<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] Consider $$ \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y}  = \frac{65}{36}$$ and set $a=\left( \frac{3}{2} \right)^{x-y}$. So, you have $a-\frac 1a=\frac{65}{36}$, that is to say $a=\frac{9}{4}$. So $$\left( \frac{3}{2} \right)^{x-y}=\frac{9}{4}=\left( \frac{3}{2} \right)^2$$ so $x-y=2$.
For the remaining, the same as in lab bhattacharjee'answer.
No ninja inspection required.
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