document.write( "Question 82504This question is from textbook Algebra 1
\n" ); document.write( ": 2.A solution containing 30% acid is to be mixed with a solution containing 50% acid to make 200L of a solution containing 40% acid. How much of each solution should be used?
\n" ); document.write( "Then there is a chart to put the work into
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Algebra.Com's Answer #59098 by checkley75(3666)\"\" \"About 
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.3x+.5(200-x)=.4*200
\n" ); document.write( ".3x+100-.5x=80
\n" ); document.write( "-.2x=-20
\n" ); document.write( "x=-20/-.2
\n" ); document.write( "x=100 answer for the 30% acid solution.
\n" ); document.write( "200-100=100 is the answer for the 50% solution also.
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