Answer: C
Step-by-step explanation:
4p + 1 > -7
⇔ 4p > -8
⇔ p > -2
=> S1 = (-2; +∞)
6p + 3 < 33
⇔ 6p < 30
⇔ p < 5
=> S2 = (-∞; 5)
=> S = S1 ∪ S2 = (-2; +∞) ∪ (-∞; 5) = R
=> p ∈ R
