document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=13241>Question 13241</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Chelsea sold 2 cars for $6,000 each, making 20% on car#1 & losing 20% on car#2. Did he make money, lose money or break even? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #6632 by </B><A HREF=/tutors/aboutme.mpl?userid=bam878s><B>bam878s(77)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bam878s><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=6632\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Chelsea made a total of 12,000 dollars in sales (6,000 + 6,000)<BR>\n" );
document.write( "For car 1, note he made 20% profit. So the price, call it x, was the price Chelsea paid for the car in the beginning. We have 6000 = x + x(.20).  We can factor out an x on the right hand side of this equation to get 6000 = x(1 + .20).  So to solve for x (the price Chelsea paid for car 1) divide 6000 by 1.20 to get x = 5,000.\r<BR>\n" );
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document.write( "For car 2, note he lost 20% on the car.  Let y be the price Chelsea paid for car 2.  Then 6,000 = y - y(.20).  Factoring out a y on each side yields 6000 = y(1-.20) or 6,000 = y(.80).  To solve for y we divide 6,000 by .80 to get $7,500. \r<BR>\n" );
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document.write( "So the total price Chelsea paid for the cars was x + y = $5,000 + $7,500 = $12,500.  However he only sold the cars for $6,000 each and gained a total of $12,000.  So, in the long run, Chelsea actually lost $500. </SPAN>\n" );
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