SOLUTION: A picture’s length is 6in less than twice its width. A uniform frame of 2in surrounds the picture. The area of the picture and frame together is 352in^2. Find the dimensions of

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Question 232979: A picture’s length is 6in less than twice its width. A uniform frame of 2in surrounds the picture. The area of the picture and frame together is 352in^2. Find the dimensions of the picture.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A picture’s length is 6in less than twice its width.
A uniform frame of 2in surrounds the picture.
The area of the picture and frame together is 352in^2.
Find the dimensions of the picture.
:
Let x = the picture's width
then
(2x-6) = the picture's length (6" less than twice the width)
;
The 2" frame adds 4" to each dimension, therefore
x + 4 = overall width
and
(2x-6) + 4 = (2x-2) = overall length
;
The overall area equation
(x+4)*(2x-2) = 352
FOIL
2x^2 - 2x + 8x- 8 = 352
:
2x^2 + 6x - 8 - 352 = 0
:
2x^2 + 6x - 360 = 0
Factors to
(2x+30)(x-12) = 0
:
Positive solution
x = 12" is the width of the picture
and
2(12)-6 = 18" is the length of the picture
;
:
Check solutions by finding the total area:
(18+4) * (12+4) = 352