# How do you solve (x+3)/(x+4)<0?

Aug 7, 2016

$x \in \left(- 4 , - 3\right)$

#### Explanation:

There are three intervals within which the sign of the quotient does not change:

$\left(- \infty , - 4\right)$

$\left(- 4 , - 3\right)$

$\left(- 3 , \infty\right)$

We do not need to consider the points $x = - 4$ and $x = - 3$ as the former gives an undefined quotient and the latter a zero quotient. So they are not in the solution set.

$\textcolor{w h i t e}{}$
Case $\boldsymbol{x \in \left(- \infty , - 4\right)}$

Both $\left(x + 3\right) < 0$ and $\left(x + 4\right) < 0$, so the quotient is positive.

So this interval is not part of the solution set.

$\textcolor{w h i t e}{}$
Case $\boldsymbol{x \in \left(- 4 , - 3\right)}$

In this interval, $\left(x + 3\right) < 0$ but $\left(x + 4\right) > 0$, so the quotient is negative.

So this interval is part of the solution set.

$\textcolor{w h i t e}{}$
Case $\boldsymbol{x \in \left(- 3 , \infty\right)}$

In this interval, $\left(x + 3\right) > 0$ and $\left(x + 4\right) > 0$, so the quotient is positive.

So this interval is not part of the solution set.

$\textcolor{w h i t e}{}$
Conclusion

The solution set is the interval $\left(- 4 , - 3\right)$