Solve for x
x\in \left(-\infty,\frac{1}{3}\right)\cup \left(\frac{2}{5},\infty\right)
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3x-1>0 2-5x<0
For the quotient to be negative, 3x-1 and 2-5x have to be of the opposite signs. Consider the case when 3x-1 is positive and 2-5x is negative.
x>\frac{2}{5}
The solution satisfying both inequalities is x>\frac{2}{5}.
2-5x>0 3x-1<0
Consider the case when 2-5x is positive and 3x-1 is negative.
x<\frac{1}{3}
The solution satisfying both inequalities is x<\frac{1}{3}.
x>\frac{2}{5}\text{; }x<\frac{1}{3}
The final solution is the union of the obtained solutions.
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