Answer:
[tex]\huge\boxed{x\geq-9}[/tex]
Step-by-step explanation:
[tex]\dfrac{-x}{3}-2\leq1\qquad|\text{add 2 to both sides}\\\\\dfrac{-x}{3}-2+2\leq1+2\\\\\dfrac{-x}{3}\leq3\qquad|\text{multiply both sides by 3}\\\\3\!\!\!\!\diagup\cdot\dfrac{-x}{3\!\!\!\!\diagup}\leq3\cdot3\\\\-x\leq9\qquad|\text{change the signs}\\\\x\geq-9[/tex]
