How to solve (x+6)/(3-x) < 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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$$\frac{ x+6 }{ 3-x } < 0$$

$x\in \left(-\infty,-6\right)\cup \left(3,\infty\right)$

For the quotient to be negative, $x+6$ and $3-x$ have to be of the opposite signs. Consider the case when $x+6$ is positive and $3-x$ is negative.

$$x+6>0$$ $$3-x<0$$

The solution satisfying both inequalities is $x>3$.

$$x>3$$

Consider the case when $3-x$ is positive and $x+6$ is negative.

$$3-x>0$$ $$x+6<0$$

The solution satisfying both inequalities is $x<-6$.

$$x<-6$$

The final solution is the union of the obtained solutions.

$$x>3\text{; }x<-6$$