document.write( "Question 457513: The perimeter of a rectangle is 122 inches. The length exceeds the withd by 25 inches. find the length and width.\r
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document.write( "Ok. I think I start out by
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document.write( "2l + 2w = 122
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document.write( "w+25=1\r
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document.write( "This is confusing, can someone help? \n" );
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Algebra.Com's Answer #313913 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! The perimeter of a rectangle is 122 inches. The length exceeds the withd by 25 inches. find the length and width. \n" ); document.write( "Ok. I think I start out by \n" ); document.write( "2l + 2w = 122 \n" ); document.write( "w+25=1 \n" ); document.write( "----------------------- \n" ); document.write( "Let the width be \"x\" \n" ); document.write( "Then length = \"x+25\" \n" ); document.write( "---- \n" ); document.write( "Perimeter = 2(width + length) \n" ); document.write( "122 = 2(x + x+25) \n" ); document.write( "61 = 2x+25 \n" ); document.write( "2x = 36 \n" ); document.write( "x = 18 inches (width) \n" ); document.write( "x+25 = 43 inches (length) \n" ); document.write( "============================= \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |