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