Question 934958: find three consecutive odd intergers such that twice the sumof the first and third more than the second by fifteen
Answer by algebrahouse.com(1659)   (Show Source):
You can put this solution on YOUR website! 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
For more help from me, visit: www.algebrahouse.com
|
|
|
