document.write( "Question 744252: A rectangle is 5m longer than it is width . If the length is shortened by 2 meters and the width is increased by 1 m. The area will remain the same. Find the length and width.\r
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Algebra.Com's Answer #453233 by MathTherapy(10552)\"\" \"About 
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\n" ); document.write( "A rectangle is 5m longer than it is width . If the length is shortened by 2 meters and the width is increased by 1 m. The area will remain the same. Find the length and width.\r
\n" ); document.write( "\n" ); document.write( "Your answer would be highly appreciated. Thank you!\r
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\n" ); document.write( "\n" ); document.write( "Let original width be W\r
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\n" ); document.write( "\n" ); document.write( "Then length = W + 5\r
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\n" ); document.write( "\n" ); document.write( "Original area = W(W + 5), or \"W%5E2+%2B+5W\"\r
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\n" ); document.write( "\n" ); document.write( "Shortened by 2 meters, the new length is: W + 5 - 2, or W + 3\r
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\n" ); document.write( "\n" ); document.write( "Increased by 1, the new width is: W + 1\r
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\n" ); document.write( "\n" ); document.write( "New area = (W + 3)(W + 1), or \"W%5E2+%2B+4W+%2B+3\"\r
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\n" ); document.write( "\n" ); document.write( "Since area remains the same, then: \"W%5E2+%2B+5W+=+W%5E2+%2B+4W+%2B+3\"\r
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\n" ); document.write( "\n" ); document.write( "\"W%5E2+-+W%5E2+%2B+5W+-+4W+=+3\" \r
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\n" ); document.write( "\n" ); document.write( "W, or original width = \"highlight_green%283m%29\"\r
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\n" ); document.write( "\n" ); document.write( "Original length = 3 + 5, or \"highlight_green%288m%29\"\r
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\n" ); document.write( "\n" ); document.write( "You can do the check!!\r
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\n" ); document.write( "\n" ); document.write( "Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
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