SOLUTION: the problem is: a 45ft rope is cut into 3 peices. the 2nd peice must be twice as long as the 1st peice. the 3rd peice must be 9 feet longer than three times the length of the seco

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Question 29435: the problem is: a 45ft rope is cut into 3 peices. the 2nd peice must be twice as long as the 1st peice. the 3rd peice must be 9 feet longer than three times the length of the second peice. how long should each peice be?
i have gotten as far as:
1st peice is x
2nd is 2x(x)
3rd is 9+3(2x(x).
so it goes to x+2x(x)+9+3(2x(x))= 45
Im lost from there.
Answer by Paul(988) About Me  (Show Source):
You can put this solution on YOUR website!
Let the first piece be x
LEt the second piece be 2x
Let the third piece be 9+3(2x)

Equation:
(x)+(2x)+(9+6x)=45
Simplfy

9x=36
x=4

2(4)=8
9+3(8)=33

Hence, the first piece is 4ft long the second piece is 8ft long and the third piece is 33ft long.
Paul.