Suggestion

  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: \[y < x+1, \quad y+6x < 20, \quad x = 5y-7.\]

Could we sketch the lines \(y=x+1\), \(y+6x=20\) and \(x=5y-7\)?

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

How else could we write the inequality?

Could we draw a sketch?