document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=858573>Question 858573</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1.	The perimeter of a rectangle is 38 inches. If the length of the rectangle is six inches less than four times the width, find the area 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 #517259 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=517259\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Perimeter = 2L + 2W<BR>\n" );
document.write( "2L + 2W = 38<BR>\n" );
document.write( "L = 4W - 6 [the length of the rectangle is six inches less than four times the width]<BR>\n" );
document.write( "Substitute the known value of L from the second equation into the first.<BR>\n" );
document.write( "2(4W-6) + 2W = 38<BR>\n" );
document.write( "Distribute<BR>\n" );
document.write( "8W-12+2W = 38<BR>\n" );
document.write( "Add the Ws<BR>\n" );
document.write( "10W - 12 = 38<BR>\n" );
document.write( "Add 12 to each side<BR>\n" );
document.write( "10W = 50<BR>\n" );
document.write( "Divide each side by 10<BR>\n" );
document.write( "W = 5<BR>\n" );
document.write( ".<BR>\n" );
document.write( "If W is 5, then L =(4*5)-6<BR>\n" );
document.write( "L = 20-6<BR>\n" );
document.write( "L = 14<BR>\n" );
document.write( ".<BR>\n" );
document.write( "To find the area, multiply the length times the width.<BR>\n" );
document.write( "5 * 14 = 70 square inches </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );