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?
 A: 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 $y(y+2)=120\iff y^2+2y-120=0$
$y^2+2y-120=(y+12)(y-10)$
A: 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.
A: 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.
A: 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: $\text{arcsinh}z=\log(z+\sqrt{1+z^2})$
