I began by setting up a system of linear equations:
14≡8a+b(mod27)
5≡26a+b(mod27)
and then subtracted them to get: 9≡9a(mod27). I know 9 doesn't have a multiplicative inverse modulo 27 but a=1 would solve this. I'm afraid I must be making some mistake somewhere, however, as solving this all the way through gives me b=6 and then translating the message doesn't quite make sense.
Answer
9a≡9mod27
a≡1mod3
∴a=1+3k
b≡14−8amod27
b=14−8(1+3k)+27m
∴b=6−24k+27m
Where k,m∈Z.
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