There are many possible answers to this problem. 

Let D = David's age now
Let S = Susan's age now
Let y = the number of years ago it was when David was Susan's present age, or S.

Then y years ago, David was D-y and he was also S, so we set those equal:

       D-y = S 

Also y years ago, Susan was S-y, and David's age now, D, is three times this,
so

         D = 3(S-y) 

So this is the system of equations:

     D - y = S
     D = 3(S - y)

Solve the first equation for y

y = D - S

D = 3(S - (D - S))
D = 3(S - D + S)
D = 3(2S - D)
D = 6S - 3D
4D = 6S
 D = S
 D = S
 D = 1.5S

All we can get is that David is 1 and a half times as old as Susan.

Assuming whole numbers for their ages, the youngest they could be
is David 3 and Susan 2.

1 year ago David was 2, the same age Susan is now.
1 year ago Susan was 1, and so David, being 3, is indeed 3 times as
old as she was then, since 3 is 3 times 1.

Or David could be 6 and Susan 4, 

2 years ago David was 4, the same age Susan is now.
2 years ago Susan was 2, and so David, being 6, is indeed 3 times as
old as she was then, since 6 is 3 times 2.

Or David could be 60 and Susan 40, 

20 years ago David was 40, the same age Susan is now.
20 years ago Susan was 20, and so David, being 60, is indeed 3 times as
old as she was then, since 60 is 30 times 2.

Or David could be 15 and Susan 10, 

5 years ago David was 10, the same age Susan is now.
5 years ago Susan was 5, and so David, being 15, is indeed 3 times as
old as she was then, since 15 is 3 times 5.

Or David could be 99 and Susan 66, 

33 years ago David was 66, the same age Susan is now.
33 years ago Susan was 33, and so David, being 99, is indeed 3 times as
old as she was then, since 99 is 3 times 33.

Was something else given in that problem you forgot to tell us?

Edwin