1. Two workers with a constant work rate can complete a job in 30 days. After 6 days of joint work, the first worker quits, and the second worker completes the job in 40 days from the day the first worker quits. How many days would it take for the first worker to complete the same job if he continued alone after 6 days of joint work?
Solution. Let the first worker complete the entire job in $a$ days, and the second in $b$ days. Then $x=\frac{1}{a}$ is the portion of the work the first worker completes in one day, and $y=\frac{1}{b}$ is the portion of the work the second worker completes in one day. Since both workers together complete the entire job in 30 days, we have

$$
30 x+30 y=1
$$

On the other hand, since after 6 days of joint work, the second worker works for another 40 days to complete the job, we have

$$
6 x+6 y+40 y=1
$$

Solving the system of linear equations (1) and (2), we get

$$
x=\frac{1}{75}, \quad y=\frac{1}{50}
$$

i.e., the first worker alone completes the entire job in 75 days, and the second in 50.

Let $z$ denote the number of days needed for the first worker to complete the job after 6 days of joint work. Then

$$
6 x+6 y+z x=1
$$

Substituting the values of $x$ and $y$ from (3) into (4), we get $z=60$. Thus, the first worker would need an additional 60 days to complete the job after 6 days of joint work.