Solve for x
x\in \left(-\frac{7}{3},\frac{5}{2}\right)
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3x+7>0 2x-5<0
For the product to be negative, 3x+7 and 2x-5 have to be of the opposite signs. Consider the case when 3x+7 is positive and 2x-5 is negative.
x\in \left(-\frac{7}{3},\frac{5}{2}\right)
The solution satisfying both inequalities is x\in \left(-\frac{7}{3},\frac{5}{2}\right).
2x-5>0 3x+7<0
Consider the case when 2x-5 is positive and 3x+7 is negative.
x\in \emptyset
This is false for any x.
x\in \left(-\frac{7}{3},\frac{5}{2}\right)
The final solution is the union of the obtained solutions.
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