by A5T last updated on 17/May/24
![when ∣x−4∣>3[x>7 or x<1]⇒6≥∣x−4∣−3 ⇒∣x−4∣≤9 −9≤x−4≤9⇒−5≤x≤13 ⇒−5≤x<1 or 7<x≤13...(i) when ∣x−4∣<3[1<x<7]⇒6≤∣x−4∣−3 ⇒∣x−4∣≥9⇒x≥13 or x≤−5 ⇒∅..(ii) (i)&(ii)⇒−5≤x<1 or 7<x≤13](https://www.tinkutara.com/question/Q207507.png)
$${when}\:\mid{x}−\mathrm{4}\mid>\mathrm{3}\left[{x}>\mathrm{7}\:{or}\:{x}<\mathrm{1}\right]\Rightarrow\mathrm{6}\geqslant\mid{x}−\mathrm{4}\mid−\mathrm{3} \\ $$$$\Rightarrow\mid{x}−\mathrm{4}\mid\leqslant\mathrm{9} \\ $$$$−\mathrm{9}\leqslant{x}−\mathrm{4}\leqslant\mathrm{9}\Rightarrow−\mathrm{5}\leqslant{x}\leqslant\mathrm{13} \\ $$$$\Rightarrow−\mathrm{5}\leqslant{x}<\mathrm{1}\:{or}\:\mathrm{7}<{x}\leqslant\mathrm{13}…\left({i}\right) \\ $$$${when}\:\mid{x}−\mathrm{4}\mid<\mathrm{3}\left[\mathrm{1}<{x}<\mathrm{7}\right]\Rightarrow\mathrm{6}\leqslant\mid{x}−\mathrm{4}\mid−\mathrm{3} \\ $$$$\Rightarrow\mid{x}−\mathrm{4}\mid\geqslant\mathrm{9}\Rightarrow{x}\geqslant\mathrm{13}\:{or}\:{x}\leqslant−\mathrm{5} \\ $$$$\Rightarrow\emptyset..\left({ii}\right) \\ $$$$\left({i}\right)\&\left({ii}\right)\Rightarrow−\mathrm{5}\leqslant{x}<\mathrm{1}\:{or}\:\mathrm{7}<{x}\leqslant\mathrm{13} \\ $$
Commented by hardmath last updated on 17/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professors} \\ $$