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Question 453874: Find three consecutive odd integers such that the first times the second is 1 more than 2 times the third.
Answer by algebrahouse.com(1659)   (Show Source):
You can put this solution on YOUR website! "Find three consecutive odd integers such that the first times the second is 1 more than 2 times the third."
x = 1st odd integer
x + 2 = 2nd odd integer
x + 4 = 3rd odd integer
x(x + 2) = 2(x + 4) + 1 {the first times the second is one more than 2 times the third}
x² + 2x = 2x + 8 + 1 {used distributive property}
x² + 2x = 2x + 9 {combined like terms}
x² = 9 {subtracted 2x from each side}
x = 3 or -3 {took the square root of both sides}
x + 2 = 5 or -1 {substituted 3 and -3, in for x, into x + 2}
x + 4 = 7 or 1 {substituted 3 and -3, in for x, into x + 4}
3,5,7 or -3,-1,1 are the three consecutive odd integers
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