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