document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=167572>Question 167572</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Leon, who is always in a hurry, walked up an escalator, while it was moving, at the rate of one step per second and reached the top in 28 steps. The next day he climbed two steps per second (skipping none), also while it was moving, and reached the top in 40 steps.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #123498 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=123498\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> lets call S the number of steps on the escalator:the escalator is moving \r<BR>\n" );
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document.write( "with Leon, so we must add the number of steps taken by Leon to the # of steps when it is still. d=rt so lets call d in this instance S+28 and S+40.  time is 28 sec in 1st scenario and 20 sec in the 2nd instance(40/2).\r<BR>\n" );
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document.write( "S+28=r(28)<BR>\n" );
document.write( "S+40=r(20)\r<BR>\n" );
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document.write( "solve the system:\r<BR>\n" );
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document.write( "subtract the 2nd equation from the 1st to eliminate the S terms\r<BR>\n" );
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document.write( "-12=8r---->r=-3/2--->now plug r's value into either equation I choose the 2nd equation\r<BR>\n" );
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document.write( "S+40=(-3/2)20<BR>\n" );
document.write( "S+40=-30<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight%28S=70%29 ALIGN=MIDDLE  ALT=\"highlight%28S=70%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >feet on the escalator when still<BR>\n" );
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