document.write( "Question 719353: a chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 30% and the third contains 75%. He wants to use all three solutions to obtain a mixture of 68 liters containing 50% acid, using 2 times as much of the 75% solution as the 30% solution. how many liters of each solution should be used? \n" ); document.write( "
Algebra.Com's Answer #441368 by josgarithmetic(39618)\"\" \"About 
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Let x amount of 20% acid
\n" ); document.write( "let y amount of 30% acid
\n" ); document.write( "let z amount of 75% acid\r
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\n" ); document.write( "\n" ); document.write( "\"using 2 times as much of the 75% solution as the 30% solution. \",
\n" ); document.write( "That means z/y=2/1 or \"z=2y\".\r
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\n" ); document.write( "\n" ); document.write( "Two essential starting equations:
\n" ); document.write( "\"%2820x%2B30y%2B75z%29%2F68=50\" and \"x%2By%2Bz=68\".\r
\n" ); document.write( "\n" ); document.write( "The less advanced way to solve this is to begin with the ratio between z and y and substitute into the \"two essential\" equations:\r
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\n" ); document.write( "\n" ); document.write( "The percentage equation:
\n" ); document.write( "\"2x%2B3y%2B7.5z=68%2A5\"
\n" ); document.write( "\"2x%2B3y%2B7.5%282y%29=68%2A5\"
\n" ); document.write( "\"2x%2B3y%2B15y=68%2A5\"
\n" ); document.write( "\"2x%2B18y=68%2A5\"
\n" ); document.write( "\"x%2B9y=34%2A5\"\r
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\n" ); document.write( "\n" ); document.write( "The solution summation equation:
\n" ); document.write( "\"x%2By%2B%282y%29=68\"
\n" ); document.write( "\"x%2B3y=68\"\r
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\n" ); document.write( "\n" ); document.write( "Our simpler system to solve is now in variables x and y:
\n" ); document.write( "\"highlight%28x%2B9y=170%29\"
\n" ); document.write( "\"highlight%28x%2B3y=68%29\"\r
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\n" ); document.write( "\n" ); document.write( "Subtract the second from the first, and from resulting 6y=102, obtain \"y=17\".
\n" ); document.write( "Use that result again in the second equation, x=68-3y, x=68-3*17, \"x=17\".\r
\n" ); document.write( "\n" ); document.write( "Now, go back to the given relationship between z and y, z=2y, z=2*17, \"z=34\".\r
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\n" ); document.write( "\n" ); document.write( "ANSWER SUMMARY: x=17 liters of 20%, y=17 liters of 30%, z=34 liters of 75% acid.
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