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 $x=-2y-1=1$.

  • $\begingroup$ If I'm not mistaken, there's a typo in the equation just before the definition of $f$. Shouldn't the exponent in the last term be $-3y-3$ instead of $-3y-5?$ $\endgroup$ – saulspatz Feb 6 '18 at 20:34
  • $\begingroup$ @saulspatz Yes, you are right! Thank you very much. $\endgroup$ – Robert Z Feb 6 '18 at 20:40
  • $\begingroup$ in which cases function would have more than one zeros? $\endgroup$ – Nemanja Djordjevic Feb 6 '18 at 23:02
  • $\begingroup$ @NemanjaDjordjevic I have no general statement to suggest. $f$ has to be NOT injective for sure. Take a look here: en.wikipedia.org/wiki/Zero_of_a_function $\endgroup$ – Robert Z Feb 7 '18 at 5:25

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?

  • $\begingroup$ that is exactly what I got, what's next? $\endgroup$ – Nemanja Djordjevic Feb 6 '18 at 19:40

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