Solve for x
x\in (-\infty,\frac{1}{3}]\cup (\frac{2}{3},\infty)
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2-6x\geq 0 3x-2<0
For the quotient to be ≤0, one of the values 2-6x and 3x-2 has to be ≥0, the other has to be ≤0, and 3x-2 cannot be zero. Consider the case when 2-6x\geq 0 and 3x-2 is negative.
x\leq \frac{1}{3}
The solution satisfying both inequalities is x\leq \frac{1}{3}.
2-6x\leq 0 3x-2>0
Consider the case when 2-6x\leq 0 and 3x-2 is positive.
x>\frac{2}{3}
The solution satisfying both inequalities is x>\frac{2}{3}.
x\leq \frac{1}{3}\text{; }x>\frac{2}{3}
The final solution is the union of the obtained solutions.
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