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