1) 1xu + 0xd + 1yu + 0yd = 25
2) 0xu + 1xd + 1yu + 0yd = 24
3) 1xu + 0xd + 0yu + 1yd = 37
4) 0xu + 1xd + 0yu + 1yd = 36


5) 1xu = 10
6) 1yu = 15

find xd

1) - 2) 

   1xu + 0xd + 1yu + 0yd = 25
 - 0xu + 1xd + 1yu + 0yd = 24
------------------------------
7) 1xu -1xd = 1
8) 1xu - 1= 1xd

substitute 5) to 8)

9) 10 - 1 = 1xd
9) 9 = 1xd
substitude  9) to 4)
10) 0xu + 1xd + 0yu + 1yd = 36
    9 + 1yd = 36
    1yd = 36 - 9
11) 1yd = 27

so the numbers at the back are 9 and 27

at the back of number 10 is 9
at the back of number 15 is 27