If 3x+7=2 (mod 16), then 2x+11+4x is congruent (mod 16) to what integer between 0 and 15, inclusive?
\(3x + 7 \equiv 2 \pmod{16}\\ 3x \equiv -5 \pmod{16}\\ 33x \equiv -55 \pmod{16}\\ x \equiv 9 \pmod{16}\)
Now you just take x = 9, then calculate the value of 2x + 11 + 4x. The answer is the remainder when 2x + 11 + 4x is divided by 16.
