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

45.3592

step-by-step explanation: