Chapter 3 - Solving Inequalities - Chapter Review - Page 226: 60

Basically, the intersection of $N$ and $P$ would be all elements in set $P$ because all numbers that are multiples of $6$ are also divisible by $2$. $N ∩ P$ = {$x|x$ is a multiple of $6$}

Basically, the intersection of $N$ and $P$ would be all elements in set $P$ because all numbers that are multiples of $6$ are also divisible by $2$. $N ∩ P$ = {$x|x$ is a multiple of $6$}