Find three consecutive odd integers whose sum is 117.

Odd integers, such as 1,3,5,7,...... increase by 2 each time.

Therefore, if the first is x, then:

x + x + 2 + x + 4 = 117 {their sum is 117}

3x + 6 = 117 {combined like terms}

3x = 111 {subtracted 6 from each side}

x = 37 {divided each side by 3}

Remember, x is the first odd integer.

x + 2 = 39 {substituted 37, in for x, into x + 2}

x + 4 = 41 {substituted 37, in for x, into x + 4}

Therefore, if the first is x, then:

**x = 1st consecutive odd integer**

x + 2 = 2nd consecutive odd integer

x + 4 = 3rd consecutive odd integerx + 2 = 2nd consecutive odd integer

x + 4 = 3rd consecutive odd integer

x + x + 2 + x + 4 = 117 {their sum is 117}

3x + 6 = 117 {combined like terms}

3x = 111 {subtracted 6 from each side}

x = 37 {divided each side by 3}

Remember, x is the first odd integer.

x + 2 = 39 {substituted 37, in for x, into x + 2}

x + 4 = 41 {substituted 37, in for x, into x + 4}

**37, 39, and 41**are the three consecutive odd integers**Ask Algebra House**