Solve for x
x\in (-\infty,-2)\cup [-1,\infty)
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x+1\leq 0 3x+6<0
For the quotient to be ≥0, x+1 and 3x+6 have to be both ≤0 or both ≥0, and 3x+6 cannot be zero. Consider the case when x+1\leq 0 and 3x+6 is negative.
x<-2
The solution satisfying both inequalities is x<-2.
x+1\geq 0 3x+6>0
Consider the case when x+1\geq 0 and 3x+6 is positive.
x\geq -1
The solution satisfying both inequalities is x\geq -1.
x<-2\text{; }x\geq -1
The final solution is the union of the obtained solutions.
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