- 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 \[\begin{equation*} \frac{12}{x-3} < x + 1, \quad (x \in \mathbb{R}, x \ne 3). \end{equation*}\]
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