These are chapter 25 practice problems from Elementary Algebra by Hall & Knight:
- Two vessels contain mixtures of wine and water; in one there is three times as much wine as water, in the other five times as much water as wine. Find how much must be drawn off from each to fill a third vessel which holds seven gallons, in order that its contents may be half wine and half water.
(26-1) We can start by saying that $a + b = 7$, where $a$ and $b$ are gallons taken from the first and second mixture, respectively.
(26-2) Since $\frac{1}{4}$ of $a$ and $\frac{5}{6}$ of $b$ are water, $\frac{a}{4} + \frac{5b}{6} = \frac{7}{2}$ represents the 50-50 mixture's water.
(26-3) Since $\frac{3}{4}$ of $a$ and $\frac{1}{6}$ of $b$ are wine, $\frac{3a}{4} + \frac{b}{6} = \frac{7}{2}$ represents the 50-50 mixture's wine.
(26-4) Optionally, you can combine (26-2) and (26-3) into $\frac{a}{4} + \frac{5b}{6} = \frac{3a}{4} + \frac{b}{6}$.
Solving the above equations gives you the correct answer of $a = 4$ and $b = 3$.
- There are two mixtures of wine and water, one of which contains twice as much water as wine, and the other three times as much wine as water. How much must there be taken from each to fill a pint cup, in which the water and wine shall be equally mixed?
Trying to solve this the same way as #26, we get the equations:
(27-1) $c + d = 1$
(27-2) $\frac{2c}{3} + \frac{d}{4} = \frac{1}{2}$
(27-3) $\frac{c}{3} + \frac{3d}{4} = \frac{1}{2}$
(27-4) $\frac{2c}{3} + \frac{d}{4} = \frac{c}{3} + \frac{3d}{4}$
Solving these equations gives an incorrect answer of $c = \frac{3}{5}$ and $d = \frac{2}{5}$.
To get the correct answer of $c = \frac{2}{3}$ and $d = \frac{1}{3}$, I had to remove the denominators from (27-4), and solve that with (27-1). I don't understand why this works; it switches from dealing with fractions of $c$ and $d$ to dealing with parts, and ignores how many of those parts make a whole; i.e. $\frac{2c}{3}$ means that $c$ has 3 parts, and two of those are water, while $2c$ just tells you that two parts of $c$ are water.
(27-1) $c + d = 1$
(27-4b) $2c + d = c + 3d$
If I try removing the denominators like this to solve #26, I get a wrong answer:
(26-1) $a + b = 7$
(26-4b) $a + 5b = 3a + b$
Based on these, $a = \frac{14}{3}$, $b = \frac{7}{3}$. But the correct answer to #26 is $a = 4$, $b = 3$.
These problems seem to have the same form, so it should be possible to solve them the same way. Why can't #27 be solved the same way I solved #26, and vice versa?
Note: I also know of a third way to solve these, and it works for both problems. I couldn't figure out how to turn this thought process into algebraic equations that can solve both problems, though. In #26, four gallons of the first mixture has an imbalance of -2 gallons of water, while 6 gallons of the second mixture has an imbalance of +4 gallons of water. So, for every 6 gallons of the second, you need two 4-gallon parts from the first. That means the ratio of the first and second mixtures needs to be 8:6, or 4:3. Since the target is 7 gallons and the ratio is 4:3, you need 4 gallons of the first mixture and 3 gallons of the second.
Best Answer
I got the numbers $c=\frac{2}{3}$ and $d=\frac{1}{3}$ by using the method mentioned at the end of the question's post. When I did this, I forgot to take into account that the first mixture has 3 parts (2 water, 1 wine) and the second has 4 parts (1 water, 3 wine). So, $c=\frac{2}{3}$, $d=\frac{1}{3}$ is actually wrong; $c=\frac{3}{5}$ and $d=\frac{2}{5}$ is right.
The issue wasn't that #26 and #27 couldn't be solved the same way, the issue was a careless mistake on my part.