document.write( "Question 312145: A car radiator has a 6-liter capacity. If the liquid in the radiators 40% antifreeze, how much liquid must be replaced with pure antifreeze to bring the mixture up to a 50% solutions? \n" ); document.write( "
Algebra.Com's Answer #223169 by Fombitz(32388)\"\" \"About 
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Let A be the amount of 40% solution currently in the radiator. You add B liters of pure antifreeze (100% solution).
\n" ); document.write( "1.\"40%28A%29%2B100%28B%29=50%28A%2BB%29\"
\n" ); document.write( "2.\"A%2BB=6\"
\n" ); document.write( "From eq. 2,
\n" ); document.write( "\"A=6-B\"
\n" ); document.write( "Substitute into eq. 1,
\n" ); document.write( "\"40%286-B%29%2B100%28B%29=50%286-B%2BB%29\"
\n" ); document.write( "\"240-40B%2B100B=300\"
\n" ); document.write( "\"60B=60\"
\n" ); document.write( "\"highlight%28B=1%29\"
\n" ); document.write( "Then from eq. 2,
\n" ); document.write( "\"A=6-B=6-1\"
\n" ); document.write( "\"highlight%28A=5%29\"
\n" ); document.write( "There are currently 5 liters of 40% solution. Adding 1 liter will fill the radiator and make a 50% solution. \r
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