document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=79808>Question 79808</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0618077286/algebrahomeworkh></A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> How many liters of water must be added to 50 liters of a 30% acid solution in order to produce a 20% acid solution? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #57278 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=57278\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x=amount of water that must be added\r<BR>\n" );
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document.write( "Now we know that the amount of pure water added (x) plus the amount of pure water in the 50 liter acid solution (50)(0.70){if it's 30% acid, then it must be 70% water} must equal the amount of pure water in the final mixture (50+x)(0.80).  So our equation to solve is:\r<BR>\n" );
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document.write( "x+(50)(0.70)=(0.80)(50+x)  get rid of parens\r<BR>\n" );
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document.write( "x+3.5=4+0.80x   subtract 0.80x and also 3.5 from both sides\r<BR>\n" );
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document.write( "x-0.80x+3.5-3.5=4-3.5+0.80x-0.80x  collect like terms\r<BR>\n" );
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document.write( "0.20x=0.5  divide both sides by0.20\r<BR>\n" );
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document.write( "x=2.5 liters-------------amount of pure water that needs to be added\r<BR>\n" );
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document.write( "CK\r<BR>\n" );
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document.write( "2.5+3.5=4+2<BR>\n" );
document.write( "6=6\r<BR>\n" );
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document.write( "Hope this helps---ptaylor </SPAN>\n" );
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