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?

Let $u=x-y$ and $a=\frac23$
\begin{align} 
\left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} & =  a^{-u} - a^{u}\\
&=e^{-u\log a}-e^{u \log a}\\
&=-2\sinh (u\log a)
\end{align} 
therefore \begin{align}  
\log a^u &=\text{arcsinh} \frac{-36}{72}\\
&=\log\Big(-\frac{36}{72}+\sqrt{1+(-\frac{36}{72})^2}\Big)\\
&=\log(\frac49)
\end{align} 
hence $$(\frac23)^{x-y}=(\frac23)^2$$ results in $x-y=2$. The rest is as others did.
Note: $\text{arcsinh}z=\log(z+\sqrt{1+z^2})$
