Question Number 207249 by hardmath last updated on 10/May/24
$$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{2y}^{\mathrm{2}} \:\:+\:\:\mathrm{xy}\:\:=\:\:\mathrm{37}}\\{\mathrm{y}^{\mathrm{2}} \:\:+\:\:\mathrm{2x}^{\mathrm{2}} \:\:+\:\:\mathrm{2xy}\:\:=\:\:\mathrm{26}}\end{cases}\:\:\:\:\mathrm{find}:\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:? \\ $$
Answered by A5T last updated on 10/May/24
$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}\right)=\mathrm{21}...\left({i}\right) \\ $$$${y}^{\mathrm{2}} −{x}^{\mathrm{2}} −{xy}=\mathrm{11}...\left({ii}\right) \\ $$$$\Rightarrow\left({i}\right)+\left({ii}\right):\:\mathrm{2}{y}^{\mathrm{2}} =\mathrm{32}\Rightarrow{y}^{\mathrm{2}} =\mathrm{16}\Rightarrow{y}=\underset{−} {+}\mathrm{4} \\ $$$${y}=\mathrm{4}\Rightarrow{x}^{\mathrm{2}} \underset{−} {+}\mathrm{4}{x}−\mathrm{5}=\mathrm{0}\Rightarrow{x}=−\mathrm{5}\:{or}\:\mathrm{1}\:{or}\:\mathrm{5}\:{or}\:\mathrm{1} \\ $$$${x}^{\mathrm{2}} =\mathrm{25}\:{or}\:\mathrm{1}\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{41}\:{or}\:\mathrm{17}. \\ $$
Commented by hardmath last updated on 10/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$