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How do you solve a system of equations with e^x. For example

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    $\begingroup$ Seems unlikely that it can be solved exactly, probably must use numerical methods $\endgroup$ – MPW May 13 '16 at 3:53
  • $\begingroup$ A system that is not linear is usually bad news, you probably will only find a numerical approximation. That is, you could plot the two and see where they intersect. $\endgroup$ – user275377 May 13 '16 at 3:55
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As said in comments, only numerical methods would be able to give the solutions.

If the equations are $$(x-1)^2+(y+1)^2=36\tag 1$$ $$y=1+e^{-\frac{x^2}{10}}\tag 2$$ use $y$ as given by $(2)$ and plug it in $(1)$. You then need to search for the zeros of $$f(x)=(x-1)^2+\left(2+e^{-\frac{x^2}{10}}\right)^2-36$$ Using inspection or graphing, you will notice that there are two solutions : one between $-5$ and $-4$, the other between $6$ and $7$.

So, let us use Newton method which, starting from a guess $x_0$, will update it accodring to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ For this case $$f'(x)=2 (x-1)-\frac{2}{5} e^{-\frac{x^2}{10}} \left(2+e^{-\frac{x^2}{10}}\right) x$$ Let us start with $x_0=-4.5$; then the successive iterates will be $$x_1=-4.61480$$ $$x_2=-4.61335$$ which is the solution for six significant figures. To this $x$ will correspond $y=1.11904$.

For the second root, starting with $x_0=6.5$,the successive iterates will be $$x_1=6.65483$$ $$x_2=6.65261$$ which is the solution for six significant figures. To this $x$ will correspond $y=1.01197$.

We could have started closer to the solution since, after graphing, we know that $|x|$ is quite large; this will make $y$ close to $1$; that this to say that the equation to solve is almost $(x-1)^2-32=0$ the roots of which being $1\pm 4 \sqrt{2}$ that is to say $\approx -4.65685$ and $\approx 6.65685$. Using these as starting points would make Newton method converging is almost one single iteration for six significant figures.

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