SOLUTION: find three consecutive integers such that the sum of the first and twice the second is 17 more than twice the third.

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Question 25216: find three consecutive integers such that the sum of the first and twice the second is 17 more than twice the third.
Answer by Paul(988) About Me  (Show Source):
You can put this solution on YOUR website!
Let the three consecutive numbers be, x, x+1, and x+2
Twice the second = 2(x+1)
"is" =
17 more than twice the third = 17+2(x+2)
Add them togeher.
x+2(x+1)=17+2(x+2)
x+2x+2=17+2x+4
3x+2=2x+21 --->SImplfy
x=19
19+1=20
19+2=21

Hence the three numbers are 19, 20, and 21.
Paul.