Knowing whether system of equations has a solution or not

Question

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

Answer 1

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

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

7. $n=108$

8. $p=60 / a$

9. $q=220000 / 2.22861$

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 $T_2,C_2\,$.