find 3 consecutive integers such that twice the greatest integer is 2 less than 3 times the least integer

The consecutive integers are n, n+1, n+2

2 (n+2) = 3n-2 perform indicated multiplication on left side

2n+4=3n-2 subtract 3n from both sides

-n+4=-2 subtract 4 from both sides

-n=-6 divide both sides by - 1

n=6

So the three consecutive numbers are 6,7, and 8