Solve for x
x\in \left(-\infty,-4\right)\cup \left(1,\infty\right)
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\frac{x-1}{x+4}>0
Add 2 and 2 to get 4.
x-1<0 x+4<0
For the quotient to be positive, x-1 and x+4 have to be both negative or both positive. Consider the case when x-1 and x+4 are both negative.
x<-4
The solution satisfying both inequalities is x<-4.
x+4>0 x-1>0
Consider the case when x-1 and x+4 are both positive.
x>1
The solution satisfying both inequalities is x>1.
x<-4\text{; }x>1
The final solution is the union of the obtained solutions.
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