document.write( "Question 17662: A car radiator holds 24 L of a 20% antifreeze solution. How much of the 20% solution should be drained and be replaced with pure antifreeze to give a 40% antifreeze solution? \n" ); document.write( "
Algebra.Com's Answer #8536 by rapaljer(4671)\"\" \"About 
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Begin with 24 L of .20 antifreeze
\n" ); document.write( "Drain x L of .20 antifreeze
\n" ); document.write( "Add x L of 1.00 antifreeze\r
\n" ); document.write( "\n" ); document.write( "Equals 24 L. of .40 antifreeze\r
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\n" ); document.write( "\n" ); document.write( "Equation:
\n" ); document.write( ".20(24) - .20x + 1.00x = .40(24)
\n" ); document.write( "4.80 + .80x = 9.60
\n" ); document.write( "4.80-4.80 +.80x = 9.60 - 4.80
\n" ); document.write( ".80x = 4.80
\n" ); document.write( "\"x=+4.80%2F.80+=+48%2F8+=+6\" liters\r
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\n" ); document.write( "\n" ); document.write( "R^2 at SCC
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