document.write( "Question 531021: I'm trying to figure out the formula for this word problem. If I can figure out the formula for mixtures I should be able to do the entire worksheet.\r
\n" ); document.write( "\n" ); document.write( "Sample Problem: A Chemist has a solution that is 12% acid and another solution that is 42% acid. She wants 10L of solution that is 30% acid. How much of each solution should she mix?
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Algebra.Com's Answer #350306 by oberobic(2304)\"\" \"About 
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With solution problems you need to determine how much 'pure' stuff you have or need.
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\n" ); document.write( "The chemist needs 10 liters of a 30% acid. .3*10 = 3 liters of 'pure' acid in a solvent (likely water).
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\n" ); document.write( "She has 12% acid and 42% acid to mix together.
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\n" ); document.write( "x = liters of 12% acid
\n" ); document.write( "y = liters of 42% acid
\n" ); document.write( "x+y =10
\n" ); document.write( "so
\n" ); document.write( "x = 10-y
\n" ); document.write( "or
\n" ); document.write( "y = 10 - x
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\n" ); document.write( ".12*x + .42*y = .3*10
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\n" ); document.write( "multiply by 100 to eliminate decimals
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\n" ); document.write( "12x + 42y = 30*10
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\n" ); document.write( "substitute for x
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\n" ); document.write( "12(10-y) + 42y = 300
\n" ); document.write( "120 -12y + 42y = 300
\n" ); document.write( "30y = 180
\n" ); document.write( "y = 6
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\n" ); document.write( "Answer: Mix 6 liters of 42% acid and 4 liters of 12% acid.
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\n" ); document.write( "Check the amount of pure acid to be sure this answer is correct.
\n" ); document.write( ".42*6 = 2.52
\n" ); document.write( ".12*4 = 0.48
\n" ); document.write( "2.52 + .48 = 3.0
\n" ); document.write( "Correct.
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\n" ); document.write( "Done.
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