Question Number 207718 by hardmath last updated on 24/May/24
$$\mid\mathrm{x}−\mathrm{3}\mid\:+\:\mid\mathrm{x}\:+\mathrm{2}\mid\:=\:\mathrm{9} \\ $$$$\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$
Answered by TonyCWX08 last updated on 24/May/24
$${There}\:{are}\:\mathrm{4}\:{possible}\:{cases}. \\ $$$${x}−\mathrm{3}+{x}+\mathrm{2}=\mathrm{9} \\ $$$$−\left({x}−\mathrm{3}\right)+{x}+\mathrm{2}=\mathrm{9} \\ $$$${x}−\mathrm{3}−\left({x}+\mathrm{2}\right)=\mathrm{9} \\ $$$$−\left({x}−\mathrm{3}\right)−\left({x}+\mathrm{2}\right)=\mathrm{9} \\ $$$$ \\ $$$$\mathrm{2}{x}=\mathrm{10},\:{x}\geqslant\mathrm{3},{x}\geqslant−\mathrm{2} \\ $$$$\emptyset \\ $$$$\emptyset \\ $$$$−\mathrm{2}{x}=\mathrm{8},{x}<\mathrm{3},{x}<−\mathrm{2} \\ $$$$ \\ $$$${x}=\mathrm{5} \\ $$$${x}=−\mathrm{4} \\ $$
Commented by hardmath last updated on 24/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Answered by A5T last updated on 24/May/24
$$\mathrm{9}=\mid{x}−\mathrm{3}\mid+\mid{x}+\mathrm{2}\mid\geqslant\mid\mathrm{2}{x}−\mathrm{1}\mid \\ $$$$\Rightarrow−\mathrm{9}\leqslant\mathrm{2}{x}−\mathrm{1}\leqslant\mathrm{9}\Rightarrow−\mathrm{4}\leqslant{x}\leqslant\mathrm{5} \\ $$$${Equality}\:{occurs}\:{at}\:{the}\:{boundary}\:{points} \\ $$$$\Rightarrow{x}=−\mathrm{4}\:{or}\:\mathrm{5} \\ $$