Inequality without factor table

  • #1
Rectifier
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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.
 

Answers and Replies

  • #2
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.
 
  • #3
PeroK
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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?
 
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  • #4
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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.
 
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  • #5
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It is always negative for x > 1.
Hmm! Really? What about ##x = 2##?
 
  • #6
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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
 
  • #7
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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##?
 
  • #8
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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.
 
  • #9
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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?
 
  • #10
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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.
 
  • #11
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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.
 
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  • #12
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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
 
  • #13
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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##?
 
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  • #14
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What about ##x< -4##?
Gives a positive outcome.
 
  • #15
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Gives a positive outcome.
So, what's the overall solution?
 
  • #16
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So, what's the overall solution?
x>1 and x<-4
 
  • #17
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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##.
 
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  • #18
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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.
 
  • #19
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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?
 
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  • #20
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Which numbers satisfy both equations?
x > 1. So I should basically take the one with the smallest common solution set?
 
  • #21
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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!
 
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  • #22
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Thank you for your help. I will take try to understand that approach a bit later today. Thank you for your help once more.
 
  • #23
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x>1 and x<-4
PeroK said:
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.
 
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  • #24
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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 :)
 
  • #25
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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.
 

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