Appendix A - Appendix A Exercises - Page 1158: 28

$$\left( { - \infty , - \frac{{11}}{5}} \right) \cup \left( { - 2,\infty } \right)$$

$$\eqalign{ & \frac{{x + 1}}{{x + 2}} < 6 \cr & {\text{subtract 6 from both sides}} \cr & \frac{{x + 1}}{{x + 2}} - 6 < 0 \cr & \frac{{x + 1 - 6\left( {x + 2} \right)}}{{x + 2}} < 0 \cr & \frac{{x + 1 - 6x - 12}}{{x + 2}} < 0 \cr & \frac{{ - 5x - 11}}{{x + 2}} < 0 \cr & {\text{The critical values are}} \cr & - 5x - 11 = 0\,\,\,\,\,\,\,and\,\,\,\,\,\,\,x + 2 = 0 \cr & x = - \frac{{11}}{5}{\text{ and }}x = - 2 \cr & \cr & {\text{ These critical values partition the number line into}} \cr & {\text{three intervals:}}\,\,\,\left( { - \infty , - \frac{{11}}{5}} \right),\,\,\left( { - \frac{{11}}{5}, - 2} \right){\text{ and }}\left( { - 2,\infty } \right).{\text{ }} \cr & {\text{Testing the intervals:}} \cr & {\text{*For the interval }}\left( { - \infty , - \frac{{11}}{5}} \right){\text{ use the test value }}x = - 3 \cr & \frac{{ - 5\left( { - 3} \right) - 11}}{{ - 3 + 2}} = - 4,\,\,\,negative \cr & {\text{*For the interval }}\left( { - \frac{{11}}{5}, - 2} \right){\text{ use the test value }}x = - 2.1 \cr & \frac{{ - 5\left( { - 2.1} \right) - 11}}{{ - 2.1 + 2}} = 5,\,\,\,positive \cr & {\text{*For the interval }}\left( { - 2,\infty } \right){\text{ use the test value }}x = - 1 \cr & \frac{{ - 5\left( { - 1} \right) - 11}}{{ - 1 + 2}} = - 6,\,\,negative \cr & \cr & {\text{Therefore, }}\frac{{ - 5x - 11}}{{x + 2}} < 0{\text{ on the intervals }}\left( { - \infty , - \frac{{11}}{5}} \right) \cr & and\,\,\left( { - 2,\infty } \right) \cr & then \cr & {\text{The solution set of the inequailty }}{x^2} - 6x + 5 < 0{\text{ is}} \cr & \left( { - \infty , - \frac{{11}}{5}} \right) \cup \left( { - 2,\infty } \right) \cr & \cr & {\text{Graph}} \cr} $$