## Task B-1.1.

Determine all real numbers $p \neq 2$ such that for the solutions of the system of equations

$$
\left\{\begin{array}{r}
x+p y=3 \\
p x+4 y=6
\end{array}\right.
$$

it holds that $y-x<p$.
## Solution.

If we express $x=3-p y$ from the first equation and substitute it into the second equation, after rearranging we get

$$
y(2-p)(2+p)=3(2-p)
$$

Since $p \neq 2$, we can divide the equation by $(p-2)$, which results in $y(2+p)=3$. We conclude that $p$ must not be -2 either for the system to have solutions. In this case, we have $y=\frac{3}{p+2}, p \neq \pm 2$. From $x=3-p y$, it follows that $x=\frac{6}{2+p}, p \neq \pm 2$.

By substituting the obtained values of $x$ and $y$ into the condition $y-x>0$, we find that for the parameter $p$ that satisfies the given conditions, the following holds

$$
p>-2, p \neq 2
$$