Question 253945
Let the two consecutive odd integers be x and x+2.
Given, (1/x)+(1/x+2) = 20/99
      (x+2+x)/x(x+2) = 20/99
            99(2x+2) = 20x(x+2)
            198x+198 = 20x^2+40x
         20x^2+40x-198x-198 = 0
         20x^2-158x-198 = 0
         10x^2 - 79x-99 = 0
         10x^2-90x+11x-99 = 0
         10x(x-9)+11(x-9)=0
         (x-9)(10x+11) =0
                    x-9 = 0  or 10x+11 = 0
                      x = 9  or   10x = -11
                      x = 9  or x = - 11/10
Since x cannot be a fraction, take x = 9.
     So the given numbers are 9 and 11