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I've been stuck on this problem for hours. I keep starting and stopping because I'm not exactly sure what I'm doing. The examples the teacher worked in class were much more straight forward.

If possible, could you please help me to get started on this problem? I primarily get confused with how to handle the middle nodes. For example I know the following:

$x_1+x_3=600$

$x_2+x_5=500$

$x_3+x_6=600$

$x_4+x_7=x_6$

$x_5+x_7=500$

Am I headed down the correct path?

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Your five equations are correct. The sixth equation is $x_1=x_2+x_4$. You'll then have six linear equations in seven real unknowns, which is not enough to find a unique solution.

Question (a) is very vague, since no unique solution exists. Perhaps what is being asked is a list of the six equations. Or perhaps you can choose a variable, say $x_1$, and express the other variables in terms of $x_1$.

Question (b) is very easy. Look at the drawing and imagine $x_6$ and $x_7$ as closed pipes. It should be obvious that $x_3=600$, $x_4=0$, and $x_5=500$. (I'll leave $x_1$ and $x_2$ to you.)

Question (c) is also easy. Look at the drawing and imagine $x_6$ as a closed pipe. Then it should be obvious that $x_3=600$ and $x_1=0$. I'll leave the rest to you. (Hint: $x_2$ and $x_7$ are negative.)

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  • $\begingroup$ Perhaps the answer to question (a) is that the system is underdetermined. If so, then there is no unique solution, that is, either there is no solution, or there are an infinite number of them. Since questions (b) and (c) have solutions, then the system has an infinite number of solutions. $\endgroup$ – Joel Reyes Noche Feb 10 '14 at 15:04
  • $\begingroup$ My book gives the following answer: $x_1=s$, $x_2=4$, $x_3=600-s$, $x_4=s-t$, $x_5=500-t$, $x_6=s$, $x_7=t$ I don't know how to get my equations into this form. Any guidance there? $\endgroup$ – hax0r_n_code Feb 11 '14 at 1:08
  • $\begingroup$ I think you mean $x_2=t$. What the book did was to let $x_1=s$ and $x_7=t$, then to write the other $x_i$'s in terms of $s$ and $t$. For example, since you know that $x_1+x_3=600$ and since we let $x_1=s$, it follows that $x_3=600-s$. $\endgroup$ – Joel Reyes Noche Feb 11 '14 at 1:36
  • $\begingroup$ Now you might ask why the book chose to fix $x_1$ and $x_7$ and not the other $x_i$'s. The choice is arbitrary, but I suggest you look at the earlier examples (or questions) in the book. Perhaps they give the assumptions needed to get the answer provided. $\endgroup$ – Joel Reyes Noche Feb 11 '14 at 1:42
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    $\begingroup$ Thanks a lot for your patience. I think I understand now. $\endgroup$ – hax0r_n_code Feb 11 '14 at 3:10

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