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 = 1st consecutive odd integer x + 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 Comments are closed.

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