Question 667785
If {{{x}}} is any {{{odd}}} number, then {{{x}}} and {{{x + 2}}} are consecutive odd numbers.

The sum of the reciprocals of two consecutive odd numbers is:

{{{(1/x)+(1/(x+2))=8/15}}}


{{{(1(x+2)/x(x+2))+(1x/x(x+2))=8/15}}}


{{{((x+2+1x)/x(x+2))=8/15}}}


{{{(2x+2)/(x(x+2))=8/15}}}........cross multiply


{{{15(2x+2)=8*x(x+2)}}}


{{{30x+30=8x^2+16x}}}


{{{0=8x^2-30x+16x-30}}}


{{{8x^2-14x-30=0}}}.....replace {{{-14x}}} with {{{10x-24x}}}


{{{8x^2+10x-24x-30=0}}}....group


{{{(8x^2-24x)+(10x-30)=0}}}


{{{8x(x-3)+10(x-3)=0}}}


{{{(8x+10)(x-3)=0}}}


solutions: positive solution


if {{{x-3=0}}}......=>.....{{{x=3}}}

the numbers are:

{{{3}}} and {{{5}}} 


check:

{{{(1/x)+(1/(x+2))=8/15}}}

{{{(1/3)+(1/5)=8/15}}}

{{{(1*5+1*3)/15=8/15}}}

{{{(5+3)/15=8/15}}}

{{{8/15=8/15}}}