SOLUTION: Find three consecutive intergers such that the sum of the second and third excedds half of the first by 33

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Question 493824: Find three consecutive intergers such that the sum of the second and third excedds half of the first by 33
Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the first integer.
Then x + (x+1) + (x+2) are three consecutive integers.
We know that the sum of the second (x+1) and the third (x+2) is (x+33)/2
So %28x%2B1%29+%2B+%28x%2B2%29+=+x%2F2+%2B+33
Combine and simplify
+2x+%2B3+=+x%2F2+%2B+33
Subtract 33 on both sides.
+2x-30+=+x%2F2
Multiply by 2 on both sides
4x+-60+=+x
Combine like terms.
-60+=+-3x
Divide by -3
x+=+20
So the consecutive integers are 20,21,22.