document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=915878>Question 915878</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A grocer wants to mix two kinds of coffee. One kind sells for $2.15 per pound and the other sells for 2.45 per pound. He wants to mix a total of 18 pounds and sell it for $2.30 per pound. How many pounds of each kind should he use in the new mix? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #555801 by </B><A HREF=/tutors/aboutme.mpl?userid=AlgebraLady88><B>AlgebraLady88(44)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=AlgebraLady88><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=AlgebraLady88><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=555801\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This grocer is mixing two kinds of coffee. I always love the smell of coffee! <BR>\n" );
document.write( "We will say that x sells for $2.15 per pound and y sells for $2.45 per pound.\r<BR>\n" );
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document.write( "Altogether there are 18 pounds of coffee he is selling.<BR>\n" );
document.write( "The algebraic equation would be x + y = 18\r<BR>\n" );
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document.write( "At $2.30 per pound, he will make 18 * $2.30 = $ 41.40\r<BR>\n" );
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document.write( "So, how many of the x and y kinds of coffee should he use to make the $2.30 per pound mixture which will net him $41.40?<BR>\n" );
document.write( "The algebraic equation would be  2.15x + 2.45y = 41.40\r<BR>\n" );
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document.write( "We now have two equations :\r<BR>\n" );
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document.write( "x+y = 18<BR>\n" );
document.write( "2.15x + 2.45y = 41.40\r<BR>\n" );
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document.write( "Substituting  x= 18 -y in the second equation gives us\r<BR>\n" );
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document.write( "2.15(18-y) + 2.45y = 41.40<BR>\n" );
document.write( "38.70 - 2.15y + 2.45y = 41.40<BR>\n" );
document.write( "38.70 + 0.3y = 41.40<BR>\n" );
document.write( "0.3y = 41.40 - 38.70<BR>\n" );
document.write( "0.3y= 2.70<BR>\n" );
document.write( "y=9\r<BR>\n" );
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document.write( "If y=9, then x= 9 as well. \r<BR>\n" );
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document.write( "So, he will use 9 pounds of the $2.15 per/lb one and another 9 pounds of the $2.45per/lb one.   </SPAN>\n" );
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