document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=146027>Question 146027</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Sarah wants to mix 400 liters of a solution that is 57.5% iodine. She has two solutions availiable. One is 20% and the other is 70% iodine. How many of each should Sarah use? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #106578 by </B><A HREF=/tutors/aboutme.mpl?userid=ptaylor><B>ptaylor(2198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ptaylor><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=ptaylor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=106578\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x=amount of 70% solution needed<BR>\n" );
document.write( "Then 400-x=amount of 20% solution needed\r<BR>\n" );
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document.write( "Now we know that the amount of pure iodine in the 70% solution (0.70x) plus the amount of pure iodine in the 20% solution (0.20(400-x)) has to equal the amount of pure iodine in the final mixture (400*0.575).  So our equation to solve:\r<BR>\n" );
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document.write( "0.70x+0.20(400-x)=400*0.575  get rid of parens (distributive law)\r<BR>\n" );
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document.write( "0.70x+80-0.20x=230  subtract 80 from each side<BR>\n" );
document.write( "0.70x+80-80-0.20x=230-80  collect like terms<BR>\n" );
document.write( "0.50x=150  divide each side by 0.5\r<BR>\n" );
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document.write( "x=300 liters-------------------------amount of 70% solution needed\r<BR>\n" );
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document.write( "400-x=400-300=100 liters----------------amount of 20% solution needed\r<BR>\n" );
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document.write( "CK<BR>\n" );
document.write( "300*0.7+100*0.2=400*0.575<BR>\n" );
document.write( "210+20=230<BR>\n" );
document.write( "230=230\r<BR>\n" );
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document.write( "We can also work this problem using two unknowns:\r<BR>\n" );
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document.write( "Let x=amount of 70% solution needed<BR>\n" );
document.write( "And let y=amount of 20% solution needed\r<BR>\n" );
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document.write( "Now we are told that:<BR>\n" );
document.write( "x+y=400-------------------------------eq1<BR>\n" );
document.write( "0.7x+0.2y=400*0.575---------------------------eq2\r<BR>\n" );
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document.write( "From eq1, substitute y=400-x into eq2 and we have the same equation as before\r<BR>\n" );
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document.write( "Hope this helps---ptaylor </SPAN>\n" );
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