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

We might get some useful insight into what is happening here by drawing a sketch.

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

Again, a sketch might be helpful here, to help check that our answer is sensible. Remember that \(x-3\) might be positive or negative.