document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=890831>Question 890831</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The perimeter of a rectangle is 38.   If its length is 6 less than 4 times its width, find the dimensions of the rectangle.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #539356 by </B><A HREF=/tutors/aboutme.mpl?userid=JulietG><B>JulietG(1812)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=JulietG><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=JulietG><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=539356\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A perimeter is a fence around.  It is 2 lengths + 2 widths.<BR>\n" );
document.write( "L = 4W-6<BR>\n" );
document.write( "P = 2L + 2W<BR>\n" );
document.write( "38 = 2L + 2W<BR>\n" );
document.write( "Take the known value of L and substitute it into that equation.<BR>\n" );
document.write( "38 = 2L + 2W<BR>\n" );
document.write( "38 = 2(4W-6) + 2W<BR>\n" );
document.write( "Distribute<BR>\n" );
document.write( "38 = 8W-12+2W<BR>\n" );
document.write( "Add<BR>\n" );
document.write( "38 = 10W-12<BR>\n" );
document.write( "Add 12 to each side<BR>\n" );
document.write( "50=10W<BR>\n" );
document.write( "Divide each side by 10<BR>\n" );
document.write( "5 = W<BR>\n" );
document.write( ".<BR>\n" );
document.write( "The width is 5.  Therefore, the length is (4*5)-6, or 14.<BR>\n" );
document.write( "Perimeter is 5+5+14+14, which equals 38. </SPAN>\n" );
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