Solve for x
x\in (-\infty,-2]\cup (\frac{5}{3},\infty)
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x+2\leq 0 3x-5<0
For the quotient to be ≥0, x+2 and 3x-5 have to be both ≤0 or both ≥0, and 3x-5 cannot be zero. Consider the case when x+2\leq 0 and 3x-5 is negative.
x\leq -2
The solution satisfying both inequalities is x\leq -2.
x+2\geq 0 3x-5>0
Consider the case when x+2\geq 0 and 3x-5 is positive.
x>\frac{5}{3}
The solution satisfying both inequalities is x>\frac{5}{3}.
x\leq -2\text{; }x>\frac{5}{3}
The final solution is the union of the obtained solutions.
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