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I have the following system of equations with variables $a,m$, and I'm wondering—can this system be solved symbolically/analytically?

\begin{align} m &= 100 + \frac{ \left( 200 \frac{\ln{\frac{1}{2}}}{26.8} \right) }{\left(\dfrac{\ln{\frac{1}{2}}}{26.8} + a \right)} \\ \\ 50 &= me^{-a\left( 19.9 \right)}- \frac{ \left( 200 \frac{\ln{\frac{1}{2}}}{26.8} \right) \exp{ \left(\dfrac{\ln{\frac{1}{2}}}{26.8} \cdot 19.9 \right) }}{\left(\dfrac{\ln{\frac{1}{2}}}{26.8} + a \right)} \end{align}

(What are some ways to recognize that a problem does or does not have a symbolic solution?)

For this specific example, I understand the answers can be numerically approximated to

\begin{align} a &\approx 0.092409 \\ m &\approx 22.2674 \end{align}

Via such numerical methods as a graph:

Graph

or other computational device such as WolframAlpha (link to this problem)1. For this specific problem, WolframAlpha only provided a numerical approximation.


1. The variables in the WolframAlpha link are $ a, n $ respectively.

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By replacing the value of m in the second equation, and renaming a with x, we are left with

$\dfrac{Ax+B}{Cx+D}~e^{ux}=\dfrac{A'x+B'}{C'x+D'}$ , which, unfortunately, cannot even be expressed in terms of the

special Lambert W function.

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From the first equation, you can eliminate $m$ as a function of $a$ $$m=\frac{100 (a+3 A)}{a+A}$$ and replacing $m$ in the second equation leads to $$f(a)=-\frac{200 e^{199 A/10} A}{a+A}+\frac{100 e^{-199 a/10} (a+3 A)}{a+A}-50=0$$ where $A=-\frac{5 \log (2)}{134}$ (I used rational numbers everywhere).

This equation cannot be solved for $a$ in a symbolic manner but can be approximated using, say, Newton. Starting with $a_0=\frac{1}{10}$, the first iterate is just awful even if perfectly analytical (I shall not give the expression here) and its value is $a_1\approx 0.0919287$; contiuning iteration gives $a_2\approx 0.0924071$, $a_3\approx 0.0924090$ which is the solution for six significant figures.

You will have very good explicit approximations after one single iteration starting with $a_0=\frac{28}{303}$ or better with $a_0=\frac{885}{9577}$. From far away, the best would be (according to RIES) $$a_0=\frac{1}{5+e (\pi -1)} \approx 0.09240903749$$ or, even better $$a_0=\frac{1}{\pi }\sin ^{-1}\left(\frac{1+\sqrt{5}}{2 e \log (8)}\right) \approx 0.09240899399$$ while, for ten significant figures, the solution is $0.09240899334$.

Have fun !

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