SOLUTION: The length of a rectangle is 9 inches greater than the width. The area is 36 inches squared. Find the dimensions. Then I have 6 boxes each fo width, length and area. How do I f

Algebra ->  Algebra  -> Surface-area -> SOLUTION: The length of a rectangle is 9 inches greater than the width. The area is 36 inches squared. Find the dimensions. Then I have 6 boxes each fo width, length and area. How do I f      Log On

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Question 159760: The length of a rectangle is 9 inches greater than the width. The area is 36 inches squared. Find the dimensions. Then I have 6 boxes each fo width, length and area. How do I figure this out? Thank you.
Answer by edjones(7311) About Me  (Show Source):
You can put this solution on YOUR website!
L=w+9
A=Lw=36"^2
.
w(w+9)=36 Substitute w+9 for L.
w^2+9w=36
w^2+9w-36=0
(w+12)(w-3)=0
w=3, L=12
.
Ed