document.write( "Question 802380: How many liters of 60% alcohol solution and 30% alcohol solution must be mixed to obtain 12 liters of 40% alcohol solution?\r
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Algebra.Com's Answer #483858 by josmiceli(19441)\"\" \"About 
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Let \"+a+\" = liters of 60% alcohol solution needed
\n" ); document.write( "Let \"+b+\" = liters of 30% alcohol solution needed
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\n" ); document.write( "(1) \"+a+%2B+b+=+12+\"
\n" ); document.write( "(2) \"+%28+.6a+%2B+.3b+%29+%2F+12+=+.4+\"
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\n" ); document.write( "(2) \"+.6a+%2B+.3b+=+4.8+\"
\n" ); document.write( "(2) \"+6a+%2B+3b+=+48+\"
\n" ); document.write( "(2) \"+2a+%2B+b+=+16+\"
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\n" ); document.write( "Subtract (1) from (2)
\n" ); document.write( "(2) \"+2a+%2B+b+=+16+\"
\n" ); document.write( "(1) \"+-a+-+b+=+-12+\"
\n" ); document.write( "\"+a+=+4+\"
\n" ); document.write( "and, since
\n" ); document.write( "(1) \"+4+%2B+b+=+12+\"
\n" ); document.write( "(1) \"+b+=+8+\"
\n" ); document.write( "4 liters of 60% alcohol solution are needed
\n" ); document.write( "8 liters of 30% alcohol solution are needed
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\n" ); document.write( "check:
\n" ); document.write( "(2) \"+%28+.6a+%2B+.3b+%29+%2F+12+=+.4+\"
\n" ); document.write( "(2) \"+%28+.6%2A4+%2B+.3%2A8+%29+%2F+12+=+.4+\"
\n" ); document.write( "(2) \"+%28+2.4+%2B+2.4+%29+%2F+12+=+.4+\"
\n" ); document.write( "(2) \"+4.8%2F12+=+.4+\"
\n" ); document.write( "(2) \"+4.8+=+4.8+\"
\n" ); document.write( "OK
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