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