Let h = the hundreds digit
Let t = the tens digit
Let u = the units or ones dogit
I am a number with 2 more tens than ones.
t = u+2
I have 5 more ones than hundreds.
u = h+5
the sum of my digits is 18.
t+h+u = 18
Solve the second for h, u-5=h
Substitute (u+2) for t, and (u-5) for h, in t+h+u = 18
(u+2) + (u-5) + u = 18
u+2+u-5+u = 18
3u-3 = 18
3u = 21
u = 7
Substitute 7 for u in u-5 = h
7-5 = h
2 = h
Substitute 7 for u in t = u+2
t = 7+2
t = 9
Answer: [h][t][u] = 297
Edwin