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