Find three consecutive odd integers, such that twice the sum of the first and third is more than the second by fifteen.

**x = 1st odd consecutive odd integer**

**x + 2 = 2nd consecutive odd integer**{odd integers increase by 2}

**x + 4 = 3rd consecutive odd integer**

2(x + x + 4) = x + 2 + 15 {twice the sum of first and third is more than second by 15}

2(2x + 4) = x + 17 {combined like terms}

4x + 8 = x + 17 {used distributive property}

3x = 9 {subtracted x and subtracted 8 from each side}

x = 3 {divided each side by 3}

x + 2 = 5 {substituted 3, in for x, into x + 2}

x + 4 = 7 {substituted 3, in for x, into x + 4}

**3,5, and 7**are the three consecutive odd integers

*- Algebra House*