## Question

- Find the solution set for \(x\) given that the following three relations for \(x\), \(y\), where \(x, y \in \mathbb{R}\), are simultaneously true:
\[\begin{equation*}
y < x + 1, \quad y + 6x < 20, \quad x = 5y - 7.
\end{equation*}\]
- Find the solution set of the inequality
\[\begin{equation*}
\frac{12}{x-3} < x + 1, \quad (x \in \mathbb{R}, x \ne 3).
\end{equation*}\]