document.write( "Question 358546: how much pure acid should be mixed with 5 gallons of a 50% acid solution in order to get an 80% acid solution? \n" ); document.write( "

Algebra.Com's Answer #255884 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi,
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document.write( "Let x repesent the amount of acid added
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document.write( ".50*5gal +1.00x = .80(5+x)
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document.write( "Simplify and Solve
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document.write( ".50*5gal +1.00x = .80*5+ .8x
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document.write( ".2x = .80*5 - .50*5
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document.write( ".2x =.3*5
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document.write( ".2x = 1.5
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document.write( "x = 7.5gal
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document.write( "checking our answer
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document.write( "2.5gal + 7.5gal = .8* 12.5gal = 10gal \n" );
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