document.write( "Question 219776: A radiator contains 15 gal of a 20% antifreeze solution. How many gallons must be drained from the radiator and replaced by pure antifreeze so that the radiator will contain 15 gal of a 40% antifreeze solution? \n" ); document.write( "

Algebra.Com's Answer #165205 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let x=amount of the 20% solution that needs to be drained and replaced with pure antifreeze\r
\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure antifreeze in the initial radiator after x amount is drained (0.20(15-x)) plus the amount of pure antifreeze that is added(x) has to equal the amount of pure antifreeze in the final radiator (0.40*15), so our equation to solve is:
\n" ); document.write( "0.20(15-x)+x=0.40*15 get rid of parens
\n" ); document.write( "3-0.20x+x=6 subtract 3 from each side
\n" ); document.write( "3-3-0.20x+x=6-3 collect like terms
\n" ); document.write( "0.80x=3 divide each side by 0.80
\n" ); document.write( "x=3.75 gal----amount of the 20% solution that needs to be drained and replaced with pure antifreeze\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "0.20*(15-3.75)+3.75=0.40*15
\n" ); document.write( "2.25+3.75=6
\n" ); document.write( "6=6\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( "

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