document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=511615>Question 511615</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I am taking an Algebra course in school right now and I am having a bit of trouble with Word problems. Thank you in advance for your help. The question I have tonight is the perimeter of a rectangle is 194 inches. The length exceeds the width by 73 inches. Find the length and the width. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #342297 by </B><A HREF=/tutors/aboutme.mpl?userid=oberobic><B>oberobic(2304)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=oberobic><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=342297\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> P = perimeter<BR>\n" );
document.write( "L = length<BR>\n" );
document.write( "W = width<BR>\n" );
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document.write( "P = 2(L+W) = 2L + 2W<BR>\n" );
document.write( "P = 194<BR>\n" );
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document.write( "The length is 73 more than width.<BR>\n" );
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document.write( "L = W+73<BR>\n" );
document.write( "or<BR>\n" );
document.write( "L-73 = W<BR>\n" );
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document.write( "substitute L = W+73<BR>\n" );
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document.write( "2(L+W) = P<BR>\n" );
document.write( "2(W+73 +W) = 194<BR>\n" );
document.write( "2W + 146  + 2W = 194<BR>\n" );
document.write( "4W = 48<BR>\n" );
document.write( "W = 12<BR>\n" );
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document.write( "L = W+73<BR>\n" );
document.write( "L = 12+73<BR>\n" );
document.write( "L = 85<BR>\n" );
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document.write( "Check to see if the perimeter is correct.<BR>\n" );
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document.write( "2(L+W) = ??<BR>\n" );
document.write( "2(85+12) = 2(97) = 194<BR>\n" );
document.write( "Correct.<BR>\n" );
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document.write( "Answer: The width = 12. The length = 85.<BR>\n" );
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document.write( "Done. </SPAN>\n" );
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