document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=891033>Question 891033</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A rectangle is 5 yards longer in its length than its width.  Its perimeter is 38 feet find the dimensions of the rectangle. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #539492 by </B><A HREF=/tutors/aboutme.mpl?userid=JulietG><B>JulietG(1812)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=JulietG><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=JulietG><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=539492\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> P = 2L + 2W<BR>\n" );
document.write( "38 = 2(W+5) + 2W<BR>\n" );
document.write( "38 = 2W + 10 + 2W<BR>\n" );
document.write( "38 = 4W + 10<BR>\n" );
document.write( "28 = 4W<BR>\n" );
document.write( "7 = W<BR>\n" );
document.write( "If W = 7, then L = 12<BR>\n" );
document.write( "P=(2*12)+(2*7)<BR>\n" );
document.write( "P = 24 + 14<BR>\n" );
document.write( "38 = 38\r<BR>\n" );
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document.write( "EEK!  I did not catch that this is feet/yards.  You'll need to make the conversion first.\r<BR>\n" );
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document.write( "P = 2L + 2W<BR>\n" );
document.write( "38 = 2(W+15) + 2W (5 yards = 15 feet!)<BR>\n" );
document.write( "38 = 2W + 30 + 2W<BR>\n" );
document.write( "38 = 4W + 30<BR>\n" );
document.write( "8 = 4W<BR>\n" );
document.write( "2 = W\r<BR>\n" );
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document.write( "If the width is 2, then the length is 17<BR>\n" );
document.write( "P=(2*17) + (2*2)<BR>\n" );
document.write( "P=34+4<BR>\n" );
document.write( "38=38 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );