document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=45162>Question 45162</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A coffee merchant has coffee beans that sell for $9 per poind and $12 per point. The two types are to be mixed to create 100lb of a mixture that will sell for $11.25 per pound.How much of each type of bean should be used in the mixture? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #29967 by </B><A HREF=/tutors/aboutme.mpl?userid=Earlsdon><B>Earlsdon(6294)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Earlsdon><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=29967\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x = the required number of pounds of $9.00/lb coffee beans and 100-x = the required number od $12.00/lb coffee beans.  You can write the equation from this.\r<BR>\n" );
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document.write( "x($9.00) + (100-x)($12.00) = 100($11.25) Simplify and solve for x.<BR>\n" );
document.write( "9x + 1200-12x = 1125 Combine the x-terms and subtract 1200 from both sides.<BR>\n" );
document.write( "-3x = -75 Divide both sides by -3<BR>\n" );
document.write( "x = 25<BR>\n" );
document.write( "100-x = 75 \r<BR>\n" );
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document.write( "The mixture should contain 25 lbs of $9.00/lb coffee beans and 75 lbs of $12.00/lb coffee beans.\r<BR>\n" );
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document.write( "Check:\r<BR>\n" );
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document.write( "25 lbs($9.00/lb) + 75 lbs($12.00/lb) = $225.00 + $900.00 = $1,125.00<BR>\n" );
document.write( "100 lbs($11.25/lb) = $1,125.00 </SPAN>\n" );
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