Question 317379
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You almost have it.  It is just that your answer comes up with the wrong sign.  Doesn't make much sense to add a negative amount of water to an acid solution, now does it?


You actually set the problem up correctly, but you made a small error when you divided both sides by 50.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-50x}{50}\ \neq\ x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-50x}{50}\ =\ -x]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -x\ =\ -3.2]


and therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 3.2]


Meaning 3.2 liters of water and 0.8 liters of the 50% acid solution.


There is a much less complex way to do this one.  First off, let *[tex \LARGE x] represent the amount of acid solution needed.  Then, realizing that 50% is the same as one-half, and 10 percent of 4 liters is 0.4 liters, you can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x}{2}\ =\ 0.4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 0.8]


And we get the same answer:  0.8 liters of acid solution and 3.2 liters of water.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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