How can I solve $x^\sqrt{y} +y^\sqrt{x} =\dfrac{49}{48}$ and $\sqrt{x}+\sqrt{y} =\dfrac72$?

Question

I have tried Wolfram Alpha and Mathematica to get the solution of the below system, but no result , I have used variable change $z=\sqrt{x}+\sqrt{y}$ for simplification but no result , $$ \left\{ \begin{array}{ll} x^\sqrt{y} +y^\sqrt{x} &=\dfrac{49}{48} \\ \sqrt{x}+\sqrt{y} &=\dfrac72 \\ \end{array} \right. $$

How I can solve this?

Answer 1

It is the same as solving $a^{2b}+b^{2a}=\frac{49}{48}$ with the constraints $a,b\geq 0$ and $a+b=\frac{7}{2}$, hence it boils down to finding the solutions of $$ g(a)\stackrel{\text{def}}{=}a^{7-a}+\left(\frac{7}{2}-a\right)^{2a}=\frac{49}{48} $$ over the interval $\left[0,\frac{7}{2}\right]$. $g(a)\geq 2$ if $a\geq \frac{1}{3}$ and over $\left[0,\frac{1}{3}\right]$ the function $g(a)$ is increasing, hence there is a unique solution in a right neighbourhood of the origin. By applying few steps of Newton's method with starting point $\frac{1}{100}$ we get $a\approx 0.00824505$ hence $x\approx 0.00006798$.

Answer 2

Doing the same as Jack D'Aurizio (eliminating $y$ from the second equation) and using a Taylor expansion around $x=0$, the first equation is $$-\frac{1}{48}+\sqrt{x} \log \left(\frac{49}{4}\right)+x \left(\frac{1}{2} \log ^2\left(\frac{49}{4}\right)-\frac{4}{7}\right)+O\left(x^{3/2}\right)$$ the solution of which being $$x=\frac{7}{96 \left(-2+343 \log ^2\left(\frac{7}{2}\right)+4 \log \left(\frac{7}{2}\right) \sqrt{7350 \log ^2\left(\frac{7}{2}\right)-84} \right)}\approx 0.0000679848$$ Using this as a starting guess, Newton iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.000067984814185381173107 \\ 1 & 0.000067980800578179758426 \\ 2 & 0.000067980800636422696813 \\ 3 & 0.000067980800636422696826 \end{array} \right)$$ which is the solution for twenty significant figures.