Solve for x
x\in (-2,-\frac{1}{3}]
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3x+1\leq 0 -x-2<0
For the quotient to be ≥0, 3x+1 and -x-2 have to be both ≤0 or both ≥0, and -x-2 cannot be zero. Consider the case when 3x+1\leq 0 and -x-2 is negative.
x\in (-2,-\frac{1}{3}]
The solution satisfying both inequalities is x\in \left(-2,-\frac{1}{3}\right].
3x+1\geq 0 -x-2>0
Consider the case when 3x+1\geq 0 and -x-2 is positive.
x\in \emptyset
This is false for any x.
x\in (-2,-\frac{1}{3}]
The final solution is the union of the obtained solutions.
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