document.write( "Question 145685: Soybean meal is 12% protein; cornmeal is 6% protein. How many pounds of each should be mixed together in order to get 240-lb mixture that is 9% protein? How many pounds of the cornmeal should be in the mixture? How many pounds of the soybean meal should be in the mixture?
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Algebra.Com's Answer #106334 by ankor@dixie-net.com(22740)\"\" \"About 
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Soybean meal is 12% protein; cornmeal is 6% protein. How many pounds of each should be mixed together in order to get 240-lb mixture that is 9% protein? How many pounds of the cornmeal should be in the mixture? How many pounds of the soybean meal should be in the mixture?
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\n" ); document.write( "Since 9% is halfway between 12% and 6%, it would need 120 lb of each
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\n" ); document.write( "However, here is the algebra procedure:
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\n" ); document.write( "Let x = amt of soybean meal (12%)
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\n" ); document.write( "It says the resulting amt = 240 lb, therefore:
\n" ); document.write( "(240-x) = amt of cornmeal (6%)
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\n" ); document.write( "write the decimal equiv of the per cent protein equation:
\n" ); document.write( ".12x + .06(240-x) = .09(240)
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\n" ); document.write( ".12x + 14.4 - .06x = 21.6
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\n" ); document.write( ".12x - .06x = 21.6 - 14.4
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\n" ); document.write( ".06x = 7.2
\n" ); document.write( "x = \"7.2%2F.06\"
\n" ); document.write( "x = 120 lb of soy meal
\n" ); document.write( "Then
\n" ); document.write( "240 - 120 = 120 lb of cornmeal \n" ); document.write( "
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