Solve for x
x\in (-\infty,-\frac{7}{2}]\cup [\frac{2}{3},\infty)
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2x+7\leq 0 3x-2\leq 0
For the product to be ≥0, 2x+7 and 3x-2 have to be both ≤0 or both ≥0. Consider the case when 2x+7 and 3x-2 are both ≤0.
x\leq -\frac{7}{2}
The solution satisfying both inequalities is x\leq -\frac{7}{2}.
3x-2\geq 0 2x+7\geq 0
Consider the case when 2x+7 and 3x-2 are both ≥0.
x\geq \frac{2}{3}
The solution satisfying both inequalities is x\geq \frac{2}{3}.
x\leq -\frac{7}{2}\text{; }x\geq \frac{2}{3}
The final solution is the union of the obtained solutions.
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