The system of exponential equations How do you solve:
$$\begin{cases}
x\cdot2^{x-y}+3y\cdot2^{2x+y-1}=1 \hspace{0.1cm},\\ x\cdot2^{2x+y+1}+3y\cdot8^{x+y}=1\\
\end{cases}$$
I subtracted the equations, factorized by grouping, and got two terms equal zero. When I tried to use one of the terms, by expressing one variable through another and put it back in the first equation, it became complicated, so I gave up. 
Result is: $ (x=1, \hspace{0.1cm}y=-1) $

We obtain $$x(2^{x-y}-2^{2x+y+1})+3y(2^{2x+y-1}-2^{3x+3y})=0$$ or
$$x2^{x-y}(1-2^{x+2y+1})+3y(1-2^{x+2y+1})=0,$$ which gives
$$x2^{x-y}+3y=0$$ or
$$x+2y+1=0.$$
Can you end it now?
