Solve for x
x\in (-\infty,-3)\cup [\frac{1}{2},\infty)
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2x-1\leq 0 x+3<0
For the quotient to be ≥0, 2x-1 and x+3 have to be both ≤0 or both ≥0, and x+3 cannot be zero. Consider the case when 2x-1\leq 0 and x+3 is negative.
x<-3
The solution satisfying both inequalities is x<-3.
2x-1\geq 0 x+3>0
Consider the case when 2x-1\geq 0 and x+3 is positive.
x\geq \frac{1}{2}
The solution satisfying both inequalities is x\geq \frac{1}{2}.
x<-3\text{; }x\geq \frac{1}{2}
The final solution is the union of the obtained solutions.
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