document.write( "Question 816165: The radiator in the car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the aniotifreeze suggests that, for summer driving, optimal cooling of the engine is obtained with only 50% antifreeze. If the capacity of the radiator is 3.6 L, how much coolant should be drained and replaced with water to reduce the antifreeze concentration to the recommended level? \n" ); document.write( "
Algebra.Com's Answer #491393 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
\"+.6%2A3.6+=+2.16+\" L antifreeze
\n" ); document.write( "originally in radiator
\n" ); document.write( "Let \"+x+\" = liters of solution to
\n" ); document.write( "be drained and replaced with water
\n" ); document.write( "\"+.6x+\" = liters of antifreeze
\n" ); document.write( "drained from radiator
\n" ); document.write( "-------------------------
\n" ); document.write( "\"+%28+2.16+-+.6x+%29+%2F+3.6+=+.5+\"
\n" ); document.write( "Note that you start with 3.6 liters
\n" ); document.write( "and end with 3.6 liters
\n" ); document.write( "\"+2.16+-+.6x+=+.5%2A3.6+\"
\n" ); document.write( "\"+.6x+=+2.16+-+1.8+\"
\n" ); document.write( "\"+.6x+=+.36+\"
\n" ); document.write( "\"+x+=+.6+\"
\n" ); document.write( ".6 liters must be drained and replaced with
\n" ); document.write( "pure water
\n" ); document.write( "----------------
\n" ); document.write( "check:
\n" ); document.write( "\"+.6%2A.6+=+.36+\" liters of alcohol is drained
\n" ); document.write( "There is \"+3.6%2A.6+-+.36+\"
\n" ); document.write( "\"+2.16+-+.36+=+1.8+\" liters alcohol left
\n" ); document.write( "-----------------
\n" ); document.write( "\"+.6+\" liters of water is added to make
\n" ); document.write( "\"+1.8+\" liters alcohol and \"+3.6+-+1.8+=+1.8+\"
\n" ); document.write( "liters water
\n" ); document.write( "They are 50% - 50%
\n" ); document.write( "OK\r
\n" ); document.write( "\n" ); document.write( "
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