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