original number = 10t+u
new number = 10u+t+3

  10u+t+3 = 2(10t+u)+2
  10u+t+3 = 20t+2u+2
     8u+1 = 19t

Write 19 in terms of its nearest multiple of the smaller coefficient 8,
as, 16+3

     8u+1 = (16+3)t
     8u+1 = 16t+3t

Divide through by the smaller coefficient, 8:

   

Isolate fractions on 
   

The left side is an integer, therefore so is the right side, say integer A

      u-2t=A and 
                    3t-1=8A

             Write 8 as 9-1   

                    3t-1=(9-1)A
                    3t-1=9A-A

Divide through by 3

                 

Isolate the fractions:

                 


The right side is an integer, therefore so is the left side, say integer B

,  3A-t=B
A-1=3B
A=3B+1

3A-t=B
3(3B+1)-t=B
9B+3-t=B
8B+3=t

Since t is a digit, B=0 and t=3

Substitute in:
 
8u+1 = 19t
8u+1 = 19(3)
8u+1 = 57
  8u = 56
   u = 7

So the original number is 37.

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Checking:

Reversing the digits is 73.
Adding 3 gives 76.
76 is 2 more than 74.
74 is 2 times 37.
So it checks.

Edwin