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Question Number 91843 by jagoll last updated on 03/May/20

$$\{\begin{array}{l}\frac{1}{x}+y=2\\ x+\frac{1}{y}=3\end{array}$$$$find{x}^2+y^2$$

Commented by jagoll last updated on 03/May/20

Commented by jagoll last updated on 03/May/20

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Commented by john santu last updated on 03/May/20

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Commented by Prithwish Sen 1 last updated on 03/May/20

$$\mathrm{dividing}\mathrm{eqn}\left(\mathrm{i}\right)\mathrm{by}\left(\mathrm{ii}\right)$$$$\frac{\mathrm{y}}{\mathrm{x}}=\frac{2}{3}\Rightarrow \frac{\mathrm{x}}{3}=\frac{\mathrm{y}}{2}=\mathrm{k}\mathrm{say}\mathrm{then}\mathrm{x}=3\mathrm{k},\mathrm{y}=2\mathrm{k}$$$$\mathrm{putting}\mathrm{these}\mathrm{in}\mathrm{any}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}$$$$6{\mathbf{k}}^2-6\mathbf{k}+1=0\Rightarrow \mathbf{k}=\frac{3\pm \sqrt{3}}{6}\Rightarrow {\mathbf{k}}^2=\frac{2\pm \sqrt{3}}{6}$$$$\therefore {\mathbf{x}}^2+{\mathbf{y}}^2=(9+4).\frac{2\pm \sqrt{3}}{6}=13\frac{(2\pm \sqrt{3})}{6}$$

Commented by jagoll last updated on 03/May/20

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Answered by mr W last updated on 03/May/20

$$letx+y=u,xy=v$$$$\left(i\right)\times \left(ii\right):$$$$1+xy+\frac{1}{xy}+1=6$$$$xy+\frac{1}{xy}=4$$$$v+\frac{1}{v}=4$$$$v^2-4v+1=0$$$$\Rightarrow v=2\pm \sqrt{3}$$$$$$$$\left(i\right)+\left(ii\right):$$$$x+\frac{1}{x}+y+\frac{1}{y}=5$$$$(x+y)(1+\frac{1}{xy})=5$$$$u(1+\frac{1}{v})=5$$$$\Rightarrow u=\frac{5}{1+\frac{1}{v}}=\frac{5v}{v+1}$$$$$$$$x^2+y^2={(x+y)}^2-2xy=u^2-2v$$$$=\frac{25v^2}{{(v+1)}^2}-2v=\frac{13(2\pm \sqrt{3})}{6}$$

Commented by jagoll last updated on 03/May/20

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Answered by MJS last updated on 03/May/20

$$\mathrm{without}\mathrm{thinking},\mathrm{without}\mathrm{tricks}:$$$$\{\begin{array}{l}1+xy=2x\\ 1+xy=3y\end{array}\Rightarrow y=\frac{2}{3}x$$$$\Rightarrow 2x^2-6x+3=0\Rightarrow x=\frac{3\pm \sqrt{3}}{2}\Rightarrow y=\frac{3\pm \sqrt{3}}{3}$$

Commented by jagoll last updated on 03/May/20

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Commented by Ar Brandon last updated on 03/May/20

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