Answer
$(-4,-11/3]$ U $[0,∞)$
Work Step by Step
$p/(p+4) \leq 3p$
$p/(p+4)*(p+4) \leq 3p*(p+4)$
$p \leq 3p^2+12p$
$p-p \leq 3p^2+12p-p$
$0 \leq 3p^2+11p$
$0 \leq p(3p+11)$
$0 = p$
$3p+11=0$
$3p+11-11=0-11$
$3p=-11$
$3p/3 = -11/3$
$0 = -11/3$
The denominator is zero when $p=-4$.
We have four sections: $(−∞,-4)$, $(-4,-11/3)$, $(-11/3,0)$, and $(0,∞)$. We need to test one value for x in each section to determine if the section would be a solution set. Since we have the $\leq$ sign, we include the end points and use brackets instead of parentheses.
Let $p=-5$, $p=-15/4$, $p=-2$, $p=5$
$p=-5$
$p/(p+4) \leq 3p$
$-5/(-5+4) \leq 3*-5$
$-5/-1 \leq -15$
$5 \leq -15$ (false)
$p=-15/4$
$p/(p+4) \leq 3p$
$-3.75/(-3.75+4) \leq 3*-3.75$
$-3.75/.25 \leq -11.25$
$-15 \leq -11.25$ (true)
$p=-2$
$p/(p+4) \leq 3p$
$-2/(-2+4) \leq 3*-2$
$-2/2 \leq -6$
$-1 \leq -6$ (false)
$p=5$
$p/(p+4) \leq 3p$
$5/(5+4) \leq 3*5$
$5/9 \leq 15$ (true)
