Question 25655
Let one of the integers be "x" and the other be "y".
x+y=10
1/x -1/y= 3/8 or 8(y-x)=3xy
Solve by substitution:
 x=10-y
8[y-(10-y)]=3(10-y)y
8[2y-10]=30y-3y^2
16y-80=30y-3y^2
3y^2-14y-80=0
(3y+10)(y-8)=0
y=8 is the positive integer solution
So x=2

Cheers,
stan H.