document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=389747>Question 389747</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The length of a rectangle is 3 more than it's width. If the perimiter is 34, what is the length? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #276274 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=276274\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the width be w and the length be 3+w. The perimeter is 34, so\r<BR>\n" );
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document.write( "2(w + (3+w)) = 34. The left-hand side simplifies to 2(3 + 2w) so\r<BR>\n" );
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document.write( "2(3 + 2w) = 34\r<BR>\n" );
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document.write( "3 + 2w = 17\r<BR>\n" );
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document.write( "w = 7, 3+w = 10 --> Length = 10\r<BR>\n" );
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document.write( "Checking, 2(7 + 10) = 34\r<BR>\n" );
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