Why is $y(n) = x(n) + x(n-1)$ a low pass filter?
What you have here is the equivalent of a moving-average filter. Specifically, it is a filter of order 1, whose impuse response is 

$$h(n)=\delta(n)+\delta(n-1)$$

Taking its $Z$-transform, we get 

$$H(z)=1+z^{-1}=\frac{z+1}{z}$$

There is a pole at $z=0$ and a zero at $z=-1$. Plotting the magnitude of the frequency response $H(\omega)\triangleq H(e^{-\imath \omega})=2\vert \cos(\omega/2)\vert$, you get the following curve

 

As you can see, this clearly is a low-pass filter. You can easily calculate the cut-off frequency from here on.