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I'm trying to get a value in a web application, using some informations. I succeeded on creating this system (I need $x$):

$ \begin{equation} \left\{ \begin{matrix} x & = & \dot a + b - c - (\dot d \cdot \dot e)\\ b & = & f \cdot \dot g\\ f & = & \frac{x - \dot d - h}{1 + \dot i}\\ c & = & \dot \alpha(b)\\ h & = & x \cdot \dot e \end{matrix} \right. \end{equation} $

But it seems I can't solve it. Please, note that I know the value of the dotted variables (like $\dot p$).

My fear is that I can't solve this system. If so, is there a way to approximate the value of $x$?

Any help would be very appreciated.

Thank you!

Edit: I added two equations. I don't know it this can help, anyway I added them. Note that I know the value of $\alpha()$, but I did not put that here because it does not contain any of the variables of the system. It's there as a placeholder for an if statement.

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With given values for the dotted variables, this is a linear system for $x$, $b$, $c$, $f$, $h$, so it's a routine matter to solve it, but it's underdetermined since you have 5 unknowns but only 3 equations. This means that you can't expect a unique solution. –  Hans Lundmark Mar 18 '11 at 11:22
    
I added two equations. I don't know if that can help, anyway... –  Donovan Mar 18 '11 at 11:30
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up vote 1 down vote accepted

If you don't know $\alpha(b)$ (I don't see it dotted) you still have six unknowns and five equations. If you do know $\alpha(b)$ you have a linear system. If you insert $x=he$ for $x$ in the third equation, then substitute all the rest into the first you have

$x=a+\frac{x-d-he}{1+i}-\alpha(b)-de$

$x\left(1-\frac{1}{1+i}\right)=x\frac{i}{1+i}=a+\frac{-d-he}{1+i}-\alpha(b)-de$

$x=\frac{i+1}{i}(a-\alpha(b)-de)-i(d+he)$

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Yes, I know the value of $\alpha(b)$. Thank you! –  Donovan Mar 18 '11 at 13:20
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