How to properly time-shift a function in relation to another function I'm stuck on a time-variance question in my  homework and I think it has something to do with my understanding of how time shifting functions works.  The problem asks to determine whether the system defined below is time-variant or time-invariant. 
$$ S: x(t) \to y(t) = x(t-2) + x(2-t) \tag{1}$$ 
I solved it such that 
$$ x_{delayed}(t) = x(t - \tau)\tag{2}$$
and
$$ y_{delayed}(t) = x(t-\tau-2) +x(2-t +\tau)  \tag{3}$$

Then I delayed the original output $ y(t) $ from $(1)$ to
$$y(t-\tau) = x(t-\tau-2) + x(2-t+\tau) \tag{4}$$ 
which I believe is incorrect.  I think that instead, it should be
$$y(t-\tau) = x(t-\tau-2) + x(2-t-\tau) \tag{5}$$ 

I believe my problem comes from my understanding of how to delay a function related to another function.  Searching around didn't help much.  Ultimately my question is, should
$$ y(t) = x(-t) \tag{6}$$
$$ y(t - \tau) = x(-t + \tau\tag{7})$$
or 
$$ y(t-\tau) = x(-t-\tau)\tag{8}$$
Using $(7)$ gives my my current answer, $(4)$, while using $(8)$ gives me what I believe is the correct answer, $(5)$.  I have found proofs for both answers which only confuses me more, so which is correct and how should I time shift the function? Any help is welcome and appreciated, thanks.
See that if you are going to go from $y(t)$ to $y(t-\tau)$ then you should replace every $t$ in the first one with $t-\tau$. This means that, in your case,

$$y(t-\tau) = x(t-\tau-2) + x(2-(t+\tau)) = x(t-\tau-2) + x(2-t+\tau)$$

In other words,

$$y(t) = x(-t)\implies y(t - \tau) = x(-(t - \tau))= x(-t + \tau)$$