Question 549366
There's two ways to do this...one by thinking it out
If the sum is 90, the average must be 45. If they are 6 digits apart, one must be 3 above 45 and the other 3 below. So our numbers are 42 and 48.

Solving this algebraically we have
x+y=90
x-y=6
____
x-y=6
x=6+y
substitute this in
6+y+y=90
6+2y=90
2y=84
y=42

x+42=90

x=46