This stage requires some what-iffing.
Looking at the hundreds column, we know that E cannot be 0, 1,8 or 9, since they are already taken; it cannot be 7 because that would make N an 8, which is already taken.
Thus, E can be 2,3,4,5,6. and N can be 3,4,5,6, or 7.
Looking at the tens column, we know that N plus 8 requires a 1 to be carried over from the ones column, which means D plus E has to be greater than 10.
The highest number that D can be in the ones column is a 7, which means that E has to be at least 5 in the ones column, in order to get a sum that is greater than 10.
Thus, E can be 5 or 6. and N can be 6 or 7.
At this point, I’ll just try E = 6. If E is 6, that would make N = 7. If that is the case, then the highest D could be (in the ones column) is 5. Adding a 5 to a 6 would make Y = 1, which is already taken.
Thus, it seems as if E must be 5.
Let’s check.
If E = 5, then N would be 6. If E is 5 in the ones column, then D would have to equal 7, since D plus E has to be more than 10. If E is 5, D would have to be at least 6, but adding those two numbers together gives a 1 in the ones column which is taken. Since 8 and 9 are already taken, the only number left for D is a 7. This would make the Y a 2
The final solution would look like this:
Borden's Blather 