A basin is filled by three pipes. The first and second pipes fill the basin in 70 minutes, the first and third in 84 minutes, and the second and third in 140 minutes. How many minutes does it take to fill the basin through each pipe individually, and how many minutes does it take if all three pipes are open at the same time?
Suppose the first pipe fills the tank in $x$ minutes, the second in $y$ minutes, and the third in $z$ minutes. According to the problem, the first and second pipes together fill $\frac{1}{70}$ of the tank in one minute, so

$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{70}
$$

Similarly,

$$
\frac{1}{x}+\frac{1}{z}=\frac{1}{84}
$$

and

$$
\frac{1}{y}+\frac{1}{z}=\frac{1}{140}
$$

From $2^{\circ}$ and $3^{\circ}$,

$$
\frac{1}{x}-\frac{1}{y}=\frac{1}{84}-\frac{1}{140}=\frac{1}{210}
$$

From $1^{\circ}$ and $4^{\circ}$,

$$
\frac{2}{x}+\frac{1}{70}+\frac{1}{210}=\frac{2}{105}
$$

which means

$$
x=105 \text { minutes, }
$$

so

$$
y=210 \text { minutes }
$$

and

$$
z=420 \text { minutes. }
$$

If all three pipes fill the tank in $u$ minutes, then

$$
\frac{1}{u}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{105}+\frac{1}{210}+\frac{1}{420}=\frac{1}{60}
$$

which means if all three pipes are open, the tank will be filled in one hour.