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I've been struggling with Mathematica to solve this system of equations Given

  1. $a=14.0028$

  2. $b=1.35525$

  3. $c=0.0051$

  4. $d=0.0000472222$

  5. $W(R_1)=C_2 (c + (d + 108 C_1) R_1 + C_1 R_1^2)$

Solve:

\begin{equation} \dfrac{dC_1R_1}{C_{1}^2R_{1}^2+W(R_1)}=\dfrac{\pi}{4}\tag{1} \end{equation}

\begin{equation} \dfrac{a C_1 R_1 \sqrt{C_1 R_1 + C_2 (108 + R_1)} (C_1 R_1 (b + C_1 R_1) + W(R_1))}{\sqrt{C_1 C_2 R_1} (C_1^2 R_1^2 + W(R_1))} = 60\tag{2} \end{equation}

and

\begin{equation} \dfrac{2.22861\sqrt{C_1 R_1 + C_2 (108 + R_1)}}{\sqrt{C_1 C_2 R_1}} = 220000\tag{3} \end{equation}

Mathematica crunches away and cannot seem to return a solution (or one that I'm willing to wait for --long hours!). I can't tell if Mathematic is hung up or not able to solve the problem.

So my question is, how would I go about determining whether a system of equations like this has a solution in the first place? Under what conditions would such a system be not solvable?

thank you

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  • $\begingroup$ why do you care about this system of equations? it's so messy nobody wants to read this bro, gotta give a reason to care about it - is it important? $\endgroup$
    – terrace
    Commented Jun 3, 2017 at 21:06
  • $\begingroup$ whats the context? $\endgroup$
    – terrace
    Commented Jun 3, 2017 at 21:07
  • $\begingroup$ A solution for what? 1-4 aren't equations. Thet are simply statements assigning values to costants. You have one equation involving the variables. We can play and manipulate them until the cows come home, but what do we actually want to find out in the end. $\endgroup$
    – fleablood
    Commented Jun 4, 2017 at 0:20
  • $\begingroup$ Your first equation misses a closing parenthese. $\endgroup$ Commented Jun 4, 2017 at 4:18
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    $\begingroup$ @dxix Thanks, fixed the missing parenthesis in $(2)$. $\endgroup$ Commented Jun 6, 2017 at 2:06

1 Answer 1

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Hint (assuming the missing parenthesis in the numerator of $(2)$ is at the very end). Define:

  1. $m=\pi / 4 \cdot 1/d$

  2. $n=108$

  3. $p=60 / a$

  4. $q=220000 / 2.22861$

Let $T_1=R_1 C_1$, $T_2=R_1 C_2\,$, then the equations can be written as:

\begin{equation} \dfrac{T_1}{T_1^2+W}=m \tag{1} \end{equation}

\begin{equation} \dfrac{T_1 \sqrt{T_1 + n C_2 + T_2} \,(b T_1 + T_1^2 + W)}{\sqrt{C_2 T_1} (T_1^2 + W)} = p \tag{2} \end{equation}

\begin{equation} \dfrac{\sqrt{T_1 + n C_2 + T_2}}{\sqrt{C_2 T_1}} = q \tag{3} \end{equation}

Substituting $T_1^2+W=\cfrac{1}{m}T_1$ from $(1)$ into $(2)$ gives:

\begin{equation}\require{cancel} \dfrac{\bcancel{T_1} \sqrt{T_1 + n C_2 + T_2} \,(b T_1 + \cfrac{1}{m} T_1)}{\sqrt{C_2 T_1} \cfrac{1}{m} \bcancel{T_1}} = p \;\;\iff\;\; \dfrac{(mb + 1) T_1\,\sqrt{T_1 + n C_2 + T_2}}{\sqrt{C_2 T_1}} = p \tag{2'} \end{equation}

Dividing $(2') \div (3)$ gives:

\begin{equation} \dfrac{(mb + 1) T_1\,\cancel{\sqrt{T_1 + n C_2 + T_2}}}{\bcancel{\sqrt{C_2 T_1}}} \cdot \dfrac{\bcancel{\sqrt{C_2 T_1}}}{\cancel{\sqrt{T_1 + n C_2 + T_2}}} = \frac{p}{q} \;\;\iff\;\; T_1 = \frac{p}{q \,(mb+1)} \end{equation}

Substituting the just determined $T_1$ in $(1)$ and $(3)$ leaves two equations to be solved for $T_2,C_2\,$, which system turns out to be linear in $T_2,C_2\,$.

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  • $\begingroup$ JohnWaylandBales thank you very much for your assistance, and to @dxiv for working out the solution. You've demonstrated that the system has a solution. Is there a simple way to know that a system of equations like this has a solution in the first place, or is this is the kind of work that must be done to determine if a solution exists or not? These equations came from a real world problem -- see my response to terrace -- so I could assume it had a solution - unless I made a mistake deriving the equations. And since Mathematica was unable to solve them, I had a doubt... thank you. $\endgroup$
    – jrive
    Commented Jun 6, 2017 at 2:45
  • $\begingroup$ @JohnWaylandBales --thank you very much for the assist! $\endgroup$
    – jrive
    Commented Jun 6, 2017 at 2:46
  • $\begingroup$ @jrive First off, the above is only a partial answer. It doesn't carry the calculations all the way to the end, so it doesn't prove, for example, that the solutions for $T_2,C_2$ make physical sense e.g. that they are positive. Is there a simple way ... No simple way in general that I am aware of. It helps, however, to consolidate the unknowns into simpler variables, as to make any "patterns" more easily recognizable. That's something John Wayland Bales did well in his edits, and btw he deserves a lot of credit since I don't think I would have understood/tackled it without those. $\endgroup$
    – dxiv
    Commented Jun 6, 2017 at 2:53

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