Question Number 153256 by Lekhraj last updated on 06/Sep/21
Commented by MJS_new last updated on 06/Sep/21
$${x}<{y} \\ $$$${x}=\frac{\mathrm{25}}{\mathrm{9}{t}}\wedge{y}=\frac{\mathrm{25}{t}}{\mathrm{9}} \\ $$$${t}=\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\:\Rightarrow\:\frac{{x}}{{y}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\left(\mathrm{for}\:{y}<{x}\:\mathrm{we}\:\mathrm{get}\:\frac{{x}}{{y}}=\frac{\mathrm{5}}{\mathrm{3}}\right) \\ $$