Integers which are congruent to 99 mod 243 are always divisible by 3, so to make such an integer divisible by 30 (i.e. have an integer solution), it must be divisible by 10 as well. Finding the first such positive integer, I obtain 1800 (a value of x = 60).
Next, we wish to generalize. Since 243 is coprime to 10, we have to add 243 a total of 10 times in order to have another integer divisible by 30. We can think of this as adding 2430. Thus, any integer that is divisible by 30 and congruent to 99 mod 243 has form 1800 + 2430n for some integer n. Set 30x = 1800 + 2430n and solve for x to describe the pattern.
I hope this helped, and please let me know if you have any other questions or if I made an error.