document.write( "Question 96352: How much 70% acid solution should be mixed with 7 liters of 20% acid solution to end up with a 30% acid solution? \n" ); document.write( "
Algebra.Com's Answer #70143 by ptaylor(2198)\"\" \"About 
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Let x=amount of 70% acid solution needed\r
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\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure acid in the 70% solution (0.70x) plus the amount of pure acid in the 20% solution (0.20(7)) has to equal the amount of pure acid in the final mixture (0.30*(7+x)). So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "0.70x+0.20*7=0.30(7+x) get rid of parens\r
\n" ); document.write( "\n" ); document.write( "0.70x+1.4=2.10+0.30x subtract 0.30x and also 1.4 from both sides\r
\n" ); document.write( "\n" ); document.write( "0.70x+1.4-1.4-0.30x=2.1-1.4+0.30x-0.30x collect like terms\r
\n" ); document.write( "\n" ); document.write( "0.40x=0.70 divide both sides by 0.40\r
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\n" ); document.write( "\n" ); document.write( "x=1.75 liters---------------amount of 70% solution needed\r
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\n" ); document.write( "\n" ); document.write( "0.70*1.75+0.20*7=0.30*8.75\r
\n" ); document.write( "\n" ); document.write( "1.225+1.4=2.625\r
\n" ); document.write( "\n" ); document.write( "2.625=2.625\r
\n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor\r
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