document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=164763>Question 164763</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the perimeter of a rectangle garden is 76 ft. the length is six feet less than three times the width. find the dimensions of the garden </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #121419 by </B><A HREF=/tutors/aboutme.mpl?userid=Mathtut><B>Mathtut(3670)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Mathtut><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=121419\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The Perimeter P = 2(length(L)) + 2(width(W)) Now we are given that L = 3W-6 and that the perimeter =76ft so plugging in known values we have 76 = 2(3W-6)+ 2W    solving for the width W...   76 = 6W-12+2W simplifying further by adding 12 to each side and combining like terms we arrive at 88 = 8W.  divide each side by 8 and you conclude that W=11   Now since L=3W-6 we have 3(11)-6 which equals 27\r<BR>\n" );
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document.write( "Answer W=11<BR>\n" );
document.write( "       L=27  </SPAN>\n" );
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