Question Number 58210 by salaw2000 last updated on 19/Apr/19
$$\mathrm{find}\:\mathrm{two}\:\mathrm{possible}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left.\mathrm{1}\right)\:\:\mathrm{xy}=\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{x}−\mathrm{y} \\ $$$$\left.\mathrm{2}\right)\mathrm{xy}=\frac{\mathrm{2x}}{\mathrm{y}}=\mathrm{3}\left(\mathrm{x}−\mathrm{y}\right) \\ $$$$\left.\mathrm{3}\right)\:\:\mathrm{xy}=\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{2}\left(\mathrm{x}−\mathrm{y}\right). \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by MJS last updated on 20/Apr/19
$$\left(\mathrm{1}\right)\:{xy}=\frac{{x}}{{y}}\:\Rightarrow\:{x}=\mathrm{0}\:\vee\:{y}=\pm\mathrm{1} \\ $$$$\frac{{x}}{{y}}={x}+{y}\:\Rightarrow\:{x}=\frac{{y}^{\mathrm{2}} }{{y}−\mathrm{1}}\:\Rightarrow\:{y}\neq\mathrm{1} \\ $$$${x}=\frac{{y}^{\mathrm{2}} }{{y}−\mathrm{1}}\:\wedge\:{y}=−\mathrm{1}\:\Rightarrow\:{x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}}\wedge{y}=−\mathrm{1}\:\Rightarrow\:{xy}=\frac{{x}}{{y}}={x}−{y}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by MJS last updated on 20/Apr/19
$$\left(\mathrm{2}\right)\:{xy}=\frac{\mathrm{2}{x}}{{y}}\:\Rightarrow\:{y}=\pm\sqrt{\mathrm{2}} \\ $$$$\frac{\mathrm{2}{x}}{{y}}=\mathrm{3}\left({x}−{y}\right)\:\Rightarrow\:{x}=\frac{\mathrm{3}{y}^{\mathrm{2}} }{\mathrm{3}{y}−\mathrm{2}}\:\Rightarrow\:{x}=\frac{\mathrm{6}}{\mathrm{7}}\pm\frac{\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{7}} \\ $$$${x}=\frac{\mathrm{6}}{\mathrm{7}}+\frac{\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{7}}\wedge{y}=\sqrt{\mathrm{2}}\:\Rightarrow\:{xy}=\frac{\mathrm{2}{x}}{{y}}=\mathrm{3}\left({x}−{y}\right)=\frac{\mathrm{18}}{\mathrm{7}}+\frac{\mathrm{6}\sqrt{\mathrm{2}}}{\mathrm{7}} \\ $$$${x}=\frac{\mathrm{6}}{\mathrm{7}}+\frac{\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{7}}\wedge{y}=−\sqrt{\mathrm{2}}\:\Rightarrow\:{xy}=\frac{\mathrm{2}{x}}{{y}}=\mathrm{3}\left({x}−{y}\right)=\frac{\mathrm{18}}{\mathrm{7}}−\frac{\mathrm{6}\sqrt{\mathrm{2}}}{\mathrm{7}} \\ $$
Commented by MJS last updated on 20/Apr/19
$$\left(\mathrm{3}\right)\:\mathrm{similar} \\ $$$${x}=−\frac{\mathrm{2}}{\mathrm{3}}\wedge{y}=−\mathrm{1}\:\Rightarrow\:{xy}=\frac{{x}}{{y}}=\mathrm{2}\left({x}−{y}\right)=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${x}=\mathrm{2}\wedge{y}=\mathrm{1}\:\Rightarrow\:{xy}=\frac{{x}}{{y}}=\mathrm{2}\left({x}−{y}\right)=\mathrm{2} \\ $$
Commented by salaw2000 last updated on 20/Apr/19
$$\mathrm{thanks} \\ $$
Commented by salaw2000 last updated on 20/Apr/19
$$\mathrm{thanks} \\ $$