Solve 2x-3/x-5>0

$$\frac{ 2x-3 }{ x-5 } > 0$$

Solve for x

$$x\in \left(-\infty,\frac{3}{2}\right)\cup \left(5,\infty\right)$$

Solution Steps

For the quotient to be positive, 2x-3 and x-5 have to be both negative or both positive. Consider the case when 2x-3 and x-5 are both negative.

$$2x-3<0 x-5<0$$

The solution satisfying both inequalities is x<\frac{3}{2}.

$$x<\frac{3}{2}$$

Consider the case when 2x-3 and x-5 are both positive.

$$x-5>0 2x-3>0$$

The solution satisfying both inequalities is x>5.

$$x>5$$

The final solution is the union of the obtained solutions.

$$x<\frac{3}{2}\text{; }x>5$$