document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1058106>Question 1058106</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the length of a rectangle is twice its width .if the width is reduced by 1cm and the length is also reduced by 2cm, the area Will be  15cm^2.find the  dimension 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 #673134 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=673134\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the length of a rectangle is twice its width .if the width is reduced by 1cm and the length is also reduced by 2cm, the area Will be 15cm^2.find the dimension of the original rectangle <BR>\n" );
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document.write( "Equations:<BR>\n" );
document.write( "L = 2*W<BR>\n" );
document.write( "(W-1)(L-2) = 15<BR>\n" );
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document.write( "Substitute for \"L\" and solve for \"W\"::<BR>\n" );
document.write( "(W-1)(2W-2) = 15<BR>\n" );
document.write( "2W^2 -4W -13 = 0<BR>\n" );
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document.write( "Width = 3.74 cm<BR>\n" );
document.write( "Length = 2*W = 7.48 cm<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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