I have this system of equations that I want to solve: $$\begin{cases} a+b = 204 \\ a+b\cdot e^{-105c} = 192.6 \\ a+b\cdot e^{-180c} = 177.6\end{cases}$$
Is there a mathematical way to solve it knowing that the constant will keep changing? I am looking for a way that I can program it to do the mathematical work for me.
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If you subtract the second equation from the first you get $$ b(1 - e^{-105c}) = 204 - 192.6. $$ If you do the same for the third and first you get an equation with the same shape.
Divide one of those equations by the other and you have a single equation involving just $c$.
You will have to solve that one numerically.
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If you do what @Ethan Bolker suggested, using whole numbers, you should end with $$\frac{57 e^{105 c}}{5 \left(e^{105 c}-1\right)}=\frac{132 e^{180 c}}{5 \left(e^{180 c}-1\right)} $$ Reduce to same denominator and simplify to get $$3 e^{105 c} \left(44 e^{75 c}-25 e^{180 c}-19\right)=0$$ So, you look for the zero of function $$f(c)=44 e^{75 c}-25 e^{180 c}-19$$ that you will need to solve using, say, Newton method.
Looking at the equation, the solution has to be quite small. Expand $f(c)$ as a Taylor series to get $$f(c)=1200 c+281250 c^2+O\left(c^3\right)$$ Then, an estimate of the solution is $$c_0=-\frac{8}{1875}$$ Now, Newton iterates would be $$\left( \begin{array}{cc} n & c_n \\ 0 & -0.004266666667 \\ 1 & -0.008648470966 \\ 2 & -0.007013654476 \\ 3 & -0.006907055886 \\ 4 & -0.006906341073 \\ 5 & -0.006906341040 \end{array} \right)$$
Hint: With $$b=204-a$$ we get $$a+(204-a)e^{-105c}=\frac{1926}{10}$$ $$a+(204-a)e^{-180c}=\frac{1776}{10}$$ From the first equation above we get $$a=\frac{3 \left(321 e^{105 c}-340\right)}{5 \left(e^{105 c}-1\right)}$$ Substituting this in the second equation above we get $$\frac{3 \left(321 e^{105 c}-340\right)}{5 \left(e^{105 c}-1\right)}+e^{-180 c} \left(204-\frac{3 \left(321 e^{105 c}-340\right)}{5 \left(e^{105 c}-1\right)}\right)=\frac{888}{5}$$ This must be solved for $c$ $$c\approx -0.006906341040$$