# For what x is 1 / ( x + 3 ) > 2/(x-3)?

Oct 28, 2015

$x \in \left(- \infty , - 9\right) \cup \left(- 3 , 3\right)$

#### Explanation:

Note that $x = - 3$ and $x = 3$ are not solutions since one or the other side of the inequality is undefined for these values of $x$.

Split the remaining possibilities into cases:

Case $\boldsymbol{x \in \left(- \infty , - 3\right) \cup \left(3 , \infty\right)}$

$x + 3$ and $x - 3$ are both negative or both positive.

In either case $\left(x + 3\right) \left(x - 3\right) > 0$ so multiply both sides of the inequality by that to get:

$x - 3 > 2 x + 6$

Subtract $x + 6$ from both sides to get:

$- 9 > x$

Combining with the conditions of the case, that gives us the solution:

$x \in \left(- \infty , - 9\right)$

Case $\boldsymbol{x \in \left(- 3 , 3\right)}$

$x + 3 > 0$ and $x - 3 < 0$, so $\left(x + 3\right) \left(x - 3\right) < 0$, so multiply both sides of the inequality by this and reverse the inequality to get:

$x - 3 < 2 x + 6$

Subtract $x + 6$ from both sides to get:

$- 9 < x$

Combining with the conditions of the case, that gives us the solution:

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

Conclusion

Putting the two cases together, the original inequality holds for all

$x \in \left(- \infty , - 9\right) \cup \left(- 3 , 3\right)$