If the sum of the reciprocals of two consecutive positive numbers is 21110 then find the numbers.
Let the two consecutive positive numbers be x and x+1.
Then according to the problem,
1x+1x+1=21110
or, 2x+1x(x+1)=21110
or, 220x+110=21x2+21x
or, 21x2−199x−110=0
or, 21x2−210x+11x−110=0
or, 21x(x−10)+11(x−10)=0
or, (x−10)(21x+11)=0
or, x=10. [ Since x is positive]
So the numbers are 10,11.