document.write( "Question 836353: (Can someone please help me with this question, thanks)
\n" ); document.write( "The perimeter of a playing field for a certain sport is 326 ft. The length is 45 ft. longer than the width. Find the dimensions.\r
\n" ); document.write( "\n" ); document.write( "The width of the playing field is ? ft. \r
\n" ); document.write( "\n" ); document.write( "The length of the playing field is ? ft.
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Algebra.Com's Answer #504151 by LinnW(1048)\"\" \"About 
You can put this solution on YOUR website!
Set W = width, L = length
\n" ); document.write( "2W + 2L = perimeter
\n" ); document.write( "2W + 2L = 326 for our problem
\n" ); document.write( "L = W + 45
\n" ); document.write( "Substitute (W + 45) for L in 2W + 2L = 326
\n" ); document.write( "2W + 2(W + 45) = 326
\n" ); document.write( "2W + 2W + 90 = 326
\n" ); document.write( "4W + 90 = 326
\n" ); document.write( "add -90 to each side
\n" ); document.write( "4W = 236
\n" ); document.write( "divide each side by 4
\n" ); document.write( "W = 59
\n" ); document.write( "Since L = W + 45
\n" ); document.write( "L = 59 + 45 = 104
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