Solve for x
x\in (\frac{1}{2},4]
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3x-12\leq 0 1-2x<0
For the quotient to be ≥0, 3x-12 and 1-2x have to be both ≤0 or both ≥0, and 1-2x cannot be zero. Consider the case when 3x-12\leq 0 and 1-2x is negative.
x\in (\frac{1}{2},4]
The solution satisfying both inequalities is x\in \left(\frac{1}{2},4\right].
3x-12\geq 0 1-2x>0
Consider the case when 3x-12\geq 0 and 1-2x is positive.
x\in \emptyset
This is false for any x.
x\in (\frac{1}{2},4]
The final solution is the union of the obtained solutions.
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