document.write( "Question 218757: The length of a picture frame is 3 in. greater than the width. The perimiter is less than 52in. Describe the dimentions of the frame.
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Algebra.Com's Answer #164692 by MathTherapy(10551)\"\" \"About 
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The length of a picture frame is 3 in. greater than the width. The perimiter is less than 52in. Describe the dimentions of the frame.\r
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\n" ); document.write( "\n" ); document.write( "Let the width of the frame be W\r
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\n" ); document.write( "\n" ); document.write( "Since its length is 3 inches greater than its width, then its length = W + 3\r
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\n" ); document.write( "\n" ); document.write( "Since the perimeter is less than 52 inches, then we'll have:\r
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\n" ); document.write( "\n" ); document.write( "2(W) + 2(W + 3) < 52\r
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\n" ); document.write( "\n" ); document.write( "2W + 4W + 6 < 52\r
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\n" ); document.write( "\n" ); document.write( "6W + 6 < 52\r
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\n" ); document.write( "\n" ); document.write( "6W < 46\r
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\n" ); document.write( "\n" ); document.write( "W <\"46%2F6\" or <\"23%2F2\", or <\"highlight_green%2811.5%29\"\r
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\n" ); document.write( "\n" ); document.write( "Since width < 11.5 inches, and length is 3 more then width, then length < (11.5 + 3), or < \"highlight_green%2814.5%29\"\r
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\n" ); document.write( "\n" ); document.write( "Since width, or W < 11.5, and since length, or L < 14.5, then let width, or W = 11, and length, or L = 14\r
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\n" ); document.write( "\n" ); document.write( "2W + 2L < Perimeter\r
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\n" ); document.write( "\n" ); document.write( "2(11) + 2(14) < 52\r
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\n" ); document.write( "\n" ); document.write( "22 + 28 < 52\r
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\n" ); document.write( "\n" ); document.write( "50 < 52 (TRUE)
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