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Question Number 99045 by bobhans last updated on 18/Jun/20

$$\{\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2=10\\ {\mathrm{x}}^2-5\mathrm{xy}+{6\mathrm{y}}^2=0\end{array}$$ $$\mathrm{find}\mathrm{x}\mathrm{\&}\mathrm{y}$$

Answered by bemath last updated on 18/Jun/20

$$\iff {\mathrm{x}}^2-5\mathrm{xy}+{6\mathrm{y}}^2=0$$ $$(\mathrm{x}-2\mathrm{y})(\mathrm{x}-3\mathrm{y})=0$$ $$\{\begin{array}{l}\mathrm{x}=2\mathrm{y}\\ \mathrm{x}=3\mathrm{y}\end{array}$$ $$\mathrm{case}\left(1\right)\mathrm{x}=2\mathrm{y}\Rightarrow {5\mathrm{y}}^2=10,\mathrm{y}=\pm \sqrt{2}$$ $$\mathrm{case}\left(2\right)\mathrm{x}=3\mathrm{y}\Rightarrow {10\mathrm{y}}^2=10,\mathrm{y}=\pm 1$$ $$\mathrm{solution}(\pm 2\sqrt{2},\pm \sqrt{2});(\pm 3,\pm 1)$$

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