How to solve (x+3)/(x-5) > 0

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

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

$x\in \left(-\infty,-3\right)\cup \left(5,\infty\right)$

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

$$x+3<0$$ $$x-5<0$$

The solution satisfying both inequalities is $x<-3$.

$$x<-3$$

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

$$x-5>0$$ $$x+3>0$$

The solution satisfying both inequalities is $x>5$.

$$x>5$$

The final solution is the union of the obtained solutions.

$$x<-3\text{; }x>5$$