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)\"\" \"About 
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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
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\n" ); document.write( "\n" ); document.write( "\"0%2A+x+%2B+1.00+%2A+1+=+.60%28x%2B1%29\"
\n" ); document.write( "1 = .60x + .60\r
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\n" ); document.write( "\n" ); document.write( "Subtract .60 from each side:
\n" ); document.write( "1-.60 = .60x
\n" ); document.write( ".40 = .60x\r
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\n" ); document.write( "\n" ); document.write( "Divide both sides by .60:\r
\n" ); document.write( "\n" ); document.write( "\".40%2F.60+=+%28.60x%29%2F.60+\"\r
\n" ); document.write( "\n" ); document.write( "\"x+=+.4%2F.6+=+4%2F6+=+2%2F3\" gallon.\r
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\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
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