Conditional expectation on Gaussian random variables

Since $X$ and $N$ are independent and normal, any combinations of $X$ and $N$ are normal. As any combination of $X+N$ and $\frac{1}{\sigma_x^2}X - \frac{1}{\sigma_n^2}N$ is also a combination of $X$ and $N$, that is, any combinations of $X+N$ and $\frac{1}{\sigma_x^2}X - \frac{1}{\sigma_n^2}N$ are normal, then $X+N$ and $\frac{1}{\sigma_x^2}X - \frac{1}{\sigma_n^2}N$ are jointly normal. Moreover, note that \begin{align*} E\bigg((X+N)\Big( \frac{1}{\sigma_x^2}X - \frac{1}{\sigma_n^2}N\Big) \bigg) &= 0. \end{align*} Then $X+N$ and $\frac{1}{\sigma_x^2}X - \frac{1}{\sigma_n^2}N$ are independent. Consequently, \begin{align*} E\big( (X+N) \mid S\big) &= S,\\ E\bigg( \Big( \frac{1}{\sigma_x^2}X - \frac{1}{\sigma_n^2}N\Big) \mid S\bigg) &= 0. \end{align*} Solving this system, we obtain that \begin{align} E(X\mid S) = \frac{\sigma_x^2}{\sigma_x^2+\sigma_n^2}S. \end{align} In other words, \begin{align} E(X\mid S=s) = \frac{\sigma_x^2}{\sigma_x^2+\sigma_n^2}s. \end{align}