SOLUTION: The length of a rectangle is 3yd less than double the width, and the area of the rectangle is 44yd^2. Find the dimensions of the rectangle?

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: The length of a rectangle is 3yd less than double the width, and the area of the rectangle is 44yd^2. Find the dimensions of the rectangle?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 809005: The length of a rectangle is 3yd less than double the width, and the area of the rectangle is 44yd^2. Find the dimensions of the rectangle?
Answer by vheroli(126) About Me  (Show Source):
You can put this solution on YOUR website!
Where:
L= length
W= width
x= variable
A= area
Formulas:
A= L*W
L= 2x-3
W= x
Solution:
44= (2x-3) * (x): Distribute x
44=+2x%5E2+-+3x : Notice that the equation could be factor out.
+2x%5E2+-+3x+-+44+=+0+ : Factor out
= (x+4) (2x-11) : Reject the factor (x+4) because it is negative.
Therefore we will use the factor (2x-11) to get the answer:
2x-11=0
2x=11
x= 11/2
x= 5.5
Substitute the x to get the answers from our previous Formulas:
L= 2x-3
L= 2(5.5)-3
L= 8 yd

W= x
W= 5.5 yd