Answer
The required inequality is $\frac{x-3}{x+4}\ge 0$.
Work Step by Step
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:
$(-\infty,-4)\cup [3,\infty)$
