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