document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=499343>Question 499343</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The manager of a coffee shop has one type of coffee that sells for $6 per pound and another type that sells for $13 per pound. The manager wishes to mix 30 pounds of the $13 coffee to get a mixture that will sell for $9 per pound. How many pounds of the $6 coffee should be used? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #337598 by </B><A HREF=/tutors/aboutme.mpl?userid=oberobic><B>oberobic(2304)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=oberobic><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=337598\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x = lb of one coffee<BR>\n" );
document.write( "13x = value of that coffee<BR>\n" );
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document.write( "30-x = lb of other coffee<BR>\n" );
document.write( "6(30-x) = value of that coffee<BR>\n" );
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document.write( "30 = lb of coffee in final mix<BR>\n" );
document.write( "9(30) = 270 = value of the final mix<BR>\n" );
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document.write( "13x + 6(30-x) = 270<BR>\n" );
document.write( "13x + 180 -6x = 270<BR>\n" );
document.write( "7x = 90<BR>\n" );
document.write( "x = 90/7 lb<BR>\n" );
document.write( "x = 12 6/7 lb of the more expensive coffee<BR>\n" );
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document.write( "That means you will add<BR>\n" );
document.write( "30-x = 17 1/7 lb of the cheaper coffee<BR>\n" );
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