Question Number 146535 by mathdanisur last updated on 13/Jul/21
$$\begin{cases}{\frac{\mathrm{15}}{{x}^{\mathrm{2}} +{xy}}\:+\:\frac{\mathrm{3}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{5},\mathrm{5}}\\{\frac{\mathrm{12}}{{x}^{\mathrm{2}} +{xy}}\:-\:\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{3}}\end{cases}\:\:\:\Rightarrow\:\:\mid{x}+{y}\mid=? \\ $$
Answered by liberty last updated on 14/Jul/21
$$\:\begin{cases}{\frac{\mathrm{30}}{{x}^{\mathrm{2}} +{xy}}+\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}=\mathrm{11}}\\{\frac{\mathrm{12}}{{x}^{\mathrm{2}} +{xy}}−\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}=\mathrm{3}}\end{cases} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\rightarrow\frac{\cancel{\mathrm{42}}\:^{\mathrm{3}} }{{x}^{\mathrm{2}} +{xy}}=\cancel{\mathrm{14}}\:^{\mathrm{1}} \: \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{xy}=\mathrm{3}\:\: \\ $$$$\Rightarrow{y}^{\mathrm{2}} +{xy}=\mathrm{6}\: \\ $$$$\Rightarrow\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\:\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{9}\: \\ $$$$\begin{cases}{{x}=\mathrm{3}−{y}\:{or}}\\{{x}=−\mathrm{3}−{y}}\end{cases}\Rightarrow\mid{x}+{y}\mid=\sqrt{\left({x}+{y}\right)^{\mathrm{2}} }=\mathrm{3} \\ $$
Commented by mathdanisur last updated on 14/Jul/21
$${thanks}\:{Ser}\:{cool} \\ $$
Answered by Olaf_Thorendsen last updated on 14/Jul/21
$$\begin{cases}{\frac{\mathrm{15}}{{x}^{\mathrm{2}} +{xy}}+\frac{\mathrm{3}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{5},\mathrm{5}\:\:\:\:\:\left(×\mathrm{2}\right)}\\{\frac{\mathrm{12}}{{x}^{\mathrm{2}} +{xy}}−\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{3}}\end{cases} \\ $$$$\begin{cases}{\frac{\mathrm{30}}{{x}^{\mathrm{2}} +{xy}}+\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{11}\:\:\:\:\:\left(\mathrm{1}\right)}\\{\frac{\mathrm{12}}{{x}^{\mathrm{2}} +{xy}}−\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{3}\:\:\:\:\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\::\:\frac{\mathrm{42}}{{x}^{\mathrm{2}} +{xy}}\:=\:\mathrm{14} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{xy}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} +{xy}\:=\:\mathrm{3}\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{1}\right)\::\:\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{11}−\frac{\mathrm{30}}{{x}^{\mathrm{2}} +{xy}} \\ $$$$\left(\mathrm{1}\right)\::\:\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{11}−\mathrm{30}×\frac{\mathrm{1}}{\mathrm{3}}\:=\:\mathrm{1} \\ $$$${y}^{\mathrm{2}} +{xy}\:=\:\mathrm{6}\:\:\:\:\:\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{3}\right)+\left(\mathrm{4}\right)\::\:{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} +{xy}\:=\:\mathrm{3}+\mathrm{6} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} \:=\:\mathrm{9} \\ $$$$\mid{x}+{y}\mid\:=\:\mathrm{3} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 14/Jul/21
$${thanks}\:{Ser}\:{cool} \\ $$