Trivial question. Inequality solution.

[Nazariy]

Hello.

I am lost a bit mathematically.

I have inequality of the following initial form:

​

It becomes:

​

I conceptually understand that for inequality to be exactly 0, we need x=4, also x cannot equal -1. If x is less than -1, we get a positive result, so not a solution. Thus the solution set has to be:

​

But is there a more mathematical approach to this? I mean If I were to have some more complicated form in denominator or numerator or both, I would probably not be able to solve it.

Thank you!

 

[Quaid]

If I were to have some more complicated form in denominator or numerator or both, I would probably not be able to solve it.

With additional factors (in the denominator and/or numerator), we often create what's called a "sign chart". You go through the same process, but multiple times; the chart is to organize your results.

Check out THIS lesson.

Cheers :cool:

 

[pka]

I have inequality of the following initial form:

View attachment 3717
​

It becomes:

View attachment 3718 Thus the solution set has to be:

View attachment 3719​

Here is the reason that works. There are three regions of interest in that question.

They are \(\displaystyle ( - \infty , - 1) \cup ( - 1,4] \cup [4,\infty )\).

We can take a number sample from each region. But only numbers from \(\displaystyle ( - 1,4] \) make the expression true.
Therefore that is the solution set.

Does that qualify as a mathematical reason?

 

[Nazariy]

With additional factors (in the denominator and/or numerator), we often create what's called a "sign chart". You go through the same process, but multiple times; the chart is to organize your results.

Check out THIS lesson.

Cheers :cool:

Yes indeed, I know this concept, let me check how it works for fractions, thank you.

 

[Nazariy]

Here is the reason that works. There are three regions of interest in that question.

They are \(\displaystyle ( - \infty , - 1) \cup ( - 1,4] \cup [4,\infty )\).

We can take a number sample from each region. But only numbers from \(\displaystyle ( - 1,4] \) make the expression true.
Therefore that is the solution set.

Does that qualify as a mathematical reason?

Yes! This is perfect! Thanks!

 

[Nazariy]

With additional factors (in the denominator and/or numerator), we often create what's called a "sign chart". You go through the same process, but multiple times; the chart is to organize your results.

Check out THIS lesson.

Cheers :cool:

Actually, since pka made me think about those regions of interest, I no longer need that lesson I just construct that sign chart based on those three regions, I now know how to approach problems like this. Awesome!