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
