document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=366601>Question 366601</A>: <I><SPAN STYLE=\"word-break: break-all;\"> if the dimension of a rectangle are such that the length is 3 inches more than the width. if the length were doubled and the width decrease by 1 inch, the area would be 50in square. what is the length and the width </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #261501 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=261501\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let L = the length of rectangle in inches<BR>\n" );
document.write( "    W = the width of rectangle in inches\r<BR>\n" );
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document.write( "L = W + 3\r<BR>\n" );
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document.write( "2L(W-1) = 50\r<BR>\n" );
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document.write( "Substitute L=W+3 in 2nd equation to express the 2nd equation in terms of W<BR>\n" );
document.write( "2(W+3)(W-1) = 50<BR>\n" );
document.write( "(2W+6)(W-1) = 50<BR>\n" );
document.write( "2W^2-2W+6W-6-50=0<BR>\n" );
document.write( "Combine like terms and divide by 2<BR>\n" );
document.write( "W^2+2W-28=0\r<BR>\n" );
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document.write( "Solve by quadratic formula<BR>\n" );
document.write( "W = -2 +/- [4-4(1)(-28)]/2\r<BR>\n" );
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document.write( "W = 4.385<BR>\n" );
document.write( "L = 4.385 +3 = 7.385 </SPAN>\n" );
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