Solve the inequality $LaTeX: \displaystyle \frac{8}{x^{2} - 16}<\frac{4}{x^{2} - 3 x - 4}$

Getting zero on one side and factoring gives $LaTeX: \displaystyle \frac{8}{\left(x - 4\right) \left(x + 4\right)} - \frac{4}{\left(x - 4\right) \left(x + 1\right)}< 0$ . This gives the least common denominator as $LaTeX: \displaystyle \left(x - 4\right) \left(x + 1\right) \left(x + 4\right)$ . Building each fraction to get the common denominator gives $LaTeX: \displaystyle \frac{8 x + 8 - (4 x + 16)}{\left(x - 4\right) \left(x + 1\right) \left(x + 4\right)} < 0$ . Simplifying gives $LaTeX: \displaystyle \frac{4 x - 8}{\left(x - 4\right) \left(x + 1\right) \left(x + 4\right)}<0$ . The inequality can change signs at the zeros of the numerator, $LaTeX: \displaystyle \left\{2\right\}$ , or the zeros of the denominator $LaTeX: \displaystyle \left\{-4, -1, 4\right\}$ . Making a sign chart gives: This gives the solution $LaTeX: \displaystyle \left(-4, -1\right) \cup \left(2, 4\right)$