document.write( "Question 131750: The lenght of a rectangle is 3 cm greater than the width. If each dimension is increased by 2cm, the area is increased by 26 cm squared. Find the orginal dimensions of the rectangle.\r
\n" ); document.write( "\n" ); document.write( "Please send the problem in linear form to Kathy.Christensen@wowway.com\r
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Algebra.Com's Answer #96180 by rapaljer(4671)\"\" \"About 
You can put this solution on YOUR website!
Let x= width
\n" ); document.write( "x+3= length
\n" ); document.write( "x(x+3) = original area\r
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\n" ); document.write( "\n" ); document.write( "x+2= increased width
\n" ); document.write( "x+3+2=x+5= increased length
\n" ); document.write( "(x+2)(x+5)= larger area
\n" ); document.write( "x^2+7x+10 = larger area\r
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\n" ); document.write( "\n" ); document.write( "Original area + 26 = Larger area
\n" ); document.write( "\"x%5E2%2B3x%2B26=x%5E2%2B7x%2B10\"\r
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\n" ); document.write( "\n" ); document.write( "Subtract \"x%5E2\" from each side
\n" ); document.write( "\"3x%2B26=7x%2B10\"
\n" ); document.write( "\"-4x=-16\"
\n" ); document.write( "\"x=4+cm\" width
\n" ); document.write( "\"x%2B3=7+cm\" length\r
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\n" ); document.write( "\n" ); document.write( "Original rectangle= 4x7=28 sq cm\r
\n" ); document.write( "\n" ); document.write( "Increased rectangle = 6x9=54 sq cm\r
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\n" ); document.write( "\n" ); document.write( "Check: Difference in the area is 26 sq cm.\r
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\n" ); document.write( "\n" ); document.write( "R^2
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