Solve for x
x\in (-\frac{1}{2},2]
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4-2x\leq 0 2x+1<0
For the quotient to be ≥0, 4-2x and 2x+1 have to be both ≤0 or both ≥0, and 2x+1 cannot be zero. Consider the case when 4-2x\leq 0 and 2x+1 is negative.
x\in \emptyset
This is false for any x.
4-2x\geq 0 2x+1>0
Consider the case when 4-2x\geq 0 and 2x+1 is positive.
x\in (-\frac{1}{2},2]
The solution satisfying both inequalities is x\in \left(-\frac{1}{2},2\right].
x\in (-\frac{1}{2},2]
The final solution is the union of the obtained solutions.
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