document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=415260>Question 415260</A>: <I><SPAN STYLE=\"word-break: break-all;\"> what is the length and width of a triangular sign if the length exceeds the width by 15 inches and the perimeter is 158 inches? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #291175 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=291175\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Since the perimeter is the sum of all the sides, 2l + w = 158 (1)<BR>\n" );
document.write( "Also l - w = 15 (2)<BR>\n" );
document.write( "Solving for l in (2) and substituting in (1) we have 2*(w+15) + w = 158<BR>\n" );
document.write( "This reduces to 3w + 30 = 158, or w = 128/3 = 42 2/3<BR>\n" );
document.write( "Putting this into (2) we have l - 42 2/3 = 15 or l = 57 2/3. </SPAN>\n" );
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