## Precalculus (6th Edition) Blitzer

The required inequality is $\frac{x-3}{x+4}\ge 0$.
Consider the inequality is $\frac{x-3}{x+4}\ge 0$. Multiply both sides by ${{\left( x+4 \right)}^{2}},x\ne -4$. ${{\left( x+4 \right)}^{2}}\frac{x-3}{x+4}\ge 0\times {{\left( x+4 \right)}^{2}}$ Solve further: $\left( x+4 \right)\left( x-3 \right)\ge 0$ Which implies the solution: $(-\infty,-4)\cup [3,\infty)$