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