document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #172069 by ankor@dixie-net.com(22740)\"\" \"About 
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A picture’s length is 6in less than twice its width.
\n" ); document.write( "A uniform frame of 2in surrounds the picture.
\n" ); document.write( "The area of the picture and frame together is 352in^2.
\n" ); document.write( "Find the dimensions of the picture.
\n" ); document.write( ":
\n" ); document.write( "Let x = the picture's width
\n" ); document.write( "then
\n" ); document.write( "(2x-6) = the picture's length (6\" less than twice the width)
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\n" ); document.write( "The 2\" frame adds 4\" to each dimension, therefore
\n" ); document.write( "x + 4 = overall width
\n" ); document.write( "and
\n" ); document.write( "(2x-6) + 4 = (2x-2) = overall length
\n" ); document.write( ";
\n" ); document.write( "The overall area equation
\n" ); document.write( "(x+4)*(2x-2) = 352
\n" ); document.write( "FOIL
\n" ); document.write( "2x^2 - 2x + 8x- 8 = 352
\n" ); document.write( ":
\n" ); document.write( "2x^2 + 6x - 8 - 352 = 0
\n" ); document.write( ":
\n" ); document.write( "2x^2 + 6x - 360 = 0
\n" ); document.write( "Factors to
\n" ); document.write( "(2x+30)(x-12) = 0
\n" ); document.write( ":
\n" ); document.write( "Positive solution
\n" ); document.write( "x = 12\" is the width of the picture
\n" ); document.write( "and
\n" ); document.write( "2(12)-6 = 18\" is the length of the picture
\n" ); document.write( ";
\n" ); document.write( ":
\n" ); document.write( "Check solutions by finding the total area:
\n" ); document.write( "(18+4) * (12+4) = 352
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