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

find two possible number such that  1)  xy=(x/y)=x−y  2)xy=((2x)/y)=3(x−y)  3)  xy=(x/y)=2(x−y).

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

Commented by MJS last updated on 20/Apr/19

(1) xy=(x/y) ⇒ x=0 ∨ y=±1  (x/y)=x+y ⇒ x=(y^2 /(y−1)) ⇒ y≠1  x=(y^2 /(y−1)) ∧ y=−1 ⇒ x=−(1/2)  x=−(1/2)∧y=−1 ⇒ xy=(x/y)=x−y=(1/2)

$$\left(\mathrm{1}\right)\:{xy}=\frac{{x}}{{y}}\:\Rightarrow\:{x}=\mathrm{0}\:\vee\:{y}=\pm\mathrm{1} \\ $$$$\frac{{x}}{{y}}={x}+{y}\:\Rightarrow\:{x}=\frac{{y}^{\mathrm{2}} }{{y}−\mathrm{1}}\:\Rightarrow\:{y}\neq\mathrm{1} \\ $$$${x}=\frac{{y}^{\mathrm{2}} }{{y}−\mathrm{1}}\:\wedge\:{y}=−\mathrm{1}\:\Rightarrow\:{x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}}\wedge{y}=−\mathrm{1}\:\Rightarrow\:{xy}=\frac{{x}}{{y}}={x}−{y}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by MJS last updated on 20/Apr/19

(2) xy=((2x)/y) ⇒ y=±(√2)  ((2x)/y)=3(x−y) ⇒ x=((3y^2 )/(3y−2)) ⇒ x=(6/7)±((9(√2))/7)  x=(6/7)+((9(√2))/7)∧y=(√2) ⇒ xy=((2x)/y)=3(x−y)=((18)/7)+((6(√2))/7)  x=(6/7)+((9(√2))/7)∧y=−(√2) ⇒ xy=((2x)/y)=3(x−y)=((18)/7)−((6(√2))/7)

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

Commented by MJS last updated on 20/Apr/19

(3) similar  x=−(2/3)∧y=−1 ⇒ xy=(x/y)=2(x−y)=(2/3)  x=2∧y=1 ⇒ xy=(x/y)=2(x−y)=2

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

Commented by salaw2000 last updated on 20/Apr/19

thanks

$$\mathrm{thanks} \\ $$

Commented by salaw2000 last updated on 20/Apr/19

thanks

$$\mathrm{thanks} \\ $$

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