Resolver em $\mathbb{R}$: $|2x + 1| - |x - 3| = 6$.

$|2x + 1| = 6 + |x - 3|$

$p:\ 2x + 1 = 6 + |x - 3|\ \vee\ q:\ 2x + 1 = -6 - |x - 3|$

$p:\ |x - 3| = 2x - 5\ \Rightarrow\ x - 3 = 2x - 5\ \vee\ 5 - 2x = x - 3\ \Rightarrow$

$\Rightarrow\ x = 2\ \vee\ x = \dfrac{8}{3}$

$q:\ |x - 3| = -7 - 2x\ \Rightarrow\ x - 3 = -7 - 2x\ \vee\ x - 3 = 7 + 2x\ \Rightarrow$

$\Rightarrow\ x = -\dfrac{4}{3}\ \vee\ x = -10$

$\fbox{$S = \left\{2, \dfrac{8}{3}, -\dfrac{4}{3}, -10\right\}$}$