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I wonder whether the system of equations and inequations below have a solution. If there are solutions, what are they? A numerical solution is also desired. $$\begin{cases} \frac{c_1}{1-x_1}+\frac{c_2}{1-x_2}+\frac{c_3}{1-x_3}=0\\ \frac{c_1}{1-x_4}+\frac{c_2}{1-x_5}+\frac{c_3}{1-x_6}=0\\ c_1\ln{\frac{x_1}{1-x_1}}+c_2\ln{\frac{x_2}{1-x_2}} +c_3\ln{\frac{x_3}{1-x_3}}=0\\ c_1\ln{\frac{x_4}{1-x_4}}+c_2\ln{\frac{x_5}{1-x_5}} +c_3\ln{\frac{x_6}{1-x_6}}=0\\ \frac{x_1(1-x_1)}{x_4(1-x_4)}=\frac{x_2(1-x_2)}{x_5(1-x_5)}=\frac{x_3(1-x_3)}{x_6(1-x_6)}>1 \end{cases}$$ where $ c_1,c_2,c_3 $ are constants, and $x_i\in(0,1), i=1,2,3,4,5,6$.

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A simple set of solutions are any $(c_{1},c_{2},c_{3})$ such that $c_{1}+c_{2}+c_{3}=0$, $x_{1}=x_{2}=x_{3}=\frac{1}{2}$, $x_{4}=x_{5}=x_{6}<\frac{1}{2}$.

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  • $\begingroup$ Thanks for your answer. Now I know when $c_1+c_2+c_3=0$, there are solutions. What if $c_1+c_2+c_3!=0$? $\endgroup$ Feb 8, 2014 at 2:22
  • $\begingroup$ @YangzheLau: That is the Mathematica command. The Latex is \neq to get, $c_1+c_2+c_3 \neq 0$. $\endgroup$ Feb 15, 2018 at 2:34

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