What set of integers fall in the solution set of the following inequality:1/3x<3x+8<-5x

Answer:-2Step-by-step explanation:[tex]\dfrac{1}{3}x<3x+8<-5x\qquad\text{multiply all sides by 3}\\\\3\!\!\!\!\diagup^1\cdot\dfrac{1}{3\!\!\!\!\diagup_1}x<3\cdot3x+3\cdot8<3\cdot(-5x)\\\\x<9x+24<-15x\\\\\text{Let split it into two inequalities}\\\\(1)\qquad x<9x+24\ \text{and}\qquad (2)\qquad 9x+24<-15x[/tex][tex](1)\\x<9x+24\qquad\text{subtract}\ 9x\ \text{from both sides}\\\\-8x<24\qquad\text{change the signs}\\\\8x>-24\qquad\text{divide both sides by 8}\\\\\boxed{x>-3}\\\\(2)\\9x+24<-15x\qquad\text{add}\ 15x\ \text{to both sides}\\\\24x+24<0\qquad\text{subtract 24 from both sides}\\\\24x<-24\qquad\text{divide both sides by 24}\\\\\boxed{x<-1}\\\\\text{From (1) and (2) we have}\\\\-3<x<-1\to\text{There is only one integer in this solution: -2}[/tex]