Question-160302

Question Number 160302 by cortano last updated on 27/Nov/21

Answered by Rasheed.Sindhi last updated on 27/Nov/21

{ ((8x^2 −8y^2 =24)),((3x^2 +3xy+6y^2 =24)) :}⇒5x^2 −3xy−14y^2 =0 ⇒5x^2 −10xy+7xy−14y^2 =0 5x(x−2y)+7y(x−2y)=0 (x−2y)(5x+7y)=0 x=2y ∣ x=−((7y)/5) x=2y x^2 −y^2 =3⇒4y^2 −y^2 =3⇒y=±1 ⇒x=2(±1)=±2 (x,y)=(1,2),(−1,−2) x=−((7y)/5) (−((7y)/5))^2 −y^2 =3⇒49y^2 −25y^2 =75 ⇒24y^2 =75⇒y^2 =((25)/8)⇒y=±(5/(2(√2))) ⇒x=−(7/5)(±(5/(2(√2))))=∓(7/( 2(√2))) (x,y)=((7/( 2(√2))),−(5/(2(√2)))),(−(7/( 2(√2))),(5/(2(√2)))) (x,y)=(1,2),(−1,−2),((7/( 2(√2))),−(5/(2(√2)))),(−(7/( 2(√2))),(5/(2(√2))))

$$\begin{cases}{\mathrm{8}{x}^{\mathrm{2}} −\mathrm{8}{y}^{\mathrm{2}} =\mathrm{24}}\\{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{xy}+\mathrm{6}{y}^{\mathrm{2}} =\mathrm{24}}\end{cases}\Rightarrow\mathrm{5}{x}^{\mathrm{2}} −\mathrm{3}{xy}−\mathrm{14}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{5}{x}^{\mathrm{2}} −\mathrm{10}{xy}+\mathrm{7}{xy}−\mathrm{14}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{5}{x}\left({x}−\mathrm{2}{y}\right)+\mathrm{7}{y}\left({x}−\mathrm{2}{y}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\left({x}−\mathrm{2}{y}\right)\left(\mathrm{5}{x}+\mathrm{7}{y}\right)=\mathrm{0} \\ $$$${x}=\mathrm{2}{y}\:\mid\:{x}=−\frac{\mathrm{7}{y}}{\mathrm{5}} \\ $$$$\underline{{x}=\mathrm{2}{y}} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{3}\Rightarrow\mathrm{4}{y}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{3}\Rightarrow{y}=\pm\mathrm{1} \\ $$$$\Rightarrow{x}=\mathrm{2}\left(\pm\mathrm{1}\right)=\pm\mathrm{2} \\ $$$$\left({x},{y}\right)=\left(\mathrm{1},\mathrm{2}\right),\left(−\mathrm{1},−\mathrm{2}\right) \\ $$$${x}=−\frac{\mathrm{7}{y}}{\mathrm{5}} \\ $$$$\left(−\frac{\mathrm{7}{y}}{\mathrm{5}}\right)^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{3}\Rightarrow\mathrm{49}{y}^{\mathrm{2}} −\mathrm{25}{y}^{\mathrm{2}} =\mathrm{75} \\ $$$$\Rightarrow\mathrm{24}{y}^{\mathrm{2}} =\mathrm{75}\Rightarrow{y}^{\mathrm{2}} =\frac{\mathrm{25}}{\mathrm{8}}\Rightarrow{y}=\pm\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow{x}=−\frac{\mathrm{7}}{\mathrm{5}}\left(\pm\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)=\mp\frac{\mathrm{7}}{\:\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\left({x},{y}\right)=\left(\frac{\mathrm{7}}{\:\mathrm{2}\sqrt{\mathrm{2}}},−\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{2}}}\right),\left(−\frac{\mathrm{7}}{\:\mathrm{2}\sqrt{\mathrm{2}}},\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{2}}}\right) \\ $$$$ \\ $$$$\left({x},{y}\right)=\left(\mathrm{1},\mathrm{2}\right),\left(−\mathrm{1},−\mathrm{2}\right),\left(\frac{\mathrm{7}}{\:\mathrm{2}\sqrt{\mathrm{2}}},−\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{2}}}\right),\left(−\frac{\mathrm{7}}{\:\mathrm{2}\sqrt{\mathrm{2}}},\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{2}}}\right) \\ $$$$ \\ $$

Commented by cortano last updated on 27/Nov/21

$$\mathrm{yes}\:\:\mathrm{nice} \\ $$

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