Solve for x
x\in (-\infty,-\frac{1}{2})\cup [1,\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<-\frac{1}{2}
The solution satisfying both inequalities is x<-\frac{1}{2}.
x-1\geq 0 2x+1>0
Consider the case when x-1\geq 0 and 2x+1 is positive.
x\geq 1
The solution satisfying both inequalities is x\geq 1.
x<-\frac{1}{2}\text{; }x\geq 1
The final solution is the union of the obtained solutions.
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