Solving for $x$ in $x + \frac{s}{s^2+4} = \frac{2x}{s^2+4}$

Question

How would I go about solving for $x$ in this equation?

$$x + \frac{s}{s^2+4} = \frac{2x}{s^2+4}$$

Answer 1

$$x+\frac{s}{s^2+4}=\frac{2x}{s^2+4}\Longleftrightarrow x-\frac{2x}{s^2+4}=-\frac{s}{s^2+4}\Longleftrightarrow$$ $$x\left[1-\frac{2}{s^2+4}\right]=-\frac{s}{s^2+4}\Longleftrightarrow x=\frac{-\frac{s}{s^2+4}}{1-\frac{2}{s^2+4}}\Longleftrightarrow \color{red}{x=-\frac{s}{2+s^2}}$$

But notice then that: $s^2+2\ne0$ and $s^2+4\ne0$

Answer 2

$$x+\frac{s}{s^2+4}=\frac{2x}{s^2+4}$$ $$\frac{s^2x+4x+s}{s^2+4}=\frac{2x}{s^2+4}$$ For real $s$, $s^2+4\ne0$. Hence, $$s^2x+4x+s=2x$$ $$x(s^2+2)+s=0$$ $$x=-\frac{s}{s^2+2}$$