document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1004583>Question 1004583</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A person wishes to mix coffee worth $12 per lb with coffee worth $6 per lb to get 90 lb of a mixture worth $8 per lb. How many pounds of the $12 and the $6 coffees will be needed? \r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #621018 by </B><A HREF=/tutors/aboutme.mpl?userid=addingup><B>addingup(3677)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=addingup><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=addingup><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=621018\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x+y= 90<BR>\n" );
document.write( "Subtract x on both sides:<BR>\n" );
document.write( "y= 90-x In the next equation we will substitute y with this value:<BR>\n" );
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document.write( "We want to get 90lbs of coffee at $8 per pound, we want 90*8= 720:<BR>\n" );
document.write( "12x+6y= 8(90) In this equation, substitute the value of y with 90-y:<BR>\n" );
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document.write( "12x+6(90-x)= 720<BR>\n" );
document.write( "12x+540-6x= 720<BR>\n" );
document.write( "6x= 180<BR>\n" );
document.write( "x= 30<BR>\n" );
document.write( "You need 30lb of $12 coffee with 90-30= 60 lbs of the $6 coffee </SPAN>\n" );
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