Solve for x
x\in (-\infty,1)\cup [\frac{3}{2},\infty)
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3-2x\geq 0 x-1<0
For the quotient to be ≤0, one of the values 3-2x and x-1 has to be ≥0, the other has to be ≤0, and x-1 cannot be zero. Consider the case when 3-2x\geq 0 and x-1 is negative.
x<1
The solution satisfying both inequalities is x<1.
3-2x\leq 0 x-1>0
Consider the case when 3-2x\leq 0 and x-1 is positive.
x\geq \frac{3}{2}
The solution satisfying both inequalities is x\geq \frac{3}{2}.
x<1\text{; }x\geq \frac{3}{2}
The final solution is the union of the obtained solutions.
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