Question 537385
three consecutive odd intergers whose sum is 117


x = first odd integer
x + 2 = 2nd odd integer {odd integers increase by 2 each time}
x + 4 = 3rd odd integer


x + x + 2 = x + 4 = 117 {their sum is 117}
3x + 6 = 117 {combined like terms}
3x = 111 {subtracted 6 from both sides}
x = 37 {divided both sides by 3}
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
<br>For more help from me, visit: <a href = "http://www.algebrahouse.com/" target = "_blank">www.algebrahouse.com</a><br><br><br>