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Question Number 84770 by jagoll last updated on 16/Mar/20

$${\mathrm{x}}^2+{\mathrm{y}}^2=30$$ $$\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}=2$$ $$\mathrm{find}\mathrm{the}\mathrm{solution}\mathrm{x}\mathrm{\&}\mathrm{y}?$$

Commented byTony Lin last updated on 16/Mar/20

$$\frac{x+y}{xy}=2$$ $$x+y=2xy$$ $$x^2+y^2$$ $$={(x+y)}^2-2xy$$ $$={(x+y)}^2-(x+y)=30$$ $$[(x+y)-6][(x+y)+5]=0$$ $$\Rightarrow \{\begin{array}{l}x+y=6\\ xy=3\end{array}\{\begin{array}{l}x+y=-5\\ xy=-\frac{5}{2}\end{array}$$ $$x^2-6x+3=0$$ $$x=3\pm \sqrt{6}$$ $$y=\frac{3}{x}=3\mp \sqrt{6}$$ $$2x^2+10x-5=0$$ $$x=\frac{-5\pm \sqrt{35}}{2}$$ $$y=\frac{-5\mp \sqrt{35}}{2}$$

Answered by jagoll last updated on 16/Mar/20

Answered by behi83417@gmail.com last updated on 16/Mar/20

$$\mathrm{x}+\mathrm{y}=\mathrm{p},\mathrm{xy}=\mathrm{q}\Rightarrow \{\begin{array}{l}{\mathrm{p}}^2-2\mathrm{q}=30\\ \frac{\mathrm{p}}{\mathrm{q}}=2\Rightarrow \mathrm{p}=2\mathrm{q}\end{array}$$ $$\Rightarrow {\left(2\mathrm{q}\right)}^2-2\mathrm{q}-30=0\Rightarrow 2\mathrm{q}=\frac{1\pm \sqrt{121}}{2}$$ $$\Rightarrow 2\mathrm{q}=6,-5\Rightarrow \{\begin{array}{l}\mathrm{q}=3,-\frac{5}{2}\\ \mathrm{p}=6,-5\end{array}$$ $$\Rightarrow {\mathrm{z}}^2-6\mathrm{z}+3=0\Rightarrow {\mathrm{z}}_{1,2}=\mathrm{x},\mathrm{y}=\frac{6\pm \sqrt{24}}{2}=3\pm \sqrt{6}$$ $$\Rightarrow {\mathrm{z}}^2+5\mathrm{z}-\frac{5}{2}=0\Rightarrow {\mathrm{z}}_{3,4}=\mathrm{x},\mathrm{y}=\frac{-5\pm \sqrt{35}}{2}$$

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