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