SOLUTION: i need help on finding three consecutive odd intergers whose sum is 117

Question 537385: i need help on finding three consecutive odd intergers whose sum is 117
Found 2 solutions by algebrahouse.com, Alan3354:
Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
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

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Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
117/3 = 39, the middle number
--> 37, 39 & 41
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