Question 673332
Let the three consecutive integers be {{{n}}}, {{{n+1}}}, and {{{n+2}}}.
The greatest integer is {{{n+2}}}.
Twice the greatest integer is {{{2(n+2)}}}.
The least integer is {{{n}}}.
Three times the least integer is {{{3n}}}.
Then, 2 less than 3 times the least integer is {{{3n-2}}}
Twice the greatest integer is 2 less than 3 times the least integer
translates as {{{2(n+2)=3n-2}}}
{{{2(n+2)=3n-2}}} --> {{{2n+4=3n-2}}} --> {{{2n+4-2n=3n-2-2n}}} --> {{{4=n-2}}} --> {{{4+2=n-2+2}}} --> {{{6=n}}} or {{{highlight(n=6)}}}
So the three consecutive integers are {{{highlight(6)}}}, {{{highlight(7)}}}, and {{{highlight(8)}}}.