SOLUTION: the length of the cover of a road atlas is 4 in. more than the width. The area is 165 in^2. Fnd the dimensions of the cover answer: 15 inch by 11 inch anyone can gi

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: the length of the cover of a road atlas is 4 in. more than the width. The area is 165 in^2. Fnd the dimensions of the cover answer: 15 inch by 11 inch anyone can gi      Log On

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Question 193755: the length of the cover of a road atlas is 4 in. more than the width. The area is 165 in^2. Fnd the dimensions of the cover


answer: 15 inch by 11 inch anyone can give a complete break down on why the anwer is 15 inch by 11 inch

Found 2 solutions by stanbon, RAY100:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
the length of the cover of a road atlas is 4 in. more than the width. The area is 165 in^2. Fnd the dimensions of the cover
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Let the width be x inches.
Then the length is x+4 inches.
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Equation:
Area = L*W
165 = x(x+4)
x^2 + 4x - 165 = 0
(x-11)(x+15) = 0
x = 11 inches (width)
x+4 = 15 inches (length)
===============================
Cheers,
Stan H.

Answer by RAY100(1637) About Me  (Show Source):
You can put this solution on YOUR website!
Area = Length * Width
but L=w+4
,
A= (w+4) * (w) =w^2 + 4w = 165
or w^2 +4w - 165 = 0
factoring
(w+15)(w-11)=0
w=-15( not realistic)
or w= 11 ,,,,,ok
then L= w+4 = 11 +4 = 15,,,ok
check 11*15 = 165,,,ok