The problem works out to a set of three equations:
b + c + d = 22
a + c + e = 22
a + b + c + d + e = 30

Solving for c = 14, leaving d = 8 – b and e = 8 – a. In other words, c must be 14, but the other two numbers just have to add up to 8. The requirement that they be unique rules out 4 + 4, so you’re left to choose from the following combations for b + d and a + e:
0 + 8
1 + 7
2 + 6
3 + 5