Solve for x
x\in [-\frac{2}{9},2)
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9x+2\geq 0 x-2<0
For the quotient to be ≤0, one of the values 9x+2 and x-2 has to be ≥0, the other has to be ≤0, and x-2 cannot be zero. Consider the case when 9x+2\geq 0 and x-2 is negative.
x\in [-\frac{2}{9},2)
The solution satisfying both inequalities is x\in \left[-\frac{2}{9},2\right).
9x+2\leq 0 x-2>0
Consider the case when 9x+2\leq 0 and x-2 is positive.
x\in \emptyset
This is false for any x.
x\in [-\frac{2}{9},2)
The final solution is the union of the obtained solutions.
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