document.write( "Question 184199This question is from textbook
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\n" ); document.write( "\n" ); document.write( "How many liters of a 30% alcohol solution must be mixed with 80 liters of a 80% solution to get a 40% solution.\r
\n" ); document.write( "\n" ); document.write( "I've tried the following, but I'm not sure it's right at all. \r
\n" ); document.write( "\n" ); document.write( "a+b=80L
\n" ); document.write( "30a+80b=3200
\n" ); document.write( "Multiply by -30 to use the elimination format, this give you:\r
\n" ); document.write( "\n" ); document.write( "-30a - 30b = -2400L
\n" ); document.write( "30a + 80b = 3200L\r
\n" ); document.write( "\n" ); document.write( " 50b = 800
\n" ); document.write( " b=16L, which makes a = 64L \r
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Algebra.Com's Answer #138239 by ankor@dixie-net.com(22740)\"\" \"About 
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How many liters of a 30% alcohol solution must be mixed with 80 liters
\n" ); document.write( " of a 80% solution to get a 40% solution.
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\n" ); document.write( "I would do it this way:
\n" ); document.write( "Let x = amt of 30% solution:
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\n" ); document.write( "Then when the two solutions are mixed together:
\n" ); document.write( "(x+80) = amt of 40% solution
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\n" ); document.write( "a mixture equation using decimal equiv of per cent
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\n" ); document.write( "30% amt + 80% amt = 40% amt
\n" ); document.write( ".3x + .8(80) = .4(x+80)
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\n" ); document.write( ".3x + 64 = .4x + 32
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\n" ); document.write( ".3x -.4x = 32 - 64
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\n" ); document.write( "-.1x = -32
\n" ); document.write( "x = \"%28-32%29%2F%28-.1%29\"
\n" ); document.write( "x = +320 liters of 30 % solution required
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