# How do you solve: (x^2-9)/(x^2-1) < 0?

Oct 20, 2015

$x \in \left(- 3 , - 1\right) \cup \left(1 , 3\right)$

#### Explanation:

$\frac{{x}^{2} - 9}{{x}^{2} - 1} < 0$

Right from the start, you know that any solution set that you might come up with cannot include the values of $x$ that will make the denominator equal to zero.

More specifically, you need to have

${x}^{2} - 1 \ne 0 \implies x \ne \pm 1$

Now, in order for this inequality to be true, you need to have

${x}^{2} - 9 < 0 \text{ }$ and $\text{ } {x}^{2} - 1 > 0$

or

${x}^{2} - 9 > 0 \text{ }$ and $\text{ } {x}^{2} - 1 < 0$

For the fist set of conditions to be true, you need to have

$\left\{\begin{matrix}{x}^{2} - 9 < 0 \implies x < \pm 3 \implies x \in \left(- 3 3\right) \\ {x}^{2} - 1 > 0 \implies x > \pm 1 \implies x \in \left(- \infty - 1\right) \cup \left(1 + \infty\right)\end{matrix}\right.$

This means that you need $x \in \left(- 3 , - 1\right) \cup \left(1 , 3\right)$.

For the second set of conditions, you need to have

$\left\{\begin{matrix}{x}^{2} - 9 > 0 \implies x > \pm 3 \implies x \in \left(- \infty - 3\right) \cup \left(3 + \infty\right) \\ {x}^{2} - 1 < 0 \implies x < \pm 1 \implies x \in \left(- 1 1\right)\end{matrix}\right.$

This time, those two intervals will not produce a valid solution set, or $x \in \emptyset$.

The only option left to you is $x \in \left(- 3 , - 1\right) \cup \left(1 , 3\right)$. The values of $x$ that belong to this interval will make the numerator negative and the denominator positive, which in turn will make the fraction negative.

graph{(x^2 - 9)/(x^2 - 1) [-18.02, 18.01, -9.01, 9.01]}