document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=796753>Question 796753</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find how many quarts of 6% butterfat milk and 2% butterfat milk should be mixed to yield 40 quarts of 3% butterfat milk </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #481552 by </B><A HREF=/tutors/aboutme.mpl?userid=wilft1><B>wilft1(217)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=wilft1><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=wilft1><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=481552\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> two equations for this\r<BR>\n" );
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document.write( "x + y = 40<BR>\n" );
document.write( "number of liters of each type that add up to 40 quarts<BR>\n" );
document.write( "6x + 2y = 3<BR>\n" );
document.write( "how much of each percentage type to make the 3 percent\r<BR>\n" );
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document.write( "start off by multiplying 3 by 40 and putting that answer in the bottom equation\r<BR>\n" );
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document.write( "x  + y  = 40<BR>\n" );
document.write( "6x + 2y = 120\r<BR>\n" );
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document.write( "now, we are going to solve this using the addition method, we want to get rid of the y then solve for x, so lets multiply the top line by -2\r<BR>\n" );
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document.write( "-2x -2y = -80<BR>\n" );
document.write( "6x  + 2y = 120\r<BR>\n" );
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document.write( "add the two lines together\r<BR>\n" );
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document.write( "4x = 40\r<BR>\n" );
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document.write( "divide both sides by 4\r<BR>\n" );
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document.write( "x = 10\r<BR>\n" );
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document.write( "we now know the value of x, plug that number into our original equation and solve for y\r<BR>\n" );
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document.write( "x + y = 40<BR>\n" );
document.write( "6(10) + 2y = 120<BR>\n" );
document.write( ".....................<BR>\n" );
document.write( "x + y = 40<BR>\n" );
document.write( "60  + 2y  = 120<BR>\n" );
document.write( "..................<BR>\n" );
document.write( "x  + y  = 40<BR>\n" );
document.write( "2y  = 60<BR>\n" );
document.write( ".................<BR>\n" );
document.write( "x  + y  = 40<BR>\n" );
document.write( "y  = 30<BR>\n" );
document.write( ".............\r<BR>\n" );
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document.write( "x = 1 and y = 30\r<BR>\n" );
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document.write( "we now know that it will take 10 quarts of 6 percent and 30 quarts of 2 percent to make the 3 percent solution </SPAN>\n" );
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