4B. In the set of real numbers, solve the system

$$
\left\{\begin{array}{l}
\frac{3 x y}{x+y}=5 \\
\frac{2 x z}{x+z}=3 \\
\frac{y z}{y+z}=4
\end{array}\right.
$$
Solution. Since $x \neq 0, y \neq 0, z \neq 0$, the given system is equivalent to

$$
\left\{\begin{array} { l } 
{ \frac { 3 } { \frac { 1 } { y } + \frac { 1 } { x } } = 5 } \\
{ \frac { 2 } { \frac { 1 } { z } + \frac { 1 } { x } } = 3 , \text { or equivalently, } } \\
{ \frac { 1 } { \frac { 1 } { y } + \frac { 1 } { z } } = 4 }
\end{array} \left\{\begin{array}{l}
\frac{1}{y}+\frac{1}{x}=\frac{3}{5} \\
\frac{1}{z}+\frac{1}{x}=\frac{2}{3} \\
\frac{1}{z}+\frac{1}{y}=\frac{1}{4}
\end{array}\right.\right.
$$

Then, if we subtract the second equation from the first, we get $\frac{1}{y}-\frac{1}{x}=-\frac{1}{15}$. From

$\frac{1}{y}-\frac{1}{z}=-\frac{1}{15}$ and $\frac{1}{y}+\frac{1}{z}=\frac{1}{4}$ we get $\frac{1}{y}=\frac{11}{120}$ and $\frac{1}{z}=\frac{19}{120}$. Then $\frac{1}{x}=\frac{3}{5}-\frac{11}{120}=\frac{61}{120}$. Finally, $x=\frac{120}{61}, y=\frac{120}{11}$ and $z=\frac{120}{19}$.

## Third Year

1AB. For real numbers $x, y \in(0, \pi)$, the equality

$$
\cos 2 x \cos y-\cos 2 y \cos x=\cos y-\cos x
$$

is satisfied. Prove that $x=y$.