Is y[n] = nx[n] − x[n−1] time invariant

A system is time invariant if delaying the input by $k$ only delays the output by $k$: $x[n-k] \rightarrow y[n-k]$.

Given: $y[n] = n\,x[n] - x[n-1]$

Step 1 – Output for the delayed input $x_1[n] = x[n-k]$ (replace $x$ only, not the $n$ multiplying it):

$$y_1[n] = n\,x[n-k] - x[n-1-k]$$

Step 2 – Delayed output (replace every $n$ by $n-k$):

$$y[n-k] = (n-k)\,x[n-k] - x[n-k-1]$$

Step 3 – Compare:

$$y_1[n] - y[n-k] = n\,x[n-k] - (n-k)\,x[n-k] = k\,x[n-k] \ne 0$$

Since $y_1[n] \ne y[n-k]$ (for $k \ne 0$), the system is time variant.

Reason: the coefficient $n$ multiplying $x[n]$ changes with time, so the system's gain is different at different instants. The second term $-x[n-1]$ alone is time invariant, but the first term makes the whole system time variant.

Numerical check: take $x[n] = \delta[n]$: $y[n] = 0\cdot\delta[n] - \delta[n-1] = -\delta[n-1]$. For $x[n] = \delta[n-1]$: $y_1[n] = n\,\delta[n-1] - \delta[n-2] = \delta[n-1] - \delta[n-2]$, which is not $y[n-1] = -\delta[n-2]$.