Properties of Spectral Transformations - Allocation (decomposition into even and odd part)
Let's say that your signal is composed of two parts: even and odd:

$$s(t)=s_e(t)+s_o(t)$$

We also know following properties of this type of functions: 


Even: $f(-x)=f(x)$
Odd:  $f(-x)=-f(x)$


Let's calculate the time inversion of your signal $s(-t)$ and apply above properties:

$$s(-t)=s_e(-t)+s_o(-t)=s_e(t)-s_o(t) $$

So now let's do the trick and add the: $s(t)$ and $s(-t)$:

$$\require{cancel} s(t)+s(-t) = \color{blue}{s_e(t)}+ \cancel{\color{red}{s_o(t)}} + \color{blue}{s_e(t)}-\cancel{\color{red}{s_o(t)}}=\color{blue}{2s_e(t)} $$

Solve it for $\color{blue}{s_e(t)}$, and you will get:

$$\boxed{\color{blue}{s_e(t)}=\dfrac{s(t)+s(-t)}{2}} $$

In the end let's subtract $s(t)$ and $s(-t)$:

$$s(t)-s(-t) = \cancel{\color{blue}{s_e(t)}} + \color{red}{s_o(t)} - \cancel{\color{blue}{s_e(t)}} + \color{red}{s_o(t)}=\color{red}{2s_o(t)} $$

Rearrange and you will get:

$$\boxed{\color{red}{s_o(t)}=\dfrac{s(t)-s(-t)}{2}} $$