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) $
By dividing the second equation by $2^{x+2y+1}$ we get $$\begin{cases} x\cdot2^{x-y}+3y\cdot2^{2x+y-1}=1 \\ x\cdot2^{x-y}+3y\cdot2^{2x+y-1}=2^{-(x+2y+1)} \end{cases}$$ which implies that $x=-2y-1$. Hence from the first equation we get $$-(2y+1)\cdot2^{-3y-1}+3y\cdot2^{-3y-3}=1$$ that is $$f(y):=5y+4+8\cdot 2^{3y}=0$$ Now note that $f$ is a strictly increasing function (it is the sum of the strictly increasing functions $g_1(y)=5y+4$ and $g_2(y)=8\cdot 2^{3y}$). Therefore $f$ is injective and $f(-1)=0$ implies that $y=-1$ is the only zero of $f$. Finally $x=-2y-1=1$.