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)![]() ![]() ![]() You can put this solution on YOUR website! 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( "------------------- \n" ); document.write( "Equation: \n" ); document.write( "anti + anti = anti \n" ); document.write( "0.50x + 0.25*90 = 0.40(x+90) \n" ); document.write( "----------------------------- \n" ); document.write( "Multiply thru by 100 to get: \n" ); document.write( "50x + 25*90 = 40x + 40*90 \n" ); document.write( "10x = 15*90 \n" ); document.write( "--- \n" ); document.write( "x = 135 gallons (amt. of 50% solution needed) \n" ); document.write( "================================================ \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "=========== \n" ); document.write( " |