H = 3D - 17

 
T = 2H + 12

 
D = H - 7
 

 
-3D +  H     = -17
    - 2H + T =  12
  D -  H     =  -7
 
Solve that and get D=12, H=19, T=50

----------------------

Let either one of their ages be A.
Let the 5 consecutive integers be I1,I2,I3,I4,I5
Let the remainders when these integers are divided
by A be R1,R2,R3,R4,R5

If none of the consecutive integers is A or
or a multiple of A, then the 5 remainders will 
themselves be 5 consecutive integers.

Let's see if there exist 5 consecutive integers 
which the remainders could be, so as to have the sum 32.






No, that is not possible, for n doesn't come out
to an integer.

That means one of 5 consectutive integers must
be a multiple of A, say kA

Whichever one of the 5 is kA, The remainder left when
dividing by A will of course be 0.  

Any of the 5 remainders of members of the 5 that are 
less than kA, will be consecutive integers, the largest 
of which must be A-1, then the one below that will be
A-2, and so on.

Any of the 5 remainders that are greater than the multiple
of A will be consecutive integers 1,2,3, and so on

Then we have 5 cases to consider, that is, when kA comes
1st, 2nd, 3rd, 4th, and 5th.

         I1    I2    I3    I4    I5     R1   R2   R3   R4  R5   Sum  
case 1.  kA   kA+1  kA+2  kA+3  kA+4     0    1    2    3   4    10
case 2. kA-1   kA   kA+1  kA+2  kA+3    A-1   0    1    2   3   A+5
case 3. kA-2  kA-1   kA   kA+1  kA+2    A-2  A-1   0    1   2    2A
case 4. kA-3  kA-2  kA-1   kA   kA+1    A-3  A-2  A-1   0   1  3A-5    
case 5. kA-4  kA-3  kA-2  kA-1   kA     A-4  A-3  A-2  A-1  0  4A-10
 
We set each of those sums on the far right equal to 32 and see which,
if any, have an integer solution:

Case 1. This is out because 10 does not equal 32.

Case 2.  A+5 = 32
           A = 27

That's a whole number, so one of their ages can be 27.

Case 3.  2A = 32 
          A = 16

That's a whole number, so one of their ages can be 16.

Case 4. 3A-5 = 32
        2A   = 37
         A   = 27/2

That is not a whole number, so Case 4 is out.

Case 5. 4A-10= 32
        4A   = 42  
         A   = 21/2

That is not a whole number, so Case 5 is out.

So Alvin is 16 and Susan is 27, making the sum of their ages 43.

Now there is no way to tell whether Susan's 5 consecutive
integers were 

26,27,28,29,30, or 53,54,55,56,57, or 80,81,82,83,84 or etc.

There is also no way to tell whether Alvin's 5 consecutive
integers were 

14,15,16,17,18, or 30,31,32,33,34, or 46,47,48,49,50 or etc.
 
However he must be 16 and she must be 27.

Edwin