Question 840588
tens digit greater than  their unit digit
<pre>t>u</pre>the sum of their digits equal to twice their difference<pre>t+u = 2(t-u)
t+u = 2t-2u
 3u = t 
So the system is:

{{{system(t > u,3u = t)}}}

Since 1 &#8806; t &#8806; 9

      1 &#8806; 3u &#8806; 9
     1/3 &#8806; u &#8806; 3
       1 &#8806; u &#8806; 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</pre>