Two workers, $A$ and $B$, completed a task assigned to them as follows. First, only $A$ worked for $\frac{2}{3}$ of the time it would take $B$ to complete the entire task alone. Then, $B$ took over from $A$ and finished the work. In this way, the work took 2 hours longer than if they had started working together and completed it collaboratively. If they had worked together in the latter manner, $A$ would have done half as much work as $B$ actually left for $B$ to complete. How many hours would it take for $A$ and $B$ to complete the work individually?
I. Solution: Let the given work be a unit. Suppose that

$A$ would complete the work in $x$ hours, so in 1 hour $\frac{1}{x}$ work is done,

$B$ would complete the work in $y$ hours, so in 1 hour $\frac{1}{y}$ work is done.

Together, therefore, in 1 hour $\frac{1}{x}+\frac{1}{y}=\frac{x+y}{x y}$ work is done, so the entire work would be completed in $1: \frac{x+y}{x y}=\frac{x y}{x+y}$ hours together.

$A$ worked for $\frac{2}{3} y$ hours, and during this time $\frac{2 y}{3} \cdot \frac{1}{x}=\frac{2 y}{3 x}$ work was done, and $B$ was left with $1-\frac{2 y}{3 x}=\frac{3 x-2 y}{3 x}$ work.

If they worked together, $A$ would have done $\frac{x y}{x+y} \cdot \frac{1}{x}=\frac{y}{x+y}$ work in $\frac{x y}{x+y}$ hours, and according to the problem, this is half of the work $A$ actually left for $B$, that is,

$$
\frac{2 y}{x+y}=\frac{3 x-2 y}{3 x}
$$

$B$ completed the work left for him in $y \cdot \frac{3 x-2 y}{3 x}=\frac{3 x y-2 y^{2}}{3 x}$ hours. According to the problem, the sum of the working hours of $A$ and $B$ working one after the other exceeds the working time in which they would complete the work together by 2 hours. Therefore,

$$
\frac{2 y}{3}+\frac{3 x y-2 y^{2}}{3 x}=\frac{x y}{x+y}+2
$$

Rearranging (1),

$$
3 x^{2}-5 x y-2 y^{2}=0
$$

Solving this equation for $x$ (ignoring the root $x=-\frac{y}{3}$ as it is meaningless), we get

$$
x=2 y
$$

(Of course, we get the same result by dividing by $y^{2}$ and solving the resulting quadratic equation for $\frac{x}{y}$.) Substituting this value of $x$ into (2),

$$
\frac{4 y}{3}=\frac{2 y}{3}+2
$$

from which $y=3$, and thus $x=2 y=6$.

Therefore, $A$ would complete the work alone in 6 hours, and $B$ in 3 hours.

II. Solution: Below we provide a solution in which only one unknown appears in equation (1), and inference replaces equation (2).

The second relationship mentioned in the text does not contain absolute data, only the ratio of how $A$ and $B$ divide the work, so it is expected that this ratio can also be determined from it. Suppose that in the same time, $B$ does $\lambda$ times as much work as $A$. Then, working together, the entire work is divided between $A$ and $B$ in the ratio $1: \lambda$, so $A$ does $\frac{1}{1+\lambda}$ of the work, and $B$ does $\frac{\lambda}{1+\lambda}$ of the work. In fact, $A$ worked for as long as it would take $B$ to complete $\frac{2}{3}$ of the work. Since he does $\frac{1}{\lambda}$ times the work of $B$, he actually did $\frac{2}{3 \lambda}$ of the entire work, and thus left $1-\frac{2}{3 \lambda}$ of the work for $B$. According to the text, this is twice the work that would have been done by $A$ if they worked together, so

$$
1-\frac{2}{3 \lambda}=\frac{2}{1+\lambda}, \quad(3 \lambda-2)(1+\lambda)=6 \lambda, \quad 3 \lambda^{2}-5 \lambda-2=0
$$

Ignoring the negative root as it is meaningless, we get $\lambda=2$. This means that working together, $A$ would do $\frac{1}{3}$ of the work, and $B$ would do $\frac{2}{3}$ of the work. If $A$ worked for $\frac{2}{3}$ of the time it would take $B$ to complete the entire work, then he worked for as long as it would take them to complete the work together, and during this time he did $\frac{1}{3}$ of the work. Thus, $B$ worked for 2 hours according to the first part of the text and completed $\frac{2}{3}$ of the work, so he would complete the entire work in 3 hours. Since $B$ does twice as much work as $A$ in the same time, $A$ would complete the entire work in 6 hours.