document.write( "Question 1156055: A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 40% and the third contains 85%. He wants to use all three solutions to obtain a mixture of 72 liters containing 55% acid, using 2 times as much of the 85% solution as the 40% solution. How many liters of each solution should be used? \n" ); document.write( "
Algebra.Com's Answer #815555 by josgarithmetic(39617)\"\" \"About 
You can put this solution on YOUR website!
\r\n" );
document.write( "CONC.%     VOLUME LITERS     PURE ACID\r\n" );
document.write( "  25           x              0.25x\r\n" );
document.write( "  40           y              0.4y\r\n" );
document.write( "  85          2y              1.7y          \r\n" );
document.write( "\r\n" );
document.write( "  55          72            0.55*72=39.6\r\n" );
document.write( "

\n" ); document.write( "You can have two equations.\r
\n" ); document.write( "\n" ); document.write( "\"system%28x%2B3y=72%2C0.25x%2B2.1y=39.6%29\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "They are equivalently
\n" ); document.write( "\"system%28x%2B3y=72%2Cx%2B8.4y=158.4%29\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Their difference gives \"5.4y=86.4\"\r
\n" ); document.write( "\n" ); document.write( "\"highlight_green%28y=16%29\"
\n" ); document.write( "and you can easily evaluate the other acid solution volumes. \n" ); document.write( "
\n" );