How do I solve this problem

asked Nov 22, 2019 119k views

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by Tauna

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$9p > 63\ \ \ \ |:9\\\\p > 7\\--------------\\\\(p)/(3)\leq2\ \ \ \ |\cdot3\\\\p\leq6$

Answer:

On the picture.

$p\in(-\infty;\ 6]\ \cup\ (7,\ \infty)$

Heremyas answered Nov 25, 2019

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1 vote

$\bf 9p>63\implies p>\cfrac{63}{9}\implies \boxed{p>7}\\\\\\\cfrac{p}{3}\le 2\implies \boxed{p\le 6}$

check the picture below.

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