document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=993666>Question 993666</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The store manager decides to make 40lbs of blend of coffees that he plans to sell for 10 dollars a lb. He mixes some coffee that sells for 9 dollars a lb with coffee that sells for 12 dollars a lb. How much of each should he use? (Hint- one coffee is x and the other coffee of the rest of the mixture so use 40 - x for the amount)\r<BR>\n" );
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document.write( "I already calculated that 40 lbs of coffee costing 10 dollars a lb would be 400 dollars. Now I need help on how to calculate how many pounds adding up to 40lbs. I would use of each that would bring me to 400 dollars total.<BR>\n" );
document.write( "Thank you! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #612893 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=612893\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x is the amount of coffee at 9 dollars a pound.<BR>\n" );
document.write( "y is the amount of coffee at 12 dollars a pound.\r<BR>\n" );
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document.write( "you have two equations:\r<BR>\n" );
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document.write( "x + y = 40\r<BR>\n" );
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document.write( "9x + 12y = 400\r<BR>\n" );
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document.write( "solve these equations simultaneously, and you will get:\r<BR>\n" );
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document.write( "x = 26.67<BR>\n" );
document.write( "y = 12.33\r<BR>\n" );
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document.write( "x + y = 40 becomes 26.67 + 13.33 = 40\r<BR>\n" );
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document.write( "9x + 12y = 400 becomes 9 * 26.67 + 12 * 13.33 which becomes 240.03 + 159.96 which becomes 299.99.\r<BR>\n" );
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document.write( "the difference from 400 is due to rounding.\r<BR>\n" );
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document.write( "26.67 is really 26.6666666666666........\r<BR>\n" );
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document.write( "13.33 is really 13.333333333333333........\r<BR>\n" );
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document.write( "to get this solution, you need to be able to solve the system of equations simultaneously.\r<BR>\n" );
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document.write( "here's a reference on how to do that.\r<BR>\n" );
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document.write( "<a  rel=nofollow HREF= \"http://www.regentsprep.org/regents/math/algebra/ae3/indexAE3.htm\" target = \"_blank\">http://www.regentsprep.org/regents/math/algebra/ae3/indexAE3.htm</a>\r<BR>\n" );
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document.write( "i'll use elimination method.\r<BR>\n" );
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document.write( "your two equations are:\r<BR>\n" );
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document.write( "x + y = 40<BR>\n" );
document.write( "9x + 12y = 400\r<BR>\n" );
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document.write( "multiply both sides of the first equation by 9 to get:\r<BR>\n" );
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document.write( "9x + 9y = 360<BR>\n" );
document.write( "9x + 12y = 400\r<BR>\n" );
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document.write( "subtract the first equation from the second equation to get:\r<BR>\n" );
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document.write( "3y = 40\r<BR>\n" );
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document.write( "divide both sides of this equation by 3 to get:\r<BR>\n" );
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document.write( "y = 13.33.....<BR>\n" );
document.write( "this rounds to 13.33\r<BR>\n" );
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document.write( "use that value of y to solve for x in either equation to get:\r<BR>\n" );
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document.write( "x = 26.66.....<BR>\n" );
document.write( "this rounds to 26.67\r<BR>\n" );
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document.write( " </SPAN>\n" );
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