SOLUTION: sum of 3 consecutive integers is 21 larger than twice the smallest integer. find integer

Question 557253: sum of 3 consecutive integers is 21 larger than twice the smallest integer. find integer
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Consecutive integers are each separated from the preceding number by 1. So if we let I represent the first integer, then the next higher integer (the second integer) is I + 1, and I + 2 is the third and final integer.
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The sum of these 3 consecutive integers is, therefore:
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I + I + 1 + I + 2
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Simplify this sum by adding the three I's and adding the two constants to get:
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3I + 3
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The problem tells you that this sum is 21 more than twice the smallest integer, which is I. So we multiply the smallest integer by 2 to get 2I and then add 21 to that. It will equal the sum of the 3 integers. In equation form this is:
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3I + 3 = 21 + 2I
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Get rid of the 2I on the right side by subtracting 2I from both sides to get:
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I + 3 = 21
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Finally get rid of the 3 on the left side by subtracting 3 from both sides and you have:
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I = 18
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This means that the 3 consecutive integers are 18, 19, and 20.
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The sum of these 3 integers is 18 + 19 + 20 = 57
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Twice the smallest (2 times 18) equals 36 and when you add 21 to that you also get 57.
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This tells us that our answer is correct, and the 3 consecutive integers are 18, 19, and 20.
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I hope this helps you to understand how to do this problem that involves consecutive integers.
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