SOLUTION: I have two algebra I word problems i need help setting up. I would like if you can explain the steps of setting up. Here it is: Problem #1 The length of a rectangle is 2 1/2 tim

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Question 612274: I have two algebra I word problems i need help setting up. I would like if you can explain the steps of setting up.
Here it is:
Problem #1
The length of a rectangle is 2 1/2 times its width. Its area is 90 square units. What are its dimensions? (Hint: length times width = area)
Let w= width in units.
L=(2 1/2)w= 5w/2 = length in units
Found 2 solutions by richwmiller, MathTherapy:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
now multiply length by width
(5w/2)*w=90
5w^2=45
w^2=9
w=3
5w/2=15/2=L

Answer by MathTherapy(10579) About Me  (Show Source):
You can put this solution on YOUR website!
I have two algebra I word problems i need help setting up. I would like if you can explain the steps of setting up.
Here it is:
Problem #1
The length of a rectangle is 2 1/2 times its width. Its area is 90 square units. What are its dimensions? (Hint: length times width = area)
Let w= width in units.
L=(2 1/2)w= 5w/2 = length in units

The person who responded is WRONG!! Width is NOT 3 units!

As indicated by you, let width be w
Then length (L), as you stated is 2%261%2F2w, or 2.5w
As it's given that area = 90 square units, we get: 2.5w(w) = 90
                                                     2.5w2 = 90
                                                        w%5E2+=+90%2F2.5
                                                        w%5E2+=+36
                           Width of rectangle, or 
                                          Length: 2.5(6) = 15 units