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