document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=339382>Question 339382</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A plane was originally flying at an altitude of x feet when it ascended 2000 feet and then descended 5000 feet. If the plane's altitude after these two chages was 1/3 its original altitude, then the solution of which of the following equations gives the plane's original altitude, in feet?\r<BR>\n" );
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document.write( "(A) x+2000=1/3(x-3000)<BR>\n" );
document.write( "(B) 1/3(x-3000)=x<BR>\n" );
document.write( "(C) x+3000=1/3x<BR>\n" );
document.write( "(D0 x-7000=1/3x<BR>\n" );
document.write( "(E) x-3000=1/3x\r<BR>\n" );
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document.write( "Could please provide an explanation of the above problem.\r<BR>\n" );
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document.write( "Thanks </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #243214 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=243214\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  A plane was originally flying at an altitude of x feet when it ascended 2000 feet and then descended 5000 feet. If the plane's altitude after these two chages was 1/3 its original altitude, then the solution of which of the following equations gives the plane's original altitude, in feet?<BR>\n" );
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document.write( "start: x<BR>\n" );
document.write( "ascend 2000: x+2000<BR>\n" );
document.write( "decend 5000: x+2000-5000<BR>\n" );
document.write( "result: x-3000<BR>\n" );
document.write( "Equation<BR>\n" );
document.write( "x-3000 = (1/3)x<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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document.write( "(A) x+2000=1/3(x-3000)<BR>\n" );
document.write( "(B) 1/3(x-3000)=x<BR>\n" );
document.write( "(C) x+3000=1/3x<BR>\n" );
document.write( "(D0 x-7000=1/3x<BR>\n" );
document.write( "(E) x-3000=1/3x <BR>\n" );
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