document.write( "Question 13262: How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60% antifreeze? \n" ); document.write( "
Algebra.Com's Answer #6744 by rapaljer(4671)![]() ![]() You can put this solution on YOUR website! Let x = amount of water to be added with 0% antifreeze \n" ); document.write( "1 = amount of pure antifreeze at 100% \n" ); document.write( "x+1 = amount of the mixture at 60% antifreeze\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "1 = .60x + .60\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Subtract .60 from each side: \n" ); document.write( "1-.60 = .60x \n" ); document.write( ".40 = .60x\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Divide both sides by .60:\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Check: Total antifreeze = 1 gallon in total mixture of 1 + 2/3 gallon = 5/3 gallons. 1 gallon divided by 5/3 equals 3/5 which is 60%. It checks.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "R^2 at SCC \n" ); document.write( " \n" ); document.write( " |