Question

Solve $\frac{x-3}{x^2+4}\ge \frac{5}{x^2+4}$

Answer 1

$$\begin{gathered} \frac{x-3}{x^2+4}\ge \frac{5}{x^2+4} \\ \mathrm{We}\mathrm{have}{x}^2\ge 0 \\ \Rightarrow x^2+4>0 \\ \mathrm{So},\mathrm{multiplying}\mathrm{by}\left(x^2+4\right)\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ x-3\ge 5 \\ \Rightarrow x\ge 8 \end{gathered}$$