document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1008843>Question 1008843</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the length of a rectangle is 5cm less than twice the width. The area is 75 square cm. Find the length and the width<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #625952 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=625952\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Area A = LW = 75 cm^2.  We also have<BR>\n" );
document.write( "L = 2W - 5<BR>\n" );
document.write( "Substituting this we have<BR>\n" );
document.write( "(2W - 5)W = 75<BR>\n" );
document.write( "2W^2 - 5W - 75 = 0<BR>\n" );
document.write( "(2W - 15)(W + 5) = 0<BR>\n" );
document.write( "so that<BR>\n" );
document.write( "2W = 15<BR>\n" );
document.write( "W = 7.5 cm<BR>\n" );
document.write( "and<BR>\n" );
document.write( "L = 2(7.5) - 5 = 10 cm </SPAN>\n" );
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