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?

Consider the equation $$\left( \dfrac{3}{2} \right)^{x-y} - \left( \dfrac{2}{3} \right)^{x-y} = \dfrac{65}{36}.$$ Let $u=2^{x-y}$ and $v=3^{x-y}$ then we have $36u^2+65uv-36v^2=0.$ Hence $$u=\dfrac49v,\,\,\,\text{or}\,\,\,\,u=-\dfrac94v$$ For real solutions, take the first one and then $\dfrac{u}{v}=\left(\dfrac23\right)^2\implies x-y=2.$
Now $$xy-x+y=118\implies (x+1)(y-1)=117\implies(y-1)(y+3)=9\times13=(-9)\times(-13).$$ Hence $x=12,\,\,\ y=10$ and $x=-10,\,\,\ y=-12$ are the only real solutions for given non-linear system.
