Solve for x
x\in (-\infty,-\frac{2}{3}]\cup (5,\infty)
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9x+6\leq 0 5x-25<0
For the quotient to be ≥0, 9x+6 and 5x-25 have to be both ≤0 or both ≥0, and 5x-25 cannot be zero. Consider the case when 9x+6\leq 0 and 5x-25 is negative.
x\leq -\frac{2}{3}
The solution satisfying both inequalities is x\leq -\frac{2}{3}.
9x+6\geq 0 5x-25>0
Consider the case when 9x+6\geq 0 and 5x-25 is positive.
x>5
The solution satisfying both inequalities is x>5.
x\leq -\frac{2}{3}\text{; }x>5
The final solution is the union of the obtained solutions.
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