Question Number 64837 by aliesam last updated on 22/Jul/19
$$\begin{cases}{{x}^{\sqrt{{y}}} \:+\:{y}^{\sqrt{{x}}} \:=\:\frac{\mathrm{49}}{\mathrm{48}}}\\{\sqrt{{x}}\:+\:\sqrt{{y}}\:=\:\frac{\mathrm{7}}{\mathrm{2}}}\end{cases} \\ $$$$ \\ $$$${find}\:{x}\:{and}\:{y} \\ $$$$ \\ $$
Answered by MJS last updated on 22/Jul/19
$$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{solve}\:\mathrm{approximately}\:\mathrm{I}\:\mathrm{think} \\ $$$$\left(\mathrm{2}\right)\:\:\Rightarrow\:\sqrt{{y}}=\frac{\mathrm{7}}{\mathrm{2}}−\sqrt{{x}};\:{y}={x}−\mathrm{7}\sqrt{{x}}+\frac{\mathrm{49}}{\mathrm{4}} \\ $$$$\left(\mathrm{1}\right)\:\:{x}^{\frac{\mathrm{7}}{\mathrm{2}}−\sqrt{{x}}} +\left({x}−\mathrm{7}\sqrt{{x}}+\frac{\mathrm{49}}{\mathrm{4}}\right)^{\sqrt{{x}}} −\frac{\mathrm{49}}{\mathrm{48}}=\mathrm{0} \\ $$$$\Rightarrow\:{x}\approx\mathrm{6}.\mathrm{79808}×\mathrm{10}^{−\mathrm{5}} \:\vee\:{x}\approx\mathrm{12}.\mathrm{1924} \\ $$$$\Rightarrow\:{y}\approx\mathrm{12}.\mathrm{1924}\:\vee\:{y}\approx\mathrm{6}.\mathrm{79808}×\mathrm{10}^{−\mathrm{5}} \\ $$