Answer:
Step-by-step explanation:
[tex]-2<\dfrac{x}{3}+1<5\qquad\text{subtract 1 from both sides}\\\\-2-1<\dfrac{x}{3}+1-1<5-1\\\\-3<\dfrac{x}{3}<4\qquad\text{multiply both sides by 3}\\\\(3)(-3)<3\!\!\!\!\diagup^1\cdot\dfrac{x}{3\!\!\!\!\diagup_1}<(3)(4)\\\\-9<x<12[/tex]
