Solve the system of equations for $\sqrt{xy}$ $$x + y\sqrt{x} = \frac{95}{8}$$
$$y + x\sqrt{y} = \frac{93}{8}$$
$$x, y \in \mathbb{R}$$
I can't solve this system of equations I got asked in a group. I added and substracted them to find the following equations but I don't really know what to do with them.
$$\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}-\sqrt{xy}\right) = \frac{1}{4}$$
$$\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{xy}\right) - 2\sqrt{xy} = \frac{47}{2}$$
This is all I have. I checked the answer using Wolframalpha and it was $\frac{15}{4}$ for $\sqrt{xy}$.
I have been looking for this question's solution but didn't find any related question/solution on Stack Exchange or Web. Sorry if it is duplicate, thank you in advance.
 A: Some curiosities:
(a) the polynomial in $u=\sqrt x$ is
$$ ( 2u - 5 ) ( 32u^6 + 80u^5 - 180u^4 - 418u^3 - 301u^2 + 2812u - 1805 ) $$
(b) the polynomial in $v=\sqrt y$ is
$$ ( 2v - 3 ) ( 32v^6 + 48v^5 - 300v^4 - 418v^3 + 133v^2 + 3968v - 2883 ) $$
(c) the polynomial in $w=\sqrt{xy}$ is
$$ ( 4w - 15 ) ( 1024 w^6 + 768 w^5 + 5952 w^4 - 120064 w^3 - 591600 w^2 + 1387684 w + 5203815 ) $$
I'm quite convinced there is no elegant way to find them. But I hope someone here will prove I'm wrong.
[EDIT]
John, it's not difficult (for a computer) to solve the system
$$u^2+u\,v^2-95/8$$
$$v^2+v\,u^2-93/8$$
$$w-u\,v$$
using elimination technics based on resultants or Gröbner basis. But if you want to know exactly what I did:
(a) is the Sylvester's Resultant of
$$(u) \, v^2 + (0)  \,v + (u^2-95/8)$$
$$(1) \, v^2 + (u^2)\,v + (-93/8)$$
(b) is the Sylvester's Resultant of
$$(1) \, u^2 + (v^2)\,u + (-95/8)$$
$$(v) \, u^2 + (0)  \,u + (v^2-93/8)$$
(c) is the Sylvester's Resultant of
$$(w) \, v^3 + (-95/8) \,v^2 + (0) \,v + (w^2)$$
$$(1) \, v^3 + (0)\,v^2 + (-93/8)\,v + (w^2)$$
I don't consider those as valid answers because I think everyone (including me) is expecting a solution that only relies on elementary algebra. I mean we are dealing with a system that is supposedly simple.
A: Building and expanding on Zaind and Lars Smith's answers, we start by substituting
$$ \sqrt{x} = u \tag 1 $$
$$ \sqrt{y} = v  \tag 2 $$
to give
$$ u^2+u v^2=\frac{95}{8} \tag 3 $$
$$ u^2 v+v^2=\frac{93}{8} \tag 4 $$
Note that if we have $ x, y \in \mathbb{R} $, we also have $ x \ge 0 $ and $ y \ge 0 $, so $ u \ge 0 $ and $ v \ge 0 $ as well.
Solving $ (3) $ for $ v^2 $:
$$ v^2 = \frac{95-8 u^2}{8 u} $$
This can be substituted into $(4)$ to give
$$ u^2 v+\frac{95-8 u^2}{8 u}=\frac{93}{8} \tag 5 $$
which allows solution for $ v $:
$$ v = \frac{8 u^2+93 u-95}{8 u^3} \tag 6 $$
We close the circle by substituting $ (6) $ into $ (3) $ to get:
$$ u^2+\frac{\left(8 u^2+93 u-95\right)^2}{64 u^5}=\frac{95}{8} \tag 7 $$
Multiply by $ 64 u^5 $ and expand to get:
$$ 64 u^7-760 u^5+64 u^4+1488 u^3+7129 u^2-17670 u+9025 = 0  $$
Recall we are only interested in positive roots of $ u $. Now let's see what we can find out about those roots. Wikipedia has a useful page on this subject, and gives the following upper bound on the absolute value of roots of a polynomial $ a_n u^n + a_{n-1} u^{n-1} + \ldots + a_0 $:
$$ 2 * \max \left\{ \left| \frac{a_{n-1}}{a_n} \right|, \left| \frac{a_{n-2}}{a_n} \right|^{1/2}, \ldots, \left| \frac{a_{1}}{a_n} \right|^{1/(n-1)}, \left| \frac{a_{n-1}}{a_n} \right|^{1/n} \right\} $$
In fact, if we're only interested in positive real roots, we can replace all the $ a_j $s which are positive with 0, which means we have to select the maximum value out of:
$$ \left\{\left(\frac{8835}{32}\right)^{1/6},\sqrt{\frac{95}{8}}\right\} $$
A minute with a calculator indicates that that the largest of these is $ \sqrt{\frac{95}{8}} \approx 3.446 $
Let's take a look at a plot of our polynomial from $ 0 $ to $ 3.5 $  to get a sense of where the roots might be:
It looks like one of those roots is very close to $ 5/2 $. Plugging $ 5/2 $ into the left hand side of $ (7) $ (much easier than doing all the terms in the polynomial) gives exactly $ 95/8 $, verifying that it is a solution. Plugging into $ (6) $ gives $ v = 3/2 $, and together we have $ \sqrt{ x y } = u v = 15/4 $.
Now, however, we still have the possibility of a root near $ u = 0.8 $, but substituting $ 0.8 $ into $ (6) $ gives $ v \approx  -3.78 $, which contradicts the assumption that $ v > 0 $, and we can discard it.
A: It is not very probable that being $x$ and $y$ not squared the two equations give a rational and besides they differ by $\dfrac14$ so we can put in a first try
$$(x,y)=(\frac{t^2}{4},\frac{s^2}{4})$$ so we get the system
$$t(2t+s^2)=95=5\cdot19\\s(2s+t^2)=93=3\cdot31$$ The solution $(t,s)=(5,3)$ comes out immediately.
Thus $$\sqrt{xy}=\frac{15}{4}$$
A: $$x + y\sqrt{x} = \frac{95}{8} \tag 1$$
$$y + x\sqrt{y} = \frac{93}{8} \tag 2$$
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 $\frac{93}8$
A: 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 $s^2$.
A: This is an algebraic proof by a method which can be useful when there is a known solution and the task is to prove there are no further ones. Since finding the intersection of the two original curves is messy, use a suitable third curve through the same point.
Define a third, continuous curve to be $\sqrt{x}+\sqrt{y} = \frac{143}{36}$ for $x\le \frac{20}{9}$ and to be $18\sqrt{x}+20\sqrt{y} = 75$ for $x\ge \frac{20}{9}.$
Finding the intersection of this new curve with either of the original curves gives a cubic (in $\sqrt{x}$ and $\sqrt{y}$, respectively).
In each case, the only real solution gives the intersection point $(\frac{25}{4},\frac{9}{4})$. Thus the new curve separates the two original curves except at the given point and this is therefore the unique solution to the original problem.
