Solve for x
x\in (\frac{1}{3},\frac{4}{3}]
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3x-4\leq 0 2-6x<0
For the quotient to be ≥0, 3x-4 and 2-6x have to be both ≤0 or both ≥0, and 2-6x cannot be zero. Consider the case when 3x-4\leq 0 and 2-6x is negative.
x\in (\frac{1}{3},\frac{4}{3}]
The solution satisfying both inequalities is x\in \left(\frac{1}{3},\frac{4}{3}\right].
3x-4\geq 0 2-6x>0
Consider the case when 3x-4\geq 0 and 2-6x is positive.
x\in \emptyset
This is false for any x.
x\in (\frac{1}{3},\frac{4}{3}]
The final solution is the union of the obtained solutions.
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