document.write( "Question 1036653: How much pure antifreeze must be added to 12 gallons of 10% antifreeze to make a 40% antifreeze solution? \n" ); document.write( "

Algebra.Com's Answer #651369 by Boreal(15235) $\"\"$ $\"About$

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.10x+(12-x)=.40(12). The amount of 10% antifreeze is .10x, pure antifreeze is 12 gallons- what is x, and the end result is 40% antifreeze.
\n" ); document.write( ".10x+12-x=4.8
\n" ); document.write( "-.90x=-7.2
\n" ); document.write( "x=8 gallons of .10
\n" ); document.write( "12-x=4 gallons of pure.
\n" ); document.write( "2/3 is 10%, and the resultant is 2/3s of the way from 100% down to 10%. \n" ); document.write( "

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