Question 105664: The sum of three consecutive integers is 33 more than the least of the integers. Find the integers. Found 2 solutions by elima, bucky:Answer by elima(1433) (Show Source):
You can put this solution on YOUR website! The sum of three consecutive integers is 33 more than the least of the integers. Find the integers.
x=first integer
x+1=second integer
x+2= third integer
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x+x+1+x+2=x+2+33
3x+3=x+35
3x-x=35-3
2x=32
x=16
so the integers are;
16,17,18
:)
You can put this solution on YOUR website! The three consecutive integers can be represented as:
.
x
x+1 and
x+2
.
The sum of these three is 3x + 3
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If you take 33 away from this sum, it is to equal the least of the integers which is x.
So
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3x + 3 - 33 = x
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Subtract x from both sides and you get:
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2x + 3 - 33 = 0
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Now combine the +3 and -33 and you have:
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2x - 30 = 0
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Add 30 to both sides:
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2x = 30
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Solve for x by dividing both sides by 2 to get:
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x = 15
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So the answer is that the three integers are 15, 16, and 17.
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