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