document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=400904>Question 400904</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Here is the problem:\r<BR>\n" );
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document.write( "The cooling system of a car has a capacity of 15 liters. If the system is currently filled with a mixture that is 40% antifreeze, how much of this mixture should be drained and replaced with pure antifreeze so that the system is filled with a solution of 50% antifreeze?\r<BR>\n" );
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document.write( "I have to 1)define the variables, 2)set up the equation, 3)solve the equation, and 4)answer the question. I am having problems with 1 and 2, defining the variables and setting up the equation. Once I have those I can solve them and answer the questions.\r<BR>\n" );
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document.write( "Please help! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #283726 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=283726\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The cooling system of a car has a capacity of 15 liters.<BR>\n" );
document.write( " If the system is currently filled with a mixture that is 40% antifreeze,<BR>\n" );
document.write( " how much of this mixture should be drained and replaced with pure antifreeze<BR>\n" );
document.write( " so that the system is filled with a solution of 50% antifreeze?<BR>\n" );
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document.write( "Let a = amt of pure antifreeze bring the mixture up to 50%<BR>\n" );
document.write( "Let a = also amt of existing mixture to be removed<BR>\n" );
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document.write( "The equation (pure antifreeze would be 1a)<BR>\n" );
document.write( ".40(15-a) + a = .50(15)<BR>\n" );
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