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Question 78165: the sum of 3 consecutive integrals numbers is 117. find the numbers
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! If we call the first unknown integer x, then the next greater integer is x+1, and the next
greater integer is (x+1)+1 which is x+2.
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According to the problem, the sum of these three consecutive integers is 117. In equation
form the sum is:
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x + (x+1) + (x+2) = 117
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with the parentheses just being used to help identify the three integers. Since each set
of parentheses is preceded by a plus sign, the parentheses can just be erased without
changing the terms within. This makes the equation:
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x + x + 1 + x + 2 = 117
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combine the x terms to get:
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3x + 1 + 2 = 117
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next combine the constants on the left side to get:
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3x + 3 = 117
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eliminate the +3 on the left side by subtracting 3. But if you subtract 3 on the left
side, you must also subtract 3 from the right side. When you make these two subtractions
the equation becomes:
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3x = 114
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solve for x by dividing both sides by 3 to get:
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x = 114/3 = 38
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this means that the first integer in the series is 38. So the second integer is 39 and the
next is 40.
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Check by adding 38, 39, and 40. The total of these integers is 117, just as the problem
required it to be. Our answer check ... three consecutive integers that add up to be 117.
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Hope this helps you to understand the problem and how to set up consecutive integers.
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