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Question Number 141372 by iloveisrael last updated on 18/May/21

Answered by EDWIN88 last updated on 18/May/21

$$\left(1\right)\mathrm{xy}+\frac{\mathrm{x}}{\mathrm{y}}=5$$$$\left(2\right)\mathrm{xy}-\frac{\mathrm{x}}{\mathrm{y}}=\frac{1}{5}$$$$\{\begin{array}{l}2\mathrm{xy}=\frac{26}{5}\to \mathrm{xy}=\frac{13}{5}\\ \frac{2\mathrm{x}}{\mathrm{y}}=\frac{24}{5}\to \frac{\mathrm{x}}{\mathrm{y}}=\frac{12}{5}\end{array}$$$$\Rightarrow \mathrm{xy}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\frac{13\times 12}{25}$$$$\Rightarrow {\mathrm{x}}^2=\frac{12\times 13}{25}\to \{\begin{array}{l}\mathrm{x}=\frac{2\sqrt{39}}{5}\to \mathrm{y}=\frac{13}{5\left(\frac{2\sqrt{39}}{5}\right)}=\frac{\sqrt{39}}{6}\\ \mathrm{x}=-\frac{2\sqrt{39}}{5}\to \mathrm{y}=-\frac{13}{5\left(\frac{2\sqrt{39}}{5}\right)}=-\frac{\sqrt{39}}{6}\end{array}$$

Answered by MJS_new last updated on 18/May/21

$$xy+\frac{x}{y}=5$$$$xy-\frac{x}{y}=\frac{1}{5}$$$$y=px$$$${px}^2+\frac{1}{p}=5\Rightarrow x^2=\frac{5p-1}{p^2}$$$${px}^2-\frac{1}{p}=\frac{1}{5}\Rightarrow x^2=\frac{p+5}{5p^2}$$$$\frac{5p-1}{p^2}=\frac{p+5}{5p^2}\Rightarrow p=\frac{5}{12}$$$$x^2=\frac{156}{25}$$$$x=\pm \frac{2\sqrt{39}}{5}\wedge y=\pm \frac{\sqrt{39}}{6}$$

Answered by Rasheed.Sindhi last updated on 18/May/21

$$\precsim \mathcal{A}n\mathcal{A}lternate\mathcal{W}ay\succsim$$$$x(y+\frac{1}{y})=5.........\left(i\right)$$$$x(y-\frac{1}{y})=1/5.......\left(ii\right)$$$$\left(i\right)/\left(ii\right):\frac{y^2+1}{y^2-1}=25$$$$25y^2-25-y^2-1=0$$$$24y^2=26\Rightarrow y=\pm \sqrt{\frac{13}{12}}=\frac{\pm \sqrt{39}}{6}$$$$x(\frac{\pm \sqrt{39}}{6}+\frac{6}{\pm \sqrt{39}})=5$$$$x\left(\frac{39+36}{\pm 6\sqrt{39}}\right)=5$$$$x=5\times \frac{\pm 6\sqrt{39}}{75}=\frac{\pm 2\sqrt{39}}{5}$$$$(x,y)=(\frac{\pm 2\sqrt{39}}{5},\frac{\pm \sqrt{39}}{6})$$

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