document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=98496>Question 98496</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0618386807/algebrahomeworkh>College Algebra and Trigonometry </A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> A radiator contains 6 liters of 25% antifreeze solution. How much should be drained and replaced with pure antifreeze to produce a 33% antifreeze solution?\r<BR>\n" );
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document.write( "So far, I know that the first solution contains 6 liters and I need to multiply that by 25%. Then since we are \"replacing\" pure antifreeze to the mixture that is 100% that I need to multiply to the unknown mixture of x+6 (which is my second solution) so that I can produce a 33% solution. Does this make any sense? I'm not sure how to do this. 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 #71726 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=71726\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A radiator contains 6 liters of 25% antifreeze solution. How much should be drained and replaced with pure antifreeze to produce a 33% antifreeze solution? <BR>\n" );
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document.write( "It looks like they want you to start with 6 and end up with 6.<BR>\n" );
document.write( "We could do it like this:<BR>\n" );
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document.write( "Let x = amt to be drained and replaced<BR>\n" );
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document.write( "The % antifreeze equation:<BR>\n" );
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document.write( ".25(6) - .25(x) + 1.0(x) = .33(6)<BR>\n" );
document.write( "1.5 - .25x + 1x = 1.98<BR>\n" );
document.write( "-.25x + 1x = 1.98 - 1.5<BR>\n" );
document.write( ".75x = .48<BR>\n" );
document.write( "x = .48/.75<BR>\n" );
document.write( "x = .64 liters of pure antifreeze to replace .64 liters of the 25% solution<BR>\n" );
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document.write( "Check our solution for x.<BR>\n" );
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document.write( "25% amt is now: 6 - .64 = 5.36<BR>\n" );
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document.write( ".25(5.36) + 1.0(.64) = .33(6)<BR>\n" );
document.write( "1.34 + .64 = 1.98; confirms our solution<BR>\n" );
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document.write( "Did this help?<BR>\n" );
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