Solve the system of equations for $\sqrt{xy}$

Solution 1:

$$x + y\sqrt{x} = \frac{95}{8} \tag 1$$ $$y + x\sqrt{y} = \frac{93}{8} \tag 2$$

From $(1)$, you have $y=\frac{95-8 x}{8 \sqrt{x}}$. Plug it in $(2)$ and you need to find the zero of function $$f(x)=\frac{\sqrt{\frac{95-8 x}{\sqrt{x}}} x}{2 \sqrt{2}}-\sqrt{x}+\frac{95}{8 \sqrt{x}}-\frac{93}{8}\tag 3$$ which, I agree, is quite ugly.

If you can see is that if $x \to 0^+$, $f(x)\to +\infty$.

Tring for a few values of $x$ you would have the following results $$\left( \begin{array}{cc} 1 & 2.54773 \\ 2 & 0.642631 \\ 3 & 0.289849 \\ 4 & 0.249754 \\ 5 & 0.216853 \\ 6 & 0.0656336 \\ 7 & -0.280515 \end{array} \right)$$ So, the solution is "close" to $x=6$.

To polish the root, using Newton method with $x_0=6$, we should have the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 6.00000 \\ 1 & 6.27644 \\ 2 & 6.25025 \\ 3 & 6.25000 \end{array} \right)$$ that is to say $x=\frac {25}4$.

Back to $y=\frac{95-8 x}{8 \sqrt{x}}$, this gives $y=\frac 94$.

I apologize for the typos : I used $938$ instead of $\frac{93}8$

Solution 2:

Let's do the substitution that Will suggests, $x=s^2$,$y=t^2$ and let's save writing those fractions $a=95/8,b=93/8$. Thus we have \begin{eqnarray*} s^2+st^2=a \\ t^2+ts^2=b. \end{eqnarray*} How do we solve these (for $(s,t)$ in terms of $a,b$) ? Rearrange the first equation to $s^2-a=s t^2$ and square this \begin{eqnarray*} s^4-2as^2+a^2=t^4s^2. \end{eqnarray*} Now multiply this by $t^2$ and use the second equation \begin{eqnarray*} (b-t^2)^2-2at(b-t^2)+a^2t^2=t^5(b-t^2). \end{eqnarray*} This shows how to solve these equations (in principle). So we just need to solve an equation of order $7$ ... Let's see what CAS says ...

So this gives the solution that Will mentions ... $t=5/2$ ... What about the order $6$ part ? Well have a look at the graph, https://www.desmos.com/calculator/xsj6mrglur there are two real roots but they give negative values for $t^2$ or $s^2$.