Solve for x
x\in \left(-\frac{10}{7},\frac{1}{3}\right)
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7x+10<0 2-6x<0
For the quotient to be positive, 7x+10 and 2-6x have to be both negative or both positive. Consider the case when 7x+10 and 2-6x are both negative.
x\in \emptyset
This is false for any x.
2-6x>0 7x+10>0
Consider the case when 7x+10 and 2-6x are both positive.
x\in \left(-\frac{10}{7},\frac{1}{3}\right)
The solution satisfying both inequalities is x\in \left(-\frac{10}{7},\frac{1}{3}\right).
x\in \left(-\frac{10}{7},\frac{1}{3}\right)
The final solution is the union of the obtained solutions.
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