Solve for x
x\in [-\frac{2}{3},3)
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3x+2\geq 0 2x-6<0
For the quotient to be ≤0, one of the values 3x+2 and 2x-6 has to be ≥0, the other has to be ≤0, and 2x-6 cannot be zero. Consider the case when 3x+2\geq 0 and 2x-6 is negative.
x\in [-\frac{2}{3},3)
The solution satisfying both inequalities is x\in \left[-\frac{2}{3},3\right).
3x+2\leq 0 2x-6>0
Consider the case when 3x+2\leq 0 and 2x-6 is positive.
x\in \emptyset
This is false for any x.
x\in [-\frac{2}{3},3)
The final solution is the union of the obtained solutions.
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