Solve for x
x\in (2,\frac{18}{7}]
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18-7x\leq 0 3x-6<0
For the quotient to be ≥0, 18-7x and 3x-6 have to be both ≤0 or both ≥0, and 3x-6 cannot be zero. Consider the case when 18-7x\leq 0 and 3x-6 is negative.
x\in \emptyset
This is false for any x.
18-7x\geq 0 3x-6>0
Consider the case when 18-7x\geq 0 and 3x-6 is positive.
x\in (2,\frac{18}{7}]
The solution satisfying both inequalities is x\in \left(2,\frac{18}{7}\right].
x\in (2,\frac{18}{7}]
The final solution is the union of the obtained solutions.
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