document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=939643>Question 939643</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the length of a new rectangular playing field is 5 yards longer than double the width. If the perimeter of the rectangular playing field is 334 yards, what are its dimensions </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #572610 by </B><A HREF=/tutors/aboutme.mpl?userid=laoman><B>laoman(51)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=laoman><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=laoman><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=572610\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the length of the field be l and its width be w<BR>\n" );
document.write( "From the first statement, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=l+=+2%2Aw+%2B+5 ALIGN=MIDDLE  ALT=\"l+=+2%2Aw+%2B+5\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( " Perimeter of the field is 334 yards<BR>\n" );
document.write( "And perimeter of rectangle = 2*l +2*w = 334\r<BR>\n" );
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document.write( "Now substituting for l, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%2A+%282%2Aw+%2B+5%29+%2B2%2Aw+=334 ALIGN=MIDDLE  ALT=\"2%2A+%282%2Aw+%2B+5%29+%2B2%2Aw+=334\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "4*w+10+2*w = 334<BR>\n" );
document.write( "6*w = 334-10<BR>\n" );
document.write( "6*w = 324,<BR>\n" );
document.write( "w = 324/6 = 54<BR>\n" );
document.write( "Also, l=2*w + 5= 2*54+5= 113<BR>\n" );
document.write( "Therefore, the dimensions are (l,b) = (113,54) </SPAN>\n" );
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