In 6 years, my father will be 3 times as old as I was when my father's age was equal to the sum of my age and my sister's age at that time. My current age is the same as my father's age was then. In 19 years, my father will be twice as old as my sister is now. How old are these three people today?
I. Solution: Let the age of my father, mine, and my sister's be $x, y, z$ respectively. Suppose that $u$ years ago, my father's age was equal to the sum of my age and my sister's age at that time. According to the problem,

$$
\begin{aligned}
x+6 & =3(y-u) \\
x-u & =(y-u)+(z-u) \\
y & =x-u \\
x+19 & =2 z
\end{aligned}
$$

Thus, we have 4 equations for the 4 unknowns.

From (2) and (3), we get $y=y+z-2 u$, hence

$$
z=2 u
$$

Substituting this value of $z$ into (4),

$$
x=4 u-19
$$

and thus from (3),

$$
y=x-u=3 u-19
$$

Substituting these values of $x$ and $y$ into (1),

$$
4 u-13=3(2 u-19)=6 u-57
$$

from which

$$
u=22
$$

and thus

$$
x=4 u-19=69, \quad y=3 u-19=47, \quad z=2 u=44
$$

Ádám Antal (Bp. VIII., Széchenyi g. II. o. t.)

II. Solution: We can also reach the goal with a system of equations with two unknowns.

Indeed, at a certain point in time, I was $x$ years old, and my sister was $y$ years old, while my father's age was $x+y$ years. The difference between my father's age and my age at that time was thus $y$ years. Since according to the problem, I am currently as old as my father was at the mentioned time, I am currently $x+y$ years old, which means $y$ years have passed since the mentioned time, so my father is now $x+2 y$ and my sister is $2 y$ years old.

According to the problem,

$$
\begin{array}{r}
x+2 y+6=3 x \\
x+2 y+19=4 y
\end{array}
$$

From this system of equations, $x=25, y=22$.

So currently, my father is $x+2 y=69$ years old

$$
\begin{aligned} I \(\quad x+y & =47\) years old \\ sister \(\quad 2 y & =44\) years old. \end{aligned}
$$