The problem on the test I most screwed up was this one:

Is the system $ y(t)=x(1-t) $ Time Invaiant?

The best way to solve this is to just go slow and if nessecary change a variable to make sure you get the expression correct

x(t) -->|system|-->y(t)=x(1-t)-->|delay|-->y(t-t0)=x(1-(t-t0))=x(1-t+t0)

x(t) -->|delay|-->x(t-t0)-->|system|-->y(t-t0)=x(1-t-t0)


So since the outputs are not the same, the system is not Time Invariant.

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett