document.write( "Question 213393: an automobile radiator holds 6 liters of fluid. there is currently a mixture in the radiator that is 80% antifreeze and 20% water. how much of this mixture should be drained and replaced by pure anitfreezes so that the resulting mixture is 90% antifreeze? \n" ); document.write( "
Algebra.Com's Answer #161198 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
the best way (for me) to solve this is to
\n" ); document.write( "express an equation in words that sum up the
\n" ); document.write( "whole situation.
\n" ); document.write( "The 1st thing to realize is that you start out with
\n" ); document.write( "6 liters of fluid and you end up with 6 liters
\n" ); document.write( "of fluid
\n" ); document.write( "Here it is in words:
\n" ); document.write( "((original amount of antifreeze) - (amount of antifreeze drained out) + (amount of antifreeze put back in))
\n" ); document.write( "[all divided by]
\n" ); document.write( "(total amount of fluid I end up with) = 90%
\n" ); document.write( "-------------------------------------------
\n" ); document.write( "Let \"x\"= liters of fluid I drain out and replace with pure antifreeze
\n" ); document.write( "Original amount of antifreeze = \".8%2A6+=+4.8\" liters
\n" ); document.write( "Amount of antfreeze drained out = \".8x\"
\n" ); document.write( "Amount of antifreeze put back in = \"x\"
\n" ); document.write( "------------------------------------------
\n" ); document.write( "Now I can write the equation
\n" ); document.write( "\"%284.8+-+.8x+%2B+x%29%2F6+=+.9\"
\n" ); document.write( "\"4.8+-+.8x+%2B+x+=+5.4\"
\n" ); document.write( "\".2x+=+5.4+-+4.8\"
\n" ); document.write( "\".2x+=+.6\"
\n" ); document.write( "\"x+=+3\"
\n" ); document.write( "3 liters, or 1/2 the fluid in the radiator must
\n" ); document.write( "be drained and replaced with pure antifreeze to get
\n" ); document.write( "90% antifreeze
\n" ); document.write( "check answer:
\n" ); document.write( "There must be 10% water in the final solution
\n" ); document.write( "I start with \".2%2A6+=+1.2\" liters water
\n" ); document.write( "I drain out \".2x+=+.2%2A3\" or \".6\" liters water
\n" ); document.write( "I end up with \"1.2+-+.6+=+.6\" liters water
\n" ); document.write( "\".6+=+.1%2A6\"
\n" ); document.write( "So, I do end up with 10% water
\n" ); document.write( "
\n" ); document.write( "
\n" );