SOLUTION: A 45-foot rope is to be cut into three pieces. The second piece must be twice as long as the first piece and the third peice must be 9 feet longer than 3 times the length of the se

Question 199508: A 45-foot rope is to be cut into three pieces. The second piece must be twice as long as the first piece and the third peice must be 9 feet longer than 3 times the length of the second piece. How long should each three peices be?
How do I write this problem algibraicly? How do I solve it? My Final is tomarrow please help.
-Andrew
Found 2 solutions by jim_thompson5910, Electrified_Levi:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Let x=length of first piece, y=length of second piece, and z=length of third piece

Since "The second piece must be twice as long as the first piece ", this means that and because "the third peice must be 9 feet longer than 3 times the length of the second piece", we know that

Now because these pieces come from a 45 ft rope, they must add back up to 45. So we have the equation

Start with the last equation.

Plug in

Plug in

Multiply

So all you need to do is solve to find "x" (which will help you find "y" and "z"). I'll let you do that.
Answer by Electrified_Levi(103) (Show Source): You can put this solution on YOUR website!
Hi, Hope I can help,
.
A 45-foot rope is to be cut into three pieces. The second piece must be twice as long as the first piece and the third peice must be 9 feet longer than 3 times the length of the second piece. How long should each three peices be?
How do I write this problem algibraicly? How do I solve it? My Final is tomarrow please help.
-Andrew
.
So the total amount of rope is 45 ft, so, that means the length of all three rope lengths will equal 45
.
We have to name the lengths in terms of "x"
.
You can usually name one of the variables, in this case there are three, "x", it probably doesn't matter, but we will use the easiest way
.
We will name Length 1, "x"
.
The second piece must be twice as long as the first piece,
.
If the length 1 is "x", and length 2 has to be twice as long, length 2 = "2x"
.
We will work the third piece a little slower, because it is the hardest of the three
.
the third peice must be 9 feet longer than 3 times the length of the second piece,
.
the third peice must be 9 feet longer than 3 times the length of the second piece, which is "2x"
.
the third peice must be 9 feet longer than 3 times "2x"
.
the third peice must be 9 feet longer than 3(2x)
.
the third peice must be 9 feet longer than "6x",
.
You would now add "9" to "6x"
.
the third peice must be "6x + 9",
.
Length 3 = "6x + 9"
.
.
.
Length 1 =
.
Length 2 =
.
Length 3 =
.
All of these lengths will add up to 45, we can now have our equation
.

.
We can get rid of the parentheses
.

.
Now add like terms
.
=
.
We will move "9" to the right side
.
= = , to find "x" divide each side by "9"
.
= = , you can check your answer by replacing "x" with "4" in the equation
.
= = = = ( True )
.

.
Length 1 = =
.
Length 2 = = =
.
Length 3 = = = =
.
To show you that you can make "x" any of the lengths, I will briefly do "x" as the Length 2
.
Length 2 = , if length 2 is twice length 1, then length 1 =
, Length 3, is 9 more than three times the second's length, or
.
Length 1 =
.
Length 2 =
.
Length 3 =
.
You would then add these to equal 45 again
.
, you would solve to find "x" = 8, which was length 2
.

.
Length 1 = = =
.
Length 2 = =
.
Length 3 = = = =
.
If you even wanted to have Length 3 = "x" you could, but that would be the hardest and most difficult.
.
Length 1 =
.
Length 2 =
.
Length 3 =
.
Hope I helped, Levi

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