[tex] \bf 9p>63\implies p>\cfrac{63}{9}\implies \boxed{p>7}\\\\\\\cfrac{p}{3}\le 2\implies \boxed{p\le 6} [/tex]
check the picture below.
[tex]9p > 63\ \ \ \ |:9\\\\p > 7\\--------------\\\\\dfrac{p}{3}\leq2\ \ \ \ |\cdot3\\\\p\leq6[/tex]
Answer:
On the picture.
[tex]p\in(-\infty;\ 6]\ \cup\ (7,\ \infty)[/tex]
