When are $y < x+1, y+6x < 20, x = 5y-7$ true together?

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

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