Question 243494
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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We need to know how much more pure acid we need to add  to get an 80% acid solution.
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Set the unknown to x.
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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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It will consist of 9 gallows of .5 acid solution + x gallons of 100% acid.  We can show this as
.5(9) + 1(x)
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Of course, we don't normally show 1 times anything. But I just wanted to be clear.
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Combining we have:
.8(9+x) = .5(9) + x
Multiplying what we have in parentheses
7.2 + .8x = 4.5 + x
Subtracting 4.5 from both sides
2.7 + .8x = x
Subtracting .8x from both sides
2.7 = .2x
Dividing both sides by .2
13.5 = x
x = 13.5
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So this suggests we need to add 13.5 gallons of pure acid to get to an 80% solution.
That seems like an awful lot.  But we always check our work, so that's next anyway.
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If we add 13.5 gallons to 9 gallons, we have 22.5 gallons of solution in total.
If it is 80% acid, then we would have .8(22.5) = 18 gallons of pure acid in the solution.
We started with 4.5 gallons of pure acid and added 13.5 gallons of pure acid = 18.
Done.