document.write( "Question 225689: How much pure acid should be mixed with 2 gallons of a 50% acid solution in order to get an 80% acid solution? \n" ); document.write( "

Algebra.Com's Answer #168311 by checkley77(12844) $\"\"$ $\"About$

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.50*2+X=.80(2+X)
\n" ); document.write( "1+X=1.6+.8X
\n" ); document.write( "X-.8X=1.6-1
\n" ); document.write( ".2X=.6
\n" ); document.write( "X=.6/.2
\n" ); document.write( "X=3 GALLONS OF PURE ACID IS USED.
\n" ); document.write( "PROOF:
\n" ); document.write( ".50*2+3=.80(2+3)
\n" ); document.write( "1+3=.8*5
\n" ); document.write( "4=4
\n" ); document.write( " \n" ); document.write( "

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