Solve for x
x\in [-\frac{1}{2},1)
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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 [-\frac{1}{2},1)
The solution satisfying both inequalities is x\in \left[-\frac{1}{2},1\right).
2x+1\leq 0 x-1>0
Consider the case when 2x+1\leq 0 and x-1 is positive.
x\in \emptyset
This is false for any x.
x\in [-\frac{1}{2},1)
The final solution is the union of the obtained solutions.
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