My steps were as follows:
$\ x_2[n] = x[n-k] $
$\ y[n-k] = x[n-k] + (n-k) $
$\ y_2[n] = x_2[n] + n = x[n-k]+(n-k)$
Does this mean that it is indeed time invariant?
Signal Processing Stack Exchange is a question and answer site for practitioners of the art and science of signal, image and video processing. It only takes a minute to sign up.Sign up to join this community
No; the system given by
$$ y[n] = x[n] + n $$ is time-varying, due to the added term $n$.
Your mistake is in the line : $$\ y_2[n] = x_2[n] + n = x[n-k]+(n-k) $$ which should be instead
$$\ y_2[n] = x_2[n] + n = x[n-k]+n$$ and therefore implies that
$$y_2[n] \neq y[n-k] $$.