Question 1016277
n and n+1 are these two numbers.


{{{n+(n+1)=77}}} should be all you need.
{{{2n+1=77}}}
{{{2n=76}}}
{{{n=76/2}}}
{{{n=38}}} so the other number must be  {{{n+1=39}}}.


There was given an additiona description.  Continuing to use n and n+1 because
these must still be the same numbers as described, n for x and n+1 for y.


{{{(1/3)(n+1)-(1/2)n=6}}},
or,
{{{system((1/3)y-(1/2)x=6,x+y=77)}}}  is expected to also fit the description.
THis is BAD because it shows  n=-34 which does not agree with the first part of the already
solved part of the description.  Some of your description is wrong.