SOLUTION: how many two digit numbers with their tens digit greater than their unit digit,have the sum of their digits equal to twice their differrence?

t>u

t+u = 2(t-u)
t+u = 2t-2u
 3u = t 
So the system is:



Since 1 ≦ t ≦ 9

      1 ≦ 3u ≦ 9
     1/3 ≦ u ≦ 3
       1 ≦ u ≦ 3

So u = 1, 2 or 3

If u = 1, then t = 3u = 3(1) = 3, so the number is 31
If u = 2, then t = 3u = 3(2) = 6, so the number is 62
If u = 3, then t = 3u = 3(3) = 9, so the number is 93

So there are three solutions: 31, 62, and 93

Edwin