document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=371646>Question 371646</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A rectangle is twice as long as it is wide. If its length and width are each reduced by 1 inch, the area of the rectangle is reduced by 14 square inches. Find the dimensions 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 #264710 by </B><A HREF=/tutors/aboutme.mpl?userid=amoresroy><B>amoresroy(361)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=amoresroy><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=amoresroy><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=264710\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A rectangle is twice as long as it is wide. If its length and width are each reduced by 1 inch, the area of the rectangle is reduced by 14 square inches. Find the dimensions of the original rectangle.\r<BR>\n" );
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document.write( "Let x = the width of the rectangle<BR>\n" );
document.write( "   2x = the length of the rectangle\r<BR>\n" );
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document.write( "(x-1)(2x-1) = x(2x) - 14\r<BR>\n" );
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document.write( "Solve for x<BR>\n" );
document.write( "2x^2-x-2x+1 = 2x^2 -14<BR>\n" );
document.write( "Combine like terms<BR>\n" );
document.write( "        -3x = -15<BR>\n" );
document.write( "          x = 5<BR>\n" );
document.write( "         2x = 10\r<BR>\n" );
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document.write( "The dimensions of the original rectabngle are<BR>\n" );
document.write( "width = 5 inches<BR>\n" );
document.write( "length= 10 inches </SPAN>\n" );
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