document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1015916>Question 1015916</A>: <I><SPAN STYLE=\"word-break: break-all;\"> jacob bought 280 blue and red paper cups he used 1/3 of the blue ones and 1/2 of the red ones. If he had an equal number of blue and red cups left over how many cups did he use altogether? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #632357 by </B><A HREF=/tutors/aboutme.mpl?userid=FrankM><B>FrankM(1040)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=FrankM><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=FrankM><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=632357\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> B+R=280\r<BR>\n" );
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document.write( "2/3B=1/2R\r<BR>\n" );
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document.write( "Multiply both side of eq 2 by 6 (the least common multiple)\r<BR>\n" );
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document.write( "4B=3R\r<BR>\n" );
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document.write( "Rearrange eq 1 to   R=280-B\r<BR>\n" );
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document.write( "Now substitute\r<BR>\n" );
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document.write( "4B = 3(280-B)\r<BR>\n" );
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document.write( "4B = 840 - 3B\r<BR>\n" );
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document.write( "7B = 840 \r<BR>\n" );
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document.write( "B = 120\r<BR>\n" );
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document.write( "R = 160\r<BR>\n" );
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document.write( "It adds to 280, and when he used 1/3 of B, 80 are left, the same as 1/2 of R. Done! </SPAN>\n" );
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