Solve for x
x\in (-\infty,-\frac{3}{2}]\cup [5,\infty)
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2x+3\leq 0 x-5\leq 0
For the product to be ≥0, 2x+3 and x-5 have to be both ≤0 or both ≥0. Consider the case when 2x+3 and x-5 are both ≤0.
x\leq -\frac{3}{2}
The solution satisfying both inequalities is x\leq -\frac{3}{2}.
x-5\geq 0 2x+3\geq 0
Consider the case when 2x+3 and x-5 are both ≥0.
x\geq 5
The solution satisfying both inequalities is x\geq 5.
x\leq -\frac{3}{2}\text{; }x\geq 5
The final solution is the union of the obtained solutions.
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