0 = 2 (an always false statement), that would imply "no solution".
For your second issue, let's solve the first for x...
x = 10 - 2y
sub into the second.
3(10 - 2y) + (6 + t)y = 30
30 - 6y + 6y + ty = 30.
Everything cancels except for
ty = 0
If t = 0 (as in the above), there are infinite solutions, since 0*y is true for all y.
If t is NOT zero, then ty = 0 has the UNIQUE solution when y = 0 (since we're in a field)