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