document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=741277>Question 741277</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi! I need help on this word problem (It has to do with Linear systems but i ant even figure out how to set it up)\r<BR>\n" );
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document.write( "A chemist needs 12 liters of a 25% acid solution. The solution is to be mixed from three solutions whose concentrations are 10%, 20%, and 50%. How many liters of each solution will satisfy each condition? \r<BR>\n" );
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document.write( "(a) Use 2 liters of the 50% solution.<BR>\n" );
document.write( "10% solution =___L<BR>\n" );
document.write( "20% solution =___L\r<BR>\n" );
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document.write( "(b) Use as little as possible of the 50% solution.<BR>\n" );
document.write( "10% solution  =  	 ___ L<BR>\n" );
document.write( "20% solution  =  	  ___L<BR>\n" );
document.write( "50% solution  =  	  ___L\r<BR>\n" );
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document.write( "(c) Use as much as possible of the 50% solution.<BR>\n" );
document.write( "10% solution    	  ___L<BR>\n" );
document.write( "20% solution    	  ___L<BR>\n" );
document.write( "50% solution    	  ___L\r<BR>\n" );
document.write( "<BR>\n" );
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document.write( "A little help on just setting it up would even be enough. Thank you!<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #452067 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=452067\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A chemist needs 12 liters of a 25% acid solution.<BR>\n" );
document.write( " The solution is to be mixed from three solutions whose concentrations are 10%, 20%, and 50%.<BR>\n" );
document.write( " How many liters of each solution will satisfy each condition?<BR>\n" );
document.write( "let a = amt of 10% solution<BR>\n" );
document.write( "let b = amt of 20%<BR>\n" );
document.write( "let c = amt of 50%<BR>\n" );
document.write( ":<BR>\n" );
document.write( "(a) Use 2 liters of the 50% solution.<BR>\n" );
document.write( "Then a + b = 10, therefore:<BR>\n" );
document.write( "a = (10-b)<BR>\n" );
document.write( ".10a + .20b + .50(2) = .25(12)<BR>\n" );
document.write( ".10a + .20b + 1 = 3<BR>\n" );
document.write( ".10a + .20b = 3 - 1<BR>\n" );
document.write( ".10a + .20b = 2<BR>\n" );
document.write( ".10(10-b) + .20b = 2<BR>\n" );
document.write( "1 - .1b + .2b = 2<BR>\n" );
document.write( ".1b = 2-1<BR>\n" );
document.write( ".1b = 1<BR>\n" );
document.write( "b = 10 liters of 20% solution<BR>\n" );
document.write( "therefore<BR>\n" );
document.write( "10% solution =_0_L<BR>\n" );
document.write( "20% solution =_10_L<BR>\n" );
document.write( "Check this: <BR>\n" );
document.write( ".10(0) + .20(10) + .50(2) = .25(12)<BR>\n" );
document.write( "0 + 2 + 1 = 3<BR>\n" );
document.write( ";<BR>\n" );
document.write( "(b) Use as little as possible of the 50% solution.<BR>\n" );
document.write( "Then use 0 liters of 10% solution<BR>\n" );
document.write( ".20b + .50c = .25(12)<BR>\n" );
document.write( ".20b + .50c = 3<BR>\n" );
document.write( "replace b with (12-c)<BR>\n" );
document.write( ".2(12-c) + .50c = 3<BR>\n" );
document.write( "2.4 - .2c + .5c = 3<BR>\n" );
document.write( ".3c = 3 - 2.4<BR>\n" );
document.write( ".3c = .6<BR>\n" );
document.write( "c = .6/.3<BR>\n" );
document.write( "c = 2 liters of 50% is minimum<BR>\n" );
document.write( "10% solution = _0_ L<BR>\n" );
document.write( "20% solution = _10_L<BR>\n" );
document.write( "50% solution = _2_L<BR>\n" );
document.write( ":<BR>\n" );
document.write( "(c) Use as much as possible of the 50% solution.<BR>\n" );
document.write( "Use 0 liters of 20% solution then<BR>\n" );
document.write( ".10a + .50c = .25(12)<BR>\n" );
document.write( ".10(12-c) + .50c = 3<BR>\n" );
document.write( "1.2 - .1c + .5c = 3<BR>\n" );
document.write( ".4c = 3 - 1.2<BR>\n" );
document.write( ".4c = 1.8<BR>\n" );
document.write( "c = 1.8/.4<BR>\n" );
document.write( "c = 4.5 liters of 50% is max<BR>\n" );
document.write( "10% solution _7.5__L<BR>\n" );
document.write( "20% solution __ 0__L<BR>\n" );
document.write( "50% solution _4.5__L  </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );