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Question 744978: find three consecutive odd intergers so that 3 times their sum is 5 more than 8 times the middle one
Answer by algebrahouse.com(1659)   (Show Source):
You can put this solution on YOUR website! Find three consecutive odd integers, so that 3 times their sum is 5 more than 8 times the middle one.
x = 1st odd integer
x + 2 = 2nd odd integer
x + 4 = 3rd odd integer {odd and also even integers increase by 2}
3(x + x + 2 + x + 4) = 8(x + 2) + 5 {sum is equal to 5 more than 8 times the middle one}
3(3x + 6) = 8x + 16 + 5 {combined like terms on left, used distributive property on right}
9x + 18 = 8x + 21 {combined like terms}
x = 3 {subtracted 8x and subtracted 18 from each side}
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
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