Question 789857
consecutive even integers: x, (x+2)


1/x + 1/(x+2) = 9/40
(2x + 2)/(x(x+2)) = 9/40
40(2x + 2) = 9(x(x+2))
80x + 80 = 9x^2 + 18x
0 = 9x^2 - 62x - 80
0 = (x - 8)(9x + 10)
x = 8, x = -10/9


x must be an integer, so the integers are 8 and 10.