Here, since we have numbers larger than the modulus, we can reduce them prior to applying any algebra. We have \(84 \equiv 9\text{,}\) \(38 \equiv 8\) and \(79 \equiv 4\text{.}\) Thus,
We got the 72 by adding \(0 \equiv 60 \pmod{15}\) to both sides of the congruence. Now divide both sides by 9. However, since \(\gcd(9, 15) = 3\text{,}\) we must divide the modulus by 3 as well:
So the solutions are those values which are congruent to 8, or equivalently 3, modulo 5. This means that in some sense there are 3 solutions modulo 15: 3, 8, and 13. We can write the solution: