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