Find four consecutive odd integers such that 3 times the largest decreased by 19 is twice the second integer

n, n + 2, n + 4, n + 6 - four consecutive odd integers

3 times the largest decreased by 19 is twice the second integer

3 (n + 6) - 19 = 2 (n + 2) use distributive property

(3) (n) + (3) (6) - 19 = (2) (n) + (2) (2)

3n + 18 - 19 = 2n + 4

3n - 1 = 2n + 4 add 1 to both sides

3n = 2n + 5 subtract 2n from both sides

n = 5

n + 2 = 5 + 2 = 7

n + 4 = 5 + 4 = 9

n + 6 = 5 + 6 = 11

Answer: 5, 7, 9, 11