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Question Number 30405 by scientist last updated on 22/Feb/18

x^2 +y^2 =13 x^2 −3xy+y^2 =35 find the value of x and y

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{13} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{xy}+{y}^{\mathrm{2}} =\mathrm{35} \\ $$$${find}\:{the}\:{value}\:{of}\:{x}\:{and}\:{y} \\ $$

Answered by mrW2 last updated on 22/Feb/18

13−3xy=35 ⇒xy=((13−35)/3)=−((22)/3) (x+y)^2 =x^2 +y^2 +2xy=13−((44)/3)=−(5/3) ⇒x+y=±i(√(5/3)) (x−y)^2 =x^2 +y^2 −2xy=13+((44)/3)=((83)/3) ⇒x−y=±(√((83)/3)) ⇒x=(1/2)(±(√((83)/3))±i(√(5/3))) ⇒y=(1/2)(∓(√((83)/3))±i(√(5/3))) solution 1 { ((x=(1/2)((√((83)/3))+i(√(5/3))))),((y=(1/2)(−(√((83)/3))+i(√(5/3))))) :} solution 2 { ((x=(1/2)((√((83)/3))−i(√(5/3))))),((y=(1/2)(−(√((83)/3))−i(√(5/3))))) :} solution 3 { ((x=(1/2)(−(√((83)/3))+i(√(5/3))))),((y=(1/2)((√((83)/3))+i(√(5/3))))) :} solution 4 { ((x=(1/2)(−(√((83)/3))−i(√(5/3))))),((y=(1/2)((√((83)/3))−i(√(5/3))))) :}

$$\mathrm{13}−\mathrm{3}{xy}=\mathrm{35} \\ $$$$\Rightarrow{xy}=\frac{\mathrm{13}−\mathrm{35}}{\mathrm{3}}=−\frac{\mathrm{22}}{\mathrm{3}} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}=\mathrm{13}−\frac{\mathrm{44}}{\mathrm{3}}=−\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\Rightarrow{x}+{y}=\pm{i}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}=\mathrm{13}+\frac{\mathrm{44}}{\mathrm{3}}=\frac{\mathrm{83}}{\mathrm{3}} \\ $$$$\Rightarrow{x}−{y}=\pm\sqrt{\frac{\mathrm{83}}{\mathrm{3}}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\pm\sqrt{\frac{\mathrm{83}}{\mathrm{3}}}\pm{i}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right) \\ $$$$\Rightarrow{y}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mp\sqrt{\frac{\mathrm{83}}{\mathrm{3}}}\pm{i}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right) \\ $$$${solution}\:\mathrm{1\begin{cases}{{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\frac{\mathrm{83}}{\mathrm{3}}}+{i}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right)}\\{{y}=\frac{\mathrm{1}}{\mathrm{2}}\left(−\sqrt{\frac{\mathrm{83}}{\mathrm{3}}}+{i}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right)}\end{cases}} \\ $$$${solution}\:\mathrm{2\begin{cases}{{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\frac{\mathrm{83}}{\mathrm{3}}}−{i}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right)}\\{{y}=\frac{\mathrm{1}}{\mathrm{2}}\left(−\sqrt{\frac{\mathrm{83}}{\mathrm{3}}}−{i}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right)}\end{cases}} \\ $$$${solution}\:\mathrm{3\begin{cases}{{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(−\sqrt{\frac{\mathrm{83}}{\mathrm{3}}}+{i}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right)}\\{{y}=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\frac{\mathrm{83}}{\mathrm{3}}}+{i}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right)}\end{cases}} \\ $$$${solution}\:\mathrm{4\begin{cases}{{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(−\sqrt{\frac{\mathrm{83}}{\mathrm{3}}}−{i}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right)}\\{{y}=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\frac{\mathrm{83}}{\mathrm{3}}}−{i}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right)}\end{cases}} \\ $$

Commented by Rasheed.Sindhi last updated on 22/Feb/18

√3Я¥ ₪î¢3 §îЯ!

Commented by mrW2 last updated on 22/Feb/18

Thanks Sir!

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