Solving a Nonlinear System $$\begin{align}
(1/300)a + (-1/200)b &= 5\\
(-1/300)a + ((-1/300) + (1/200))b + (-1/200)c &= -e^b\\
(-1/200)b + (1/200)c &= -e^c
\end{align}
$$
how do I solve for $a, b$ and $c$?
Thanks!
I know if I derive an equation that isolates a variable, like
$kx + e^x = 0$
I can use Newton's  method to approximate x.
But still can't figure it out.
(I am just a high school student - so if you can make your answers as easy to understand as possible)
 A: The first equation is used to substitute all $a \rightarrow \frac{3}{2}b+1500$. The two remaining equations can be collected into a 2x1 vector $f = 0$
$$f = \begin{pmatrix} \hat{e}^b-\frac{b}{300}-\frac{c}{200}-5 \\ \hat{e}^c-\frac{b}{200}+\frac{c}{200} \end{pmatrix} $$
The derivatives with respect to $b$ and $c$ for each part are
$$ f\,' =  \begin{pmatrix} \hat{e}^b-\frac{1}{300} & -\frac{1}{200} \\ -\frac{1}{200} & \hat{e}^c+\frac{1}{200} \end{pmatrix}  $$
Newton raphson with vectors is $ (b,c) \rightarrow (b,c) - {f\,'}^{-1} f $
$$ \begin{pmatrix}b\\c\end{pmatrix} \rightarrow \begin{pmatrix}b\\c\end{pmatrix} - 
\begin{pmatrix} \hat{e}^b-\frac{1}{300} & -\frac{1}{200} \\ -\frac{1}{200} & \hat{e}^c+\frac{1}{200} \end{pmatrix}^{-1} \begin{pmatrix} \hat{e}^b-\frac{b}{300}-\frac{c}{200}-5 \\ \hat{e}^c-\frac{b}{200}+\frac{c}{200} \end{pmatrix}  $$
with an initial guess of $\begin{pmatrix}b = 1\\c = 1 \end{pmatrix} $ I get the following iterations
$$ \begin{bmatrix} 1&1\\1.841661&0.003381327\\1.634514&-0.9835724\\1.608907&-1.936174\\1.607725&-2.783955\\1.607126&-3.380392 \end{bmatrix} $$
So in the end we have $ a= \frac{3}{2} 1.607126 + 1500 $, $ b = 1.607126 $, $ c = -3.380392 $.
