Thanks! this stuff is really interesting

I haven't looked at the two links you gave me yet but I think I found another answer. I am still not 100% on what I am trying to do exactly but I was playing around with different things that you said and I got the following:

I noticed that $\displaystyle 7(12) \equiv 3~mod~27$

and I noticed that 3 was a common factor of 12, so I multiplied both sides by 12 to get:

$\displaystyle (7\times 12)x \equiv (12\times 12)~mod~ 27$

$\displaystyle 3x \equiv 9~mod~27$

$\displaystyle x \equiv 3~mod~27$