I was given the following problem:
Of three workmen, the second and third can complete a job in 10 days. The first and third can do it in 12 days, and the first and second can do it in 15 days. In how many days can each of them do the job alone?
This is my work:
Let a = the first workman, b= the second workman, and c = the third workman.
$$\frac{a+b}{2}=15\rightarrow a+b=30\ \rightarrow a=30-b\\
\frac{b+c}{2}=10\rightarrow b+c=20\rightarrow c=20-b\\
\frac{a+c}{2}=12\rightarrow a+c=24$$
Substitute using above identities: $30-b+20-b=24$
$$50-2b=24\\
26=2b\\
b=13$$
Substituting in the value for b in the first two equations gives: a=17 and c=7.
Solution: The first workman takes 17 days to do the job. The second workman takes 13 days, and the third takes 7.
The obvious problem with this is that it can't be that when working together it can take one of the workmen longer to complete the work than when he works alone. However, this is what happens with workman c. And so obviously this technique does not work. Why not?
*Edit: My thought now is to divide each sum by four. This is because when they work together we need to half BOTH of their times. Otherwise if two people that take the same amount of time work together, they won't go any faster than usual. Does this make sense?
You want to set up your equations in terms of work rate. To make life easier, I'll pretend that the job is painting one house.
That is, suppose that worker 1 can paint one house in $a$ days. This means that they have a work rate of $(1/a)$ houses per day. And similarly say worker 2 has work rate $(1/b)$ houses per day.
Now, the work rate of a team is the sum of the individual work rates. Therefore your first equation should be $$\left(\frac 1a + \frac 1b\right) \text{ [houses / day] } * 15 \text{ [days] } = 1 \text{ [houses]}$$