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