Answer:
[tex]\huge\boxed{-10<x<-3\to x\in(-10;\ -3)}[/tex]
Step-by-step explanation:
[tex]-36<3x-6<-15\qquad|\text{add 6 to all sides}\\\\-36+6<3x-6+6<-15+6\\\\-30<3x<-9\qquad|\text{divide both sides by 3}\\\\\dfrac{-30}{3}<\dfrac{3x}{3}<\dfrac{-9}{3}\\\\-10<x<-3\to x\in(-10;\ -3)[/tex]
