Answer:
[tex]\large\boxed{x<-1\dfrac{1}{2}\ or\ x>9\dfrac{3}{5}\to x\in\left(-\infty,\ -1\dfrac{1}{2}\right)\ \cup\ \left(9\dfrac{3}{5},\ \infty\right)}[/tex]
Step-by-step explanation:
[tex]12x+7<-11\qquad\text{subtrct 7 from both sides}\\\\12x+7-7<-11-7\\\\12x<-18\qquad\text{divide both sides by 12}\\\\\dfrac{12x}{12}<\dfrac{-18}{12}\\\\x<-\dfrac{18:6}{12:6}\\\\x<-\dfrac{3}{2}\\\\x<-1\dfrac{1}{2}\\===========================[/tex]
[tex]5x-8>40\qquad\text{add 8 to both sides}\\\\5x-8+8>40+8\\\\5x>48\qquad\text{divide both sides by 5}\\\\\dfrac{5x}{5}>\dfrac{48}{5}\\\\x>\dfrac{48}{5}\\\\x>9\dfrac{3}{5}\\===========================[/tex]
[tex]x<-1\dfrac{1}{2}\ or\ x>9\dfrac{3}{5}[/tex]
