Question 1044351
Let's say that
{{{n}}}= one of the integers involved, with {{{n<>0}}} , so that the reciprocal of {{{n}}} exists.
{{{1/n}}}= the reciprocal of the integer above.
For consecutive integers, let's say that 
{{{n+1}}}= the other integer, also with {{{n+1<>0}}} , so
{{{1/(n+1)}}}= the reciprocal of the "other integer" above.
So, {{{n(n+1)}}}= the product of those integers.
Since neither {{{n}}} , nor {{{n+1}}} is zero,
neither is their product: {{{n(n+1)<>0}}} ,so the reciprocal exists.
{{{1/(n(n+1))}}}= the reciprocal of the product of those two consecutive integers.
What the problem says is that
{{{1/n+1/(n+1)=11(1/(n(n+1)))}}}
That is the equation to simplify and solve.
{{{1/n+1/(n+1)=11(1/(n(n+1)))}}}
{{{(n+1)/(n(n+1))+n/(n(n+1))=11/(n(n+1))}}}
{{{(n+n+1)/(n(n+1))=11/(n(n+1))}}}
{{{(2n+1)/(n(n+1))=11/(n(n+1))}}}
Since {{{n(n+1)>0}}} , we can multiply both sides of the equation times {{{n(n+1)}}} to get a simpler, equivalent equation.
{{{2n+1=11}}}
{{{2n=11-1}}}
{{{2n=10}}}
{{{n=10/2}}}
{{{highlight(n=5)}}}
{{{n+1=5+1}}}
{{{highlight(n+1=6)}}}
The two integers are {{{highlight(5)}}} and {{{highlight(6)}}} .