hmmmm let's say our numbers are "a" and "b", but we know they're consecutive integer, therefore "b" is either a-1 or a+1, hmm let's use the latter
a = first integer
a+1 = second integer
we know their reciprocals add up to 5/6, so
[tex]\begin{cases} a\\ a+1 \end{cases}\qquad \stackrel{\textit{sum of their reciprocals}}{\cfrac{1}{a}+\cfrac{1}{a+1}}=\cfrac{5}{6}[/tex]
let's use the LCD of (a)(a+1)(6) and multiply both sides by it, to do away with the denominators
[tex]6a(a+1)\left[ \cfrac{1}{a}+\cfrac{1}{a+1} \right]=6a(a+1)\left[ \cfrac{5}{6} \right] \\\\\\ 6(a+1)+6a = 5a(a+1)\implies 6a+6+6a=5a^2+5a \\\\\\ 12a+6=5a^2+5a\implies 0 = 5a^2-7a-6\implies 0 = (a-2)(5a+3) \\\\\\ \begin{cases} 2=a&\textit{\large \checkmark}\\ -\frac{3}{5}=a&\textit{not an integer} \end{cases}~\hfill \begin{cases} a &= 2\\ a+1 &= 3 \end{cases}[/tex]
