<s>[INST] Quick way to solve the system $\displaystyle \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} = \frac{65}{36}$, $xy-x+y=118$. Consider the system
$$\begin{aligned} \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} & = \frac{65}{36}, \\ xy -x +y & = 118. \end{aligned}$$
I have solved it by performing the substitutions $x-y=u$ and $xy=v$. Then I multiplied the first equation by $6^u$ and used $a^2-b^2=(a+b)(a-b)$ to find
$$(3^u+2^u)(3^u-2^u) = 65 \cdot 6^{u-2}.$$
By inspection I found $u=2$ and $v=120$. I solved the original system in $x,y$ and got the answers. Is there another quicker way to solve this without resorting to this sort of ninja inspection? I have found a second solution by solving $a^u +1/a^u = 65/36$, which assures $u=2$ but takes much more time.
Could there be a third way faster than these?
 [/INST] If $\left(\dfrac32\right)^u=a$
If $a+\dfrac1a=\dfrac{65}{36}<2$
and for $a>0, a+\dfrac1a\ge2\sqrt{a\cdot\dfrac1a}=2,$
I assume $a-\dfrac1a=\dfrac{65}{36}$
$\iff36a^2-65a-36=0$
$a=\dfrac{65\pm\sqrt{65^2+4\cdot36^2}}{2\cdot36}=\dfrac94,-\dfrac49$
For real $u,$
$\left(\dfrac32\right)^u=\dfrac94=\left(\dfrac32\right)^2\implies u=2$
Consequently, $x-y=u=2,xy=118+x-y=120$
Replacing $x$ with $y+2,$  we have $y(y+2)=120\iff y^2+2y-120=0$
$y^2+2y-120=(y+12)(y-10)$
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