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

step-by-step explanation:

∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. formed with the point d(1, 2) as its third vertex, then ∆abd is triangle.