document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=146201>Question 146201</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how much antifeeze should be added to 300 gallons of 30% antifreeze solution to raise the antifreeze solution to 40%? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #106759 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=106759\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x=amount of pure antifreeze that needs to be added\r<BR>\n" );
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document.write( "Now we know that the amount of pure antifreeze in the quantity that was added (x) plus the amount of pure antifreeze in the 300 gal (0.30*300) has to equal the amount of pure antifreeze in the final mixture(0.40(300+x)).  So our equation to solve is:\r<BR>\n" );
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document.write( "x+0.30*300=0.40(300+x) get rid of parens (distributive law)<BR>\n" );
document.write( "x+90=120+0.40x subtract 0.40x and also 90 from each side<BR>\n" );
document.write( "x+90-90-0.40x=120-90+0.40x-0.40x  collect like terms\r<BR>\n" );
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document.write( "0.60x=30  divide each side by 0.60<BR>\n" );
document.write( "x=50 gal-------------------------------amount of pure antifreeze needed\r<BR>\n" );
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document.write( "CK\r<BR>\n" );
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document.write( "50+90=0.40(350)<BR>\n" );
document.write( "140=140\r<BR>\n" );
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document.write( "Hope this helps------ptaylor </SPAN>\n" );
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