document.write( "Question 462897: How many gallons of a 50% anti-freeze solution must be mixed with 90 gallons of 25% anti-freeze to get a mixture that is 40% anti-freeze? \n" ); document.write( "
Algebra.Com's Answer #317244 by stanbon(75887)\"\" \"About 
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How many gallons of a 50% anti-freeze solution must be mixed with 90 gallons of 25% anti-freeze to get a mixture that is 40% anti-freeze?
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\n" ); document.write( "Equation:
\n" ); document.write( "anti + anti = anti
\n" ); document.write( "0.50x + 0.25*90 = 0.40(x+90)
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\n" ); document.write( "Multiply thru by 100 to get:
\n" ); document.write( "50x + 25*90 = 40x + 40*90
\n" ); document.write( "10x = 15*90
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\n" ); document.write( "x = 135 gallons (amt. of 50% solution needed)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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