# How do you solve: <mstyle displaystyle="true"> x 2

How do you solve: $\frac{{x}^{2}-9}{{x}^{2}-1}<0$ ?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

fugprurgeil
Step 1
$\frac{{x}^{2}-9}{{x}^{2}-1}<0$
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
${x}^{2}-1\ne 0⇒x\ne ±1$
Now, in order for this inequality to be true, you need to have
${x}^{2}-9<0\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}$ and $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{x}^{2}-1>0$
or
${x}^{2}-9>0\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}$ and $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{x}^{2}-1<0$
For the fist set of conditions to be true, you need to have
$\left\{\begin{array}{l}{x}^{2}-9<0⇒x<±3⇒x\in \left(-3,3\right)\\ {x}^{2}-1>0⇒x>±1⇒x\in \left(-\infty ,-1\right)\cup \left(1,+\infty \right)\end{array}$
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
$\left\{\begin{array}{l}{x}^{2}-9>0⇒x>±3⇒x\in \left(-\infty ,-3\right)\cup \left(3,+\infty \right)\\ {x}^{2}-1<0⇒x<±1⇒x\in \left(-1,1\right)\end{array}$
This time, those two intervals will not produce a valid solution set, or $x\in \varnothing$
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.