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 Comments are closed.

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