I have a statement that says:
A teacher A does a job in 6 hours, while a teacher B runs it in the twice as long as the teacher A. In how many hours would they work together?
So, I did the trivial thing:
$w^{-1} = \frac{1}{6} + \frac{1}{3}$
$w^{-1} = \frac{1}{2}$
$w = 2$
Then, they take $2$ hours to do the work together. I have solved several exercises in this way and they were fine, but why not here? , since the answer should be $ 4 $
Teacher A completes one unit of work in $6$ hours, so has a rate of completion of $\frac{1 \text{ unit of work}}{6 \text{ hours}}$. Teacher B completes one unit of work in twice $6$ hours, which is $12$ hours, so has a rate of completion of $\frac{1 \text{ unit of work}}{12 \text{ hours}}$. Together they have a rate of completion of \begin{align*} & \frac{1 \text{ unit of work}}{6 \text{ hours}} + \frac{1 \text{ unit of work}}{12 \text{ hours}} \\ =& \frac{2 \text{ units of work}}{12 \text{ hours}} + \frac{1 \text{ unit of work}}{12 \text{ hours}} \\ =& \frac{3 \text{ units of work}}{12 \text{ hours}} \\ =& \frac{1 \text{ unit of work}}{4 \text{ hours}} \text{.} \end{align*} Therefore, we expect the pair of teachers to complete the work in $4$ hours.