3
$\begingroup$

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) $

$\endgroup$
5
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.