document.write( "Question 79617: Not in book\r
\n" ); document.write( "\n" ); document.write( "A mixture of 10% acid and 90% water is added to 5 liters of pure acid. The final mixture is 40% water. How many liters of water are in the final mixture?
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Algebra.Com's Answer #57154 by ptaylor(2198)\"\" \"About 
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\n" ); document.write( "Let x=number of liters of acid/water mixture added to the 5 liters of pure acid.\r
\n" ); document.write( "\n" ); document.write( "Then0.90x=amount of pure water added to the 5 liters of pure acid\r
\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure water added to the 5 liters of pure acid equals the amount of pure water in the final mixture. So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "0.90x=0.40(5+x) get rid of parens\r
\n" ); document.write( "\n" ); document.write( "0.90x=2+0.40x subtract 0.40x from each side\r
\n" ); document.write( "\n" ); document.write( "0.90x-0.40x=2+0.40x-0.40x collect like terms\r
\n" ); document.write( "\n" ); document.write( "0.50x=2 divide both sides by 0.50\r
\n" ); document.write( "\n" ); document.write( "x=4 liters------------------amount of mixture added to the pure acid\r
\n" ); document.write( "\n" ); document.write( "0.90x=0.90*4=3.6 liters---------------------amount of pure water in final mixture\r
\n" ); document.write( "\n" ); document.write( "CK.\r
\n" ); document.write( "\n" ); document.write( "0.40(5+4)=3.6\r
\n" ); document.write( "\n" ); document.write( "3.6=3.6\r
\n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor\r
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