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I have this system of 2 equations in 4 unknowns: $$ \begin{cases} \displaystyle y_1=\pi_1 \frac{1}{1+e^{\alpha_1}}+\pi_2 \frac{1}{1+e^{\alpha_2}}+\pi_3 \frac{1}{1+e^{\alpha_3}}+\pi_4 \frac{1}{1+e^{\alpha_4}}\\ \displaystyle y_2=\pi_1 \frac{e^{\alpha_1}}{1+e^{\alpha_1}}+\pi_2 \frac{e^{\alpha_2}}{1+e^{\alpha_2}}+\pi_3 \frac{e^{\alpha_3}}{1+e^{\alpha_3}}+\pi_4 \frac{e^{\alpha_4}}{1+e^{\alpha_4}}\\ \end{cases} $$ The unknowns are $\alpha_1,\alpha_2,\alpha_3,\alpha_4$; all other terms are known.

I have never taken any course on solving systems of equations and my questions are very basics (apologies in advance): what can I say about the number of solutions? Are there infinite solutions and why? Is there any way to pin down a unique solution?

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    $\begingroup$ Add the two equations to get $y_1+y_2=\pi_1+\pi_2+\pi_3+\pi_4$. If that is false, then the system has no solution. If that is true, then one of the equations are redundant, which means you only need to solve one of the equations. $\endgroup$ – Colescu Sep 8 '16 at 14:37
  • $\begingroup$ Thank you very much. It is true. Does it mean that I have 1 equation in 4 unknowns? Is there any way to pin down $\alpha_1,\alpha_2,\alpha_3,\alpha_4$? Or, at least, can I say that there are infinite solutions? $\endgroup$ – STF Sep 8 '16 at 14:43
  • $\begingroup$ No you cannot. With four unknowns there must be infinite solutions. However if these unknowns are restricted, then things might be different. $\endgroup$ – Colescu Sep 9 '16 at 1:41
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Adding together the two equations and collecting terms gives $$y_1+y_2=\sum_{i=1}^4\frac{\pi_i}{1+e^{\alpha_i}}(1+e^{\alpha_i})=\sum_{i=1}^4{\pi_i}.$$

If false, there are no solutions. If true, you have one equation with four unknowns, which is underdetermined and this has an infinite number of solutions over the reals.

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