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