Task B-1.6. Professor Algebraic and Professor Korijenko are talking. Prof Algebraic: however fast the years pass,... I already have $25 \%$ more years, than you had when I had as many years as you have now. Prof. Korijenko: eee... if our health serves us, when I have as many years as you have now, together we will have 168 years. How old is each professor?
## Solution.

Let Prof. Algebraic be $x$ years old, and Prof. Korijenko be $y$ years old.

Prof. Algebraic is older than Prof. Korijenko by $x-y$ years, so Prof. Algebraic was $y$ years old $x-y$ years ago. At that time, Prof. Korijenko was $y-(x-y)=2 y-x$ years old.

$$
x=1.25(2 y-x)
$$

When Prof. Korijenko is $x$ years old, Prof. Algebraic will be $x+x-y=2 x-y$ years old, so we have:

$$
x+2 x-y=168
$$

We have the system of equations:

$$
\left\{\begin{aligned}
y & =0.9 x \\
3 x-y & =168
\end{aligned}\right.
$$

The solutions are $x=80, y=72$.

Prof. Algebraic is 80 years old, and Prof. Korijenko is 72.

(2 points)