Determine if system is linear time variant The system equation is given as:
$$y(n)=(n-1)x(n-1)+(n+1)x(n+1)$$
I solved that the system is time variant:
\begin{align}
y(n-k)&amp;=(n-k-1)x(n-k-1)+(n-k+1)x(n-k+1)\\
H[x(n-k)]&amp;=(n-1)x(n-k-1)+(n+1)x(n-k+1)
\end{align}
And for the linear/non-linear part, here is what I have so far:
\begin{align}
H\big[a_1x_1(n)+a_2x_2(n)\big]&amp;=(n-1)\big[a_1x_1(n-1)+a_2x_2(n-1)\big]\\&amp;+(n+1)\big[a_1x_1(n+1)+a_2x_2(n+2)\big]
\end{align}
It is ok?
To be time invariant you would have to show that:
$$y(n-k) = (n-1)x(n-k-1) + (n+1)x(n-k+1)$$
(You shift the $x$ and you get a shift in $y$).
But you can't get rid of the $(n-k-1)$ and $(n-k+1)$.
In general any time you have the time term $n$ by itself it will be time variant unless it cancels out somehow.