Task B-1.6. If $\frac{z}{x+y}=2$ and $\frac{y}{x+z}=3$, what is the value of $\frac{z}{y+z}$?
Solution. From the assumptions, we get

$$
\begin{aligned}
& z=2 x+2 y \\
& y=3 x+3 z
\end{aligned}
$$

If we express $y$ and $z$ in terms of $x$, we get $y=-\frac{9}{5} x, z=-\frac{8}{5} x$.

Substituting this into the expression $\frac{z}{y+z}$, we get

$$
\frac{z}{y+z}=\frac{-\frac{8}{5} x}{-\frac{9}{5} x-\frac{8}{5} x}=\frac{8}{17}
$$

Note that $x$ cannot be $0$, otherwise the conditions of the problem would state $\frac{z}{y}=2$ and $\frac{y}{z}=3$, which cannot both be true. Therefore, the above calculation is valid.

(2 points)