Question 387968
You have the right idea, but there's a slight problem. 




If {{{x>y}}}, then this means that {{{x-y>0}}}. So the first equation {{{x-y=6}}} implies that {{{x>y}}} (ie 'x' is the bigger number)



However, recall that if {{{x>y}}}, then {{{1/x<1/y}}}. So the equation {{{1/x-1/y=8/15}}} implies that 'x' is the smaller number, which is a contradiction.



So the second equation should be {{{1/y-1/x=8/15}}}