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Question Number 58210 by salaw2000 last updated on 19/Apr/19

$$\mathrm{find}\mathrm{two}\mathrm{possible}\mathrm{number}\mathrm{such}\mathrm{that}$$$$1)\mathrm{xy}=\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{x}-\mathrm{y}$$$$2)\mathrm{xy}=\frac{2\mathrm{x}}{\mathrm{y}}=3(\mathrm{x}-\mathrm{y})$$$$3)\mathrm{xy}=\frac{\mathrm{x}}{\mathrm{y}}=2(\mathrm{x}-\mathrm{y}).$$$$$$$$$$$$$$$$$$$$$$

Commented by MJS last updated on 20/Apr/19

$$\left(1\right)xy=\frac{x}{y}\Rightarrow x=0\vee y=\pm 1$$$$\frac{x}{y}=x+y\Rightarrow x=\frac{y^2}{y-1}\Rightarrow y\ne 1$$$$x=\frac{y^2}{y-1}\wedge y=-1\Rightarrow x=-\frac{1}{2}$$$$x=-\frac{1}{2}\wedge y=-1\Rightarrow xy=\frac{x}{y}=x-y=\frac{1}{2}$$

Commented by MJS last updated on 20/Apr/19

$$\left(2\right)xy=\frac{2x}{y}\Rightarrow y=\pm \sqrt{2}$$$$\frac{2x}{y}=3(x-y)\Rightarrow x=\frac{3y^2}{3y-2}\Rightarrow x=\frac{6}{7}\pm \frac{9\sqrt{2}}{7}$$$$x=\frac{6}{7}+\frac{9\sqrt{2}}{7}\wedge y=\sqrt{2}\Rightarrow xy=\frac{2x}{y}=3(x-y)=\frac{18}{7}+\frac{6\sqrt{2}}{7}$$$$x=\frac{6}{7}+\frac{9\sqrt{2}}{7}\wedge y=-\sqrt{2}\Rightarrow xy=\frac{2x}{y}=3(x-y)=\frac{18}{7}-\frac{6\sqrt{2}}{7}$$

Commented by MJS last updated on 20/Apr/19

$$\left(3\right)\mathrm{similar}$$$$x=-\frac{2}{3}\wedge y=-1\Rightarrow xy=\frac{x}{y}=2(x-y)=\frac{2}{3}$$$$x=2\wedge y=1\Rightarrow xy=\frac{x}{y}=2(x-y)=2$$

Commented by salaw2000 last updated on 20/Apr/19

$$\mathrm{thanks}$$

Commented by salaw2000 last updated on 20/Apr/19

$$\mathrm{thanks}$$

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