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x-2-y-2-13-x-2-3xy-y-2-35-find-the-value-of-x-and-y-




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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