Question Number 207116 by hardmath last updated on 06/May/24
$$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{x}\:−\:\mathrm{2y}}\\{\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{y}\:−\:\mathrm{2x}}\end{cases}\:\:\:\:\:\Rightarrow\:\:\:\:\:\mathrm{x}\:−\:\mathrm{y}\:=\:? \\ $$
Answered by A5T last updated on 06/May/24
$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={x}−{y}+\mathrm{2}\left({x}−{y}\right)=\mathrm{3}\left({x}−{y}\right) \\ $$$${x}−{y}\neq\mathrm{0}\Rightarrow{x}+{y}=\mathrm{3} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={x}+{y}−\mathrm{2}\left({x}+{y}\right)\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =−{x}−{y} \\ $$$$\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}=−\mathrm{3}\Rightarrow\mathrm{2}{xy}=\mathrm{12}\Rightarrow{xy}=\mathrm{6} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{4}{xy}=\mathrm{9}−\mathrm{24}=−\mathrm{15} \\ $$$$\Rightarrow{x}−{y}=\underset{−} {+}{i}\sqrt{\mathrm{15}}\:{or}\:{x}−{y}=\mathrm{0} \\ $$
Answered by MATHEMATICSAM last updated on 07/May/24
$${x}^{\mathrm{2}} \:=\:{x}\:−\:\mathrm{2}{y}\:…\:\left(\mathrm{i}\right) \\ $$$${y}^{\mathrm{2}} \:=\:{y}\:−\:\mathrm{2}{x}\:…\:\left(\mathrm{ii}\right) \\ $$$$……………….. \\ $$$$\left(\mathrm{i}\right)\:−\:\left(\mathrm{ii}\right) \\ $$$${x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:=\:{x}\:−\:{y}\:+\:\mathrm{2}\left({x}\:−\:{y}\right) \\ $$$$\Rightarrow\:\left({x}\:+\:{y}\right)\left({x}\:−\:{y}\right)\:=\:\mathrm{3}\left({x}\:−\:{y}\right) \\ $$$$\Rightarrow\:{x}\:+\:{y}\:=\:\mathrm{3}\:\:\mathrm{when}\:{x}\:−\:{y}\:\neq\:\mathrm{0} \\ $$$$\left(\mathrm{i}\right)\:+\:\left(\mathrm{ii}\right) \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:{x}\:+\:{y}\:−\mathrm{2}\:\left({x}\:+\:{y}\right) \\ $$$$\Rightarrow\:\left({x}\:+\:{y}\right)^{\mathrm{2}} \:−\:\mathrm{2}{xy}\:=\:−\:\left({x}\:+\:{y}\right) \\ $$$$\Rightarrow\:\mathrm{3}^{\mathrm{2}} \:−\:\mathrm{2}{xy}\:=\:−\mathrm{3} \\ $$$$\Rightarrow\:\mathrm{9}\:+\:\mathrm{3}\:=\:\mathrm{2}{xy} \\ $$$$\Rightarrow\:{xy}\:=\:\mathrm{6} \\ $$$$\mathrm{Now}\:\left({x}\:−\:{y}\right)^{\mathrm{2}} \:=\:\left({x}\:+\:{y}\right)^{\mathrm{2}} \:−\:\mathrm{4}{xy} \\ $$$$=\:\mathrm{3}^{\mathrm{2}} \:−\:\mathrm{4}\:×\:\mathrm{6}\:=\:\mathrm{9}\:−\:\mathrm{24}\:=\:−\:\mathrm{15} \\ $$$$\therefore\:\:{x}\:−\:{y}\:=\:\pm\:{i}\sqrt{\mathrm{15}} \\ $$$$ \\ $$$$\therefore\:\boldsymbol{{x}}\:−\:\boldsymbol{{y}}\:=\:\pm\:\boldsymbol{{i}}\sqrt{\mathrm{15}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{{x}}\:−\:\boldsymbol{{y}}\:=\:\mathrm{0}\:\left(\boldsymbol{\mathrm{Ans}}\right)\: \\ $$