Inequality without factor table

[Rectifier]

I am trying to solve this inequality without using a factor table.

The problem
$$ \frac{x+4}{x-1} > 0 $$

The attempt at a solution
As I can see $x \neq 1$. I want to muliply both sides of the expression with x-1 to get rid of it, from the fraction. But before that, I have to consider two cases where x-1 is bigger and smaller than 0 because the sign gets inverted when multiplying (or dividing) by a negative number.

Case 1:
$x-1 > 0 \\ x >1$ gives
$$ \frac{x+4}{x-1} > 0 \\ \\ x+4 > 0 \\ x > -4$$


Case 2:
$x-1 < 0 \\ x < 1$ gives
$$ \frac{x+4}{x-1} > 0 \\ \\ x+4 < 0 \\ x < -4$$


This is the place where I get stuck. I am not sure how to take all this information and produce an answer.

 

[member 587159]

I am trying to solve this inequality without using a factor table.

The problem
$$ \frac{x+4}{x-1} > 0 $$

The attempt at a solution
As I can see $x \neq 1$. I want to muliply both sides of the expression with x-1 to get rid of it, from the fraction. But before that, I have to consider two cases where x-1 is bigger and smaller than 0 because the sign gets inverted when multiplying (or dividing) by a negative number.

Case 1:
$x-1 > 0 \\ x >1$ gives
$$ \frac{x+4}{x-1} > 0 \\ \\ x+4 > 0 \\ x > -4$$


Case 2:
$x-1 < 0 \\ x < 1$ gives
$$ \frac{x+4}{x-1} > 0 \\ \\ x+4 < 0 \\ x < -4$$


This is the place where I get stuck. I am not sure how to take all this information and produce an answer.

Well, you can't just multiply both sides with x - 1 since it contains information to solve the inequality. From what you have now, x can be any real number except 4 or 1. What if x = 0? I don't know whether you know a sign table. You can use this to solve this exercise.

 

[PeroK]

This is the place where I get stuck. I am not sure how to take all this information and produce an answer.

Let's take case 1. $x > 1$ What can you say about the expression $\frac{x+4}{x-1}$ in this case?

 

[Rectifier]

Let's take case 1. $x > 1$ What can you say about the expression $\frac{x+4}{x-1}$ in this case?

It is always positive for x > 1.

 

[PeroK]

It is always negative for x > 1.

Hmm! Really? What about $x = 2$?

 

[Rectifier]

Hmm! Really? What about $x = 2$?

You were fast! Edited the answer just before you commented :D. It is positive. I cannot tell that from my expression that I get in case 1. Not sure how to interpret x > -4

 

[PeroK]

You were fast! Edited the answer just before you commented :D. It is positive.

So, that's half your answer: If $x > 1$ the expression is always positive.

What about when $x < 1$?

 

[Rectifier]

So, that's half your answer: If $x > 1$ the expression is always positive.

What about when $x < 1$?

Well, that is the tricky part! :D x = 0 gives a negative number. x = - 5 gives a positive number. Not sure how to use my case studies to come to an answer though.

 

[PeroK]

Well, that is the tricky part! :D x = 0 gives a negative number. x = - 5 gives a positive number.

Is there another critical value for $x$? Perhaps something you've already calculated?

 

[Rectifier]

Is there another critical value for $x$? Perhaps something you've already calculated?

I have a feeling that it is the interval that I calculated in case 2 but I am not sure why.

 

[PeroK]

I have a feeling that it is the interval that I calculated in case 2 but I am not sure why.

What you did in the OP was correct, but it created a logically slightly difficult solution that you weren't able to interpret. I suggest you solve the problem and I can try to explain what happened in your OP, if you like.

You actually should have used 3 cases here. The next case is $-4 \le x < 1$. Let's call this case 2.

 

[Rectifier]

What you did in the OP was correct, but it created a logically slightly difficult solution that you weren't able to interpret. I suggest you solve the problem and I can try to explain what happened in your OP, if you like.

You actually should have used 3 cases here. The next case is $-4 \le x < 1$. Let's call this case 2.

Yes, please.

In the third case gives a negative outcome.

I feel like its much easier to solve these with a sign table :D

 

[PeroK]

Yes, please.

In the third case gives a negative outcome.

Not quite. When $x = -4$ the expression is zero. But, there are no solutions for $-4 \le x < 1$.

What about $x< -4$?

 

[Rectifier]

What about $x< -4$?

Gives a positive outcome.

 

[PeroK]

Gives a positive outcome.

So, what's the overall solution?

 

[Rectifier]

So, what's the overall solution?

x>1 and x<-4

 

[PeroK]

What happened in your original post?

You started by assuming that $x > 1$ and looked for solutions. You found solutions when $x > -4$. This looked strange but it was actually quite logical. What you had was:

$x > 1$ and $x > -4$

Which, logically, is simply $x > 1$.

There's something else that might have happened in a different problem. You might have got something like:

$x > 1$ and $x < -2$

Again, that is not a problem, but would tell you that there are no solutions for $x > 1$.

 

[Rectifier]

What happened in your original post?

You started by assuming that $x > 1$ and looked for solutions. You found solutions when $x > -4$. This looked strange but it was actually quite logical. What you had was:

$x > 1$ and $x > -4$

Which, logically, is simply $x > 1$.

There's something else that might have happened in a different problem. You might have got something like:

$x > 1$ and $x < -2$

Again, that is not a problem, but would tell you that there are no solutions for $x > 1$.

Oh okay, thank you for the explanation! But why do I choose x>1 and not x > -4? I suspect that the answer is trivial.

 

[PeroK]

Oh okay, thank you for the explanation! But why do I choose x>1 and not x > -4? I suspect that the answer is trivial.

Which numbers satisfy both equations?

 

[Rectifier]

Which numbers satisfy both equations?

x > 1. So I should basically take the one with the smallest common solution set?

 

[PeroK]

x > 1. So I should basically take the one with the smallest common solution set?

Yes, but best to understand the logic. Here's the full logic of your original post:

$(x > 1 \ AND \ x> -4) \ OR \ (x < 1 \ AND \ x < -4)$

$(x > 1) \ OR \ (x < -4)$

So, now I'm going to slightly correct your overall solution:

x>1 and x<-4

Perhaps $x > 1$ or $x < -4$ is better!

If you ever do some computer programming, this sort of logical thinking can be quite useful!

 

[Rectifier]

Thank you for your help. I will take try to understand that approach a bit later today. Thank you for your help once more.

 

[Mark44]

x>1 and x<-4

Perhaps $x > 1$ or $x < -4$ is better!

If you ever do some computer programming, this sort of logical thinking can be quite useful!

The problem with x > 1 AND x < -4 is that x can't simultaneously be larger than 1 and smaller than -4. There are no numbers that satisfy both inequalities. That's why PeroK's version is better.

 

[Rectifier]

The problem with x > 1 AND x < -4 is that x can't simultaneously be larger than 1 and smaller than -4. There are no numbers that satisfy both inequalities. That's why PeroK's version is better.

Thank you for the clarification :)

 

[SammyS]

x > 1. So I should basically take the one with the smallest common solution set?

Maybe that's not the best way to look at it.

Case 1 holds only for x > 1. No matter what the result you get for this case, in this problem you got x > -4, that result is only good for x > 1.

Case 2 holds only for x < 1. No matter what the result you get for this case, in this problem you got x < -4, that result is only good for x < 1.

In the end, it's more like the intersection. Well, I suppose that is the smallest common set for each case.

 

[Sahil Kukreja]

Another way is to solve is graphically

let $f(x)= \frac{x+4}{x-1} \$
find f'(x) you will see that f'(x)always negative wherever defined
Hence f(x) is Monotonically decreasing wherever defined

you can easily draw a rough sketch of the graph using that the vertical asymptote is x=1 and root of f(x)=0 is x=-4.

you will find that f(x)>0 for x<-4 U x>1

 

[Mark44]

Another way is to solve is graphically

let $f(x)= \frac{x+4}{x-1} \$
find f'(x) you will see that f'(x)always negative wherever defined

This thread was posted in the Precalculus section, so it's very likely that the poster will not know what f'(x) means.

Hence f(x) is Monotonically decreasing wherever defined

you can easily draw a rough sketch of the graph using that the vertical asymptote is x=1 and root of f(x)=0 is x=-4.

you will find that f(x)>0 for x<-4 U x>1