Originally Posted by

**ThePerfectHacker** Let us first assume the first case,

$\displaystyle x^2-1<0$

$\displaystyle x^2<1$

$\displaystyle |x|<1$ *IMPORTANT*

Multiplying both sides by the denominator will flip the inequality sign because it is negative,

$\displaystyle x^2+8x+15\leq 0$

$\displaystyle x^2+8x \leq -15$

$\displaystyle x^2+8x+16 \leq -15+16$

$\displaystyle (x+4)^2 \leq 1$

$\displaystyle -1\leq x+4 \leq 1$

$\displaystyle 3\leq x\leq 5$

Look at the important message, we assumed that $\displaystyle |x|<1$ the only ones that work are $\displaystyle -1<x<1$. Thus, that is one solution,

$\displaystyle x=(-1,1)$