document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1021689>Question 1021689</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The width of a rectangle is 2yd less than six times the length.The perimeter of a rectangle is 92yd.Find the dimensions. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #637440 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=637440\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Perimeter P = 2L + 2W = 92.<BR>\n" );
document.write( "We also have W = 6L - 2.<BR>\n" );
document.write( "Now substitute into the first equations and get<BR>\n" );
document.write( "2L + 2(6L - 2) = 92<BR>\n" );
document.write( "2L + 12L - 4 = 92<BR>\n" );
document.write( "14L - 4 = 92<BR>\n" );
document.write( "14L = 96<BR>\n" );
document.write( "L = 96/14 = 6 6/7 yd<BR>\n" );
document.write( "W = 6(6 6/7) - 2 = 39 1/7 yd<BR>\n" );
document.write( "The dimensions are 39 1/7 yds by 6 6/7 yds. </SPAN>\n" );
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