Question 1161973
<pre>
Their answers are correct, but they all essentially used trial and error
to solve the Diophantine equation in integers.  Here is how to do it
by actually solving for t and u, without using trial and error.  

[Warning: it is much longer, but more practical in general integer problems.]

In fact, your teacher may not allow trial and error, since one of the main
uses of algebra is to avoid trial and error.

(1)        8b + 17 = 19a

The smallest coefficient of either letter in absolute value is 8.
We write every number in terms of the nearest multiple of 8.

       8b + (16+1) = (16+3)a

       8b + 16 + 1 = 16a + 3a

Divide through by 8

       b + 2 + 1/8 = 2a + 3a/8  

Isolate the fractions:

       b + 2 - 2a = 3a/8 - 1/8

The left side is an integer, so the right side is also. 
Let that integer be P.  So we have:

(2)    b + 2 - 2a = P

and

       3a/8 - 1/8 = P

Clearing of fractions:

(3)        3a - 1 = 8P

The smallest coefficient of either letter in absolute value is 3.
We write every number in terms of the nearest multiple of 3.

           3a - 1 = (6+2)P

           3a - 1 = 6P + 2P

Divide through by 3

          a - 1/3 = 2P + 2P/3  

Isolate the fractions:

           a - 2P = 2P/3 + 1/3

The left side is an integer, so the right side is also. 
Let that integer be Q.  So we have:

(4)        a - 2P = Q

and

       2P/3 + 1/3 = Q

Clear of fractions:

(5)        2P + 1 = 3Q

---------

The smallest coefficient of either letter in absolute value is 2.
We write every number in terms of the nearest multiple of 2.

           2P + 1 = (2+1)Q

           2P + 1 = 2Q + Q

Divide through by 2

          P + 1/2 = Q + Q/2  

Isolate the fractions:

            P - Q = Q/2 - 1/2

The left side is an integer, so the right side is also. 
Let that integer be R.  So we have:

(5)         P - Q = R

and

        Q/2 - 1/2 = R

Clear of fractions:

            Q - 1 = 2R

(6)             Q = 2R + 1

We finally have solved for a letter with no fraction terms, so
we can now substitute back:

Use (6) to substitute for Q in 5

(5)         P - Q = R
     P - (2R + 1) = R
       P - 2R - 1 = R
                P = 3R+1

(4)        a - 2P = Q
      a - 2(3R+1) = 2R + 1
       a - 6R - 2 = 2R + 1
(6)             a = 8R+3

Since a is a digit,

      0  a <= 9
      0 <= 8R+3 <= 9
     -3 <= 8R <= 6
   -3/8 <= R <= 6/8

Since R is an integer, R = 0

Substituting in (6)

(6)   a = 8R+3
      a = 8(0)+3
      a = 0+3
(7)   a = 3

Substitute in 

(1)   8b + 17 = 19a  
      8b + 17 = 19(3)
      8b + 17 = 57
           8b = 40
            b = 5

So Emma's age = 35

Now I will agree that their trial and error methods were much easier and
much shorter. However, in general, Diophantine equations, trial and error
would involve trying many integers.  In those cases this method would be the
only practical method. 

Edwin</pre>