How do you solve: $\frac{{x}^{2}-9}{{x}^{2}-1}<0$ ?

Crystal Wheeler
2022-07-09
Answered

How do you solve: $\frac{{x}^{2}-9}{{x}^{2}-1}<0$ ?

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fugprurgeil

Answered 2022-07-10
Author has **12** answers

Step 1

Your inequality looks like this

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

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

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

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

$\{\begin{array}{l}{x}^{2}-9<0\Rightarrow x<\pm 3\Rightarrow x\in (-3,3)\\ {x}^{2}-1>0\Rightarrow x>\pm 1\Rightarrow x\in (-\infty ,-1)\cup (1,+\infty )\end{array}$

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

For the second set of conditions, you need to have

$\{\begin{array}{l}{x}^{2}-9>0\Rightarrow x>\pm 3\Rightarrow x\in (-\infty ,-3)\cup (3,+\infty )\\ {x}^{2}-1<0\Rightarrow x<\pm 1\Rightarrow x\in (-1,1)\end{array}$

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

The only option left to you is $x\in (-3,-1)\cup (1,3)$ . 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.

Your inequality looks like this

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

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

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

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

$\{\begin{array}{l}{x}^{2}-9<0\Rightarrow x<\pm 3\Rightarrow x\in (-3,3)\\ {x}^{2}-1>0\Rightarrow x>\pm 1\Rightarrow x\in (-\infty ,-1)\cup (1,+\infty )\end{array}$

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

For the second set of conditions, you need to have

$\{\begin{array}{l}{x}^{2}-9>0\Rightarrow x>\pm 3\Rightarrow x\in (-\infty ,-3)\cup (3,+\infty )\\ {x}^{2}-1<0\Rightarrow x<\pm 1\Rightarrow x\in (-1,1)\end{array}$

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

The only option left to you is $x\in (-3,-1)\cup (1,3)$ . 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.

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