document.write( "Question 1130639: A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 35% ACID, and the third contains
\n" ); document.write( "65%. He wants to use all three solutions to obtain a mixture of 102 liters containing 40% acid, using 2 times as much of the 65% solution as the 35% solution. How many liters of each solution should be used?
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Algebra.Com's Answer #747264 by josgarithmetic(39618)\"\" \"About 
You can put this solution on YOUR website!
x of 25%
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\n" ); document.write( "z of 65%
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\n" ); document.write( "\"z%2Fy=2\"
\n" ); document.write( "\"z=2y\"\r
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\n" ); document.write( "He wants to use all three solutions to obtain a mixture of 102 liters containing 40% acid,
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\n" ); document.write( "\n" ); document.write( "\"x%2By%2B2y=102\"
\n" ); document.write( "\"x%2B3y=102\"\r
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\n" ); document.write( "\n" ); document.write( "Accounting for the desired percentages and amount of pure acid
\n" ); document.write( "\"25x%2B35y%2B65%282y%29=40%2A102\"\r
\n" ); document.write( "\n" ); document.write( "Simplify this pure-acid equation.\r
\n" ); document.write( "\n" ); document.write( "\"25x%2B%2835%2B2%2A65%29y=40%2A102\"\r
\n" ); document.write( "\n" ); document.write( "\"25x%2B165y=40%2A102\"
\n" ); document.write( "divide both sides by 5,
\n" ); document.write( "\"5x%2B33y=8%2A102\"
\n" ); document.write( "\"5x%2B33y=816\"\r
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\n" ); document.write( "\n" ); document.write( "Simpler system in variables x and y:
\n" ); document.write( "\"system%28x%2B3y=102%2C5x%2B33y=816%29\"\r
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\n" ); document.write( "\n" ); document.write( "You should get, if continue with Elimination Method,
\n" ); document.write( "\"highlight_green%28y=17%29\"
\n" ); document.write( "From that you should find that \"highlight_green%28z=34%29\".......
\n" ); document.write( "and \"highlight_green%28x=51%29\".
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