document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=186439>Question 186439</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Angelina wants to add 60% antifreeze solution to her car's radiator. About how much 60% antifreeze solution can she make by mixing 1 gallon of pure antifreeze with water? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #139725 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=139725\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x=amount of water needed<BR>\n" );
document.write( "Now we know that the amount of pure antifreeze before the water is added (1 gal) has to equal the amount of pure antifreeze after the water is added (0.60(1+x)), so:\r<BR>\n" );
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document.write( "1=0.60(1+x)<BR>\n" );
document.write( "1=0.60+0.60x  subtract 0.60 from each side<BR>\n" );
document.write( "1-0.60=0.60-0.60+0.60x collect like terms<BR>\n" );
document.write( "0.40=0.60x divide each side by 0.60<BR>\n" );
document.write( "x=2/3  gal------------------amt of water needed<BR>\n" );
document.write( "1=0.60(1 2/3)=0.60*(5/3)<BR>\n" );
document.write( "1=1\r<BR>\n" );
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document.write( "Hope this helps---ptaylor </SPAN>\n" );
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