Solve for x
x\in \left(-\frac{1}{3},\frac{1}{2}\right)
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3x+1>0 4x-2<0
For the quotient to be negative, 3x+1 and 4x-2 have to be of the opposite signs. Consider the case when 3x+1 is positive and 4x-2 is negative.
x\in \left(-\frac{1}{3},\frac{1}{2}\right)
The solution satisfying both inequalities is x\in \left(-\frac{1}{3},\frac{1}{2}\right).
4x-2>0 3x+1<0
Consider the case when 4x-2 is positive and 3x+1 is negative.
x\in \emptyset
This is false for any x.
x\in \left(-\frac{1}{3},\frac{1}{2}\right)
The final solution is the union of the obtained solutions.
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