Diophantine Equation Solution in Plain Text
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This time, I want to edit the solution's description in "vi".
The equation is:

   3       5
  ---  +  --- = 4
   x       y

What we need are all pairs of integers (x,y) that satisfy the equation. The 
equation can be solved by finding y in terms of x or x in terms of y. So, 
let us start by multiplying both sides by xy. Remember that neither x nor y 
can be zero:

     3y + 5x = 4xy ==>

==>  5x = 4xy - 3y = (4x-3)y ==>

            5x       4x - 3 + x + 3          x + 3
==>  y = -------- = ---------------- = 1 + --------
          4x - 3         4x - 3             4x - 3

And y is an integer, which means that the denominator divides the numerator.
Or 4x - 3 divides x + 3, which means that 4x - 3 divides 4(x+3)=4x+12. And of 
course
4x - 3 divides itself, thus it divides the difference between itself and 
4x+12.
So, 4x - 3 divides 4x + 12 - (4x - 3) = 15
Let us use the following table to find values of x and y if 4x - 3 divides 15:

  +========+=====+==============+
  | 4x - 3 |   x  |       y     |
  +========+======+=============+
  |     1  |  1   |           5 |
  +--------+-----+--------------+
  |    -1  |  0.5 |  Never mind |
  +--------+------+-------------+
  |     3  |  1.5 |  Never mind |
  +--------+------+-------------+
  |    -3  |  0   |  Invalid    |
  +--------+------+-------------+
  |     5  |  2   |           2 |
  +--------+------+-------------+
  |    -5  | -0.5 |  Never mind |
  +--------+------+-------------+
  |    15  |  4.5 |  Never mind |
  +--------+------+-------------+
  |   -15  | -3   |           1 |
  +========+======+=============+

From the table above, we get that the solution set for (x,y) is:

  {(1, 5), (2, 2),(-3, 1)}

Another method to find pairs satisfying the equation is to find a bounded set 
from which to check integer values of x:
If 4x - 3 divides x + 3, it means that either

  x + 3 = 0

or

  | 4x - 3 | <= | x + 3 |

We can solve the absolute value inequality by squaring both sides. And then 
you will have less values from which to select.