Find three consecutive odd integers such that the first is 33 less than twice the second

n, n + 2, n + 4 - three consecutive odd integers

the first is 33 less than twice the second

n - 33 = 2 (n + 2) use distributive property

n - 33 = (2) (n) + (2) (2)

n - 33 = 2n + 4 add 33 to both sides

n = 2n + 37 subtract 2n from both sides

-n = 37 change the signs

n = 37

n + 2 = 37 + 2 = 39

n + 4 = 37 + 4 = 41

Answer: 37, 39, 41