Solve for x
x\in (-4,2]
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2-x\leq 0 3x+12<0
For the quotient to be ≥0, 2-x and 3x+12 have to be both ≤0 or both ≥0, and 3x+12 cannot be zero. Consider the case when 2-x\leq 0 and 3x+12 is negative.
x\in \emptyset
This is false for any x.
2-x\geq 0 3x+12>0
Consider the case when 2-x\geq 0 and 3x+12 is positive.
x\in (-4,2]
The solution satisfying both inequalities is x\in \left(-4,2\right].
x\in (-4,2]
The final solution is the union of the obtained solutions.
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