MHB How to Solve a System of Equations with Square Roots?

[Albert1]

$x,y\in R$
$find \,\ solution \,\,of \,\,x\,\, and\,\, y:$
$\left\{\begin{matrix}
x+y+\dfrac{9}{x}+\dfrac{4}{y}=10---(1)\\
(x^2+9)(y^2+4)=24xy---(2)
\end{matrix}\right.$

 

[Wilmer]

$x,y\in R$
$find \,\ solution \,\,of \,\,x\,\, and\,\, y:$
$\left\{\begin{matrix}
x+y+\dfrac{9}{x}+\dfrac{4}{y}=10---(1)\\
(x^2+9)(y^2+4)=24xy---(2)
\end{matrix}\right.$

Spoiler

One real solution: x=3, y=2

 

[lfdahl]

My attempt:

Spoiler

From (2), we get ($x,y \ne 0$):\[xy\left ( x+\frac{9}{x} \right )\left ( y+\frac{4}{y} \right )=24xy\]

or

\[\left ( x+\frac{9}{x} \right )\left ( y+\frac{4}{y} \right )=24\].Let $a = x+\frac{9}{x}$ and $b = y+\frac{4}{y}$. Then the two equations read:\[a + b = 10\: \: \wedge \: \: ab = 24\]So either $a = 4$ and $b = 6$ or vice versa.Case I.
$a = 4$: The corresponding quadratic expression: $x^2-4x+9 = 0$, has no real roots and it makes no sense
to search for a y-root.Case II.
$a = 6$: The corresponding quadratic expression:$ x^2-6x+9 = (x-3)^2$.
$b = 4$: $y^2-4y+4 = (y-2)^2$.

Thus the only solution is the pair: $(x,y) = (3,2)$.

 

[Opalg]

$x,y\in R$
$find \,\ solution \,\,of \,\,x\,\, and\,\, y:$
$\left\{\begin{matrix}
x+y+\dfrac{9}{x}+\dfrac{4}{y}=10---(1)\\
(x^2+9)(y^2+4)=24xy---(2)
\end{matrix}\right.$

[sp]The left side of (2) is positive, so $x$ and $y$ are either both positive or both negative. If they are both negative then (1) cannot hold. So they are both positive.

Let $u = \sqrt x$, $v = \sqrt y.$ Then $$\Bigl(u - \dfrac3u\Bigr)^2 + \Bigl(v - \dfrac2v\Bigr)^2 = u^2 - 6 + \frac9{u^2} + v^2 - 4 + \frac4{v^2} = x+y+\dfrac{9}{x}+\dfrac{4}{y}-10 = 0.$$ Therefore $u - \dfrac3u = v - \dfrac2v = 0$, so that $x = u^2 = 3$ and $y = v^2 = 2.$[/sp]