Find five consecutive integers such that:
“the sum of the first and 3 times the fourth is equal to 39 less than twice the sum of the second, third, and fifth.”
make sure to list all five integers in your answer and show all of your

  17, 18, 19, 20, 21

Step-by-step explanation:

Let x represent the third (middle) integer. Then the first is (x-2) and the given relation is ...

  (x-2) + 3(x+1) = -39 +2((x-1)+x+(x+2))

  4x +1 = 6x -37 . . . . simplify

  38 = 2x . . . . . . . . . add 37-4x

  x = 19 . . . . . . . . . . divide by 2

The integers are ...

  17, 18, 19, 20, 21

Check

  17 + 3·20 = -39 +2(18 +19 +21)

  17 +60 = -39 +2(58)

  77 = -39 +116 . . . . . yes

∈3,600

step-by-step explanation: