Textbook Problem

The product $\left(x-2\right)\left(x+3\right)$ is positive if both factors are negative or if both factors are positive. Therefore, we can solve $\left(x-2\right)\left(x+3\right)$ as follows,

$\begin{array}{r}\left(x-2<\text{and}x+3<0\right)\\ \left(x<2\text{and}x<-3\right)\\ x<-3\end{array}$ $\begin{array}{c}or\\ or\\ or\end{array}$ $\begin{array}{r}\left(x-2>0\text{and}x+3>0\right)\\ \left(x>2\text{and}x>-3\right)\\ x<2\end{array}$

The solution set is $\left(-\infty ,-3\right)\cup \left(2,\infty \right)$. Use this type of analysis to solve each of the following.

(a) $\left(x-2\right)\left(x+7\right)>0$	(b) $\left(x-3\right)\left(x+9\right)\ge 0$
(c) $\left(x+1\right)\left(x-6\right)\le 0$	(d) $\left(x+4\right)\left(x-8\right)<0$
(e) $\frac{x+4}{x-7}>0$	(f) $\frac{x+4}{x-7}>0$