document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=158252>Question 158252</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The length of a certain rectangle is 10 meters greater than twice its width. if its length were doubled and the width were halved, the perimeter would be increased by 80 meters. Find the area of the original rectangle. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #116626 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley77><B>checkley77(12844)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley77><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=116626\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> L=2W+10<BR>\n" );
document.write( "W=W<BR>\n" );
document.write( "2(2W+10)+2W=2*2L+2W/2-80<BR>\n" );
document.write( "4W+20+2W=4(2W+10)+W-80<BR>\n" );
document.write( "6W+20=8W+40+W-80<BR>\n" );
document.write( "6W-9W=-40-20<BR>\n" );
document.write( "-3W=-60<BR>\n" );
document.write( "W=-60/-3<BR>\n" );
document.write( "W=20 FOR THE ORIGINAL WIDTH<BR>\n" );
document.write( "L=2*20+10<BR>\n" );
document.write( "L=40+10<BR>\n" );
document.write( "L=50 FOR THE ORIGINAL LENGTH.<BR>\n" );
document.write( "PROOF:<BR>\n" );
document.write( "2(2*20+10)+2*20=4(2*20+10)+2*20/2-80<BR>\n" );
document.write( "2(40+10)+40=4(40+10)+40/2-80<BR>\n" );
document.write( "2*50+40=4*50+20-80<BR>\n" );
document.write( "100+40=200+20-80<BR>\n" );
document.write( "140=220-80<BR>\n" );
document.write( "140=140<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );