...solution of a congruence with modulo notation from the reciprocal

Hey guys, new to the forum. I have been looking through my notes and I am stuck on an area of my number theory module.

I understand the modulo notation and for example that 14 ≡ 2 mod 12, etc etc.

I have the reciprocals (modulo 15):

1bar = 1

2bar = 8

3bar = No such x, also 5bar, 6bar, 9bar, 10bar, 12bar do not exist

4bar = 4

7bar = 13

8bar = 2

11bar = 11

13bar = 7

14bar = 14

I know this is just the reverse process but if someone could explain to me step by step how these solutions are found that would be great.

Thanks