Question Number 214191 by golsendro last updated on 01/Dec/24
$$\:\:\:\:\:\:\begin{cases}{\mathrm{y}^{\mathrm{3}} =\:\mathrm{x}^{\mathrm{x}+\mathrm{y}} }\\{\mathrm{y}^{\mathrm{x}+\mathrm{y}} \:=\:\mathrm{x}^{\mathrm{6}} \mathrm{y}^{\mathrm{3}} }\end{cases} \\ $$$$\:\:\:\:\cancel{\underline{ }} \\ $$
Answered by TonyCWX08 last updated on 01/Dec/24
$${I}\:{can}\:{see}\:{that}\:\left({x},{y}\right)=\left(\mathrm{1},\mathrm{1}\right) \\ $$$$\mathrm{2}\left(\mathrm{1}\right)−\mathrm{1}=\mathrm{1} \\ $$
Commented by efronzo1 last updated on 01/Dec/24
$$\mathrm{how}\:\mathrm{about}\:\left(\mathrm{x},\mathrm{y}\right)=\:\left(\mathrm{2},\mathrm{4}\right),\left(−\mathrm{3},\mathrm{9}\right) \\ $$
Commented by TonyCWX08 last updated on 01/Dec/24
$${Sorry} \\ $$$${Im}\:{an}\:{amateur}\:{compared}\:{to}\:{you}\:{guys}\:{here}. \\ $$
Answered by Rasheed.Sindhi last updated on 01/Dec/24
$$\:\:\:\:\:\:\begin{cases}{\mathrm{x}^{\mathrm{x}+\mathrm{y}} =\:\mathrm{y}^{\mathrm{3}} }\\{\mathrm{y}^{\mathrm{x}+\mathrm{y}} \:=\:\mathrm{x}^{\mathrm{6}} \mathrm{y}^{\mathrm{3}} }\end{cases}\:;\:\mathrm{2}{x}−{y}=? \\ $$$${Multiplying}\:: \\ $$$$\left({xy}\right)^{{x}+{y}} ={x}^{\mathrm{6}} {y}^{\mathrm{6}} =\left({xy}\right)^{\mathrm{6}} \\ $$$$\Rightarrow{x}+{y}=\mathrm{6}\:\Rightarrow{y}=\mathrm{6}−{x} \\ $$$$\mathrm{2}{x}−{y}\:{is}\:{not}\:{unique}: \\ $$$$\mathrm{2}{x}−{y}=\mathrm{2}{x}−\mathrm{6}+{x}=\mathrm{3}{x}−\mathrm{6}\:{for}\:{any}\:{x} \\ $$
Answered by efronzo1 last updated on 01/Dec/24
$$\:\:\mathrm{y}=\mathrm{x}^{\frac{\mathrm{x}+\mathrm{y}}{\mathrm{3}}} \:,\:\:\cancel{\nprec} \\ $$