document.write( "Question 243494: how much pure acid should be mixed with 9 gallons of a 50% acid solution in order to get an 80% acid solution \n" ); document.write( "
Algebra.Com's Answer #178389 by oberobic(2304)\"\" \"About 
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Always start by sorting out what you know and assigning variables. We know we have 9 gallons of 50% acid solution. Multiplying 9 by .5 = 4.5 gallons of pure acid in the solution.
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\n" ); document.write( "We need to know how much more pure acid we need to add to get an 80% acid solution.
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\n" ); document.write( "Set the unknown to x.
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\n" ); document.write( "Since we are adding x gallons of pure acid, the final solution will be 9+x gallons at 80% acid. We can state this algebraically as: .8(9+x)
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\n" ); document.write( "It will consist of 9 gallows of .5 acid solution + x gallons of 100% acid. We can show this as
\n" ); document.write( ".5(9) + 1(x)
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\n" ); document.write( "Of course, we don't normally show 1 times anything. But I just wanted to be clear.
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\n" ); document.write( "Combining we have:
\n" ); document.write( ".8(9+x) = .5(9) + x
\n" ); document.write( "Multiplying what we have in parentheses
\n" ); document.write( "7.2 + .8x = 4.5 + x
\n" ); document.write( "Subtracting 4.5 from both sides
\n" ); document.write( "2.7 + .8x = x
\n" ); document.write( "Subtracting .8x from both sides
\n" ); document.write( "2.7 = .2x
\n" ); document.write( "Dividing both sides by .2
\n" ); document.write( "13.5 = x
\n" ); document.write( "x = 13.5
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\n" ); document.write( "So this suggests we need to add 13.5 gallons of pure acid to get to an 80% solution.
\n" ); document.write( "That seems like an awful lot. But we always check our work, so that's next anyway.
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\n" ); document.write( "If we add 13.5 gallons to 9 gallons, we have 22.5 gallons of solution in total.
\n" ); document.write( "If it is 80% acid, then we would have .8(22.5) = 18 gallons of pure acid in the solution.
\n" ); document.write( "We started with 4.5 gallons of pure acid and added 13.5 gallons of pure acid = 18.
\n" ); document.write( "Done.
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