Question 935954: Find three consecutive integers such that the sum of the three is equal to 2 times more the sum of the first two integers
Answer by cornelius.bayler(4)   (Show Source):
You can put this solution on YOUR website! x = first
x + 1 = second
x + 2 = third {consecutive integers increase by 1}
x + x + 1 + x + 2 = 2(x + x + 1) {sum of the three is equal to 2 times the sum of first two}
3x + 3 = 2(2x + 1) {combined like terms}
3x + 3 = 4x + 2 {used distributive property}
x = 1 {subtracted 3x and subtracted 2 from each side}
x + 1 = 2 {substituted 1, in for x, into x + 1}
x + 2 = 3 {substituted 1, in for x, into x + 2}
1, 2, and 3 are the three consecutive integers
For more help from me, visit: www.algebrahouse.com
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