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