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Inequality without factor table

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  1. May 8, 2016 #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.
     
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  3. May 8, 2016 #2

    Math_QED

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    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.
     
  4. May 8, 2016 #3

    PeroK

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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?
     
    Last edited by a moderator: May 8, 2016
  5. May 8, 2016 #4

    Rectifier

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    It is always positive for x > 1.
     
  6. May 8, 2016 #5

    PeroK

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    Hmm! Really? What about ##x = 2##?
     
  7. May 8, 2016 #6

    Rectifier

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    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
     
  8. May 8, 2016 #7

    PeroK

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

    What about when ##x < 1##?
     
  9. May 8, 2016 #8

    Rectifier

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    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.
     
  10. May 8, 2016 #9

    PeroK

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    Is there another critical value for ##x##? Perhaps something you've already calculated?
     
  11. May 8, 2016 #10

    Rectifier

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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.
     
  12. May 8, 2016 #11

    PeroK

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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.
     
  13. May 8, 2016 #12

    Rectifier

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    Yes, please.

    In the third case gives a negative outcome.

    I feel like its much easier to solve these with a sign table :D
     
  14. May 8, 2016 #13

    PeroK

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

    What about ##x< -4##?
     
  15. May 8, 2016 #14

    Rectifier

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    Gives a positive outcome.
     
  16. May 8, 2016 #15

    PeroK

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    So, what's the overall solution?
     
  17. May 8, 2016 #16

    Rectifier

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    x>1 and x<-4
     
  18. May 8, 2016 #17

    PeroK

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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##.
     
  19. May 8, 2016 #18

    Rectifier

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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.
     
  20. May 8, 2016 #19

    PeroK

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    Which numbers satisfy both equations?
     
  21. May 8, 2016 #20

    Rectifier

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