QuestionIf $x+z \propto y$ and $y+z \propto x$, prove $$x+y \propto z$$

Studdy Solution
Assuming $1 - k_2 \cdot k_1 \neq 0$, solve for $y$ in terms of $z$:
$$y = \frac{-z(k_2 + 1)}{1 - k_2 \cdot k_1}$$
Substitute back to express $x$ in terms of $z$ using $x = k_1 \cdot y - z$:
$$x = k_1 \cdot \frac{-z(k_2 + 1)}{1 - k_2 \cdot k_1} - z$$
Combine $x$ and $y$:
$$x + y = \frac{-z(k_1 \cdot (k_2 + 1) + (k_2 + 1))}{1 - k_2 \cdot k_1}$$
Simplify:
$$x + y = \frac{-z(k_1 + 1)(k_2 + 1)}{1 - k_2 \cdot k_1}$$
Thus, $x + y \propto z$.