Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

calebcalebhoffman

  • 4 months ago

how would I solve for x if 2x-3/x+1<1

  • This Question is Closed
  1. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\frac{2x-3}{x+1}<1\]

  2. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I would split it into two cases

  3. calebcalebhoffman
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how would you do that

  4. myininaya
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I would first write as f(x)<0 Then find where f(x) is zero and undefined Then draw a number line with the numbers that make f zero or undefined Then test the intervals around those numbers to see if f is less than 0 or not

  5. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Since we are to multiply (x-1) on both sides, I would split it into x-1<0 and x-1>0

  6. myininaya
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you mean x+1?

  7. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes sorry

  8. calebcalebhoffman
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    SO i take x+ 1 and get \[2x-3<x+1 and 2x-3>x+1\]

  9. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes

  10. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well actually I believe we should do it case by case

  11. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    When x+1<0: 2x-3>x+1 ... When x+1>0: 2x-3<x+1 ...

  12. calebcalebhoffman
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so then my next step would to add in the 3 on both sides corect? so it would turn into \[2x>x+4\] and \[2x<x+4\]

  13. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yep

  14. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    And for presentation purpose I would write it as 2x>x+4 where x+1<0 and 2x<x+4 where x+1>0

  15. calebcalebhoffman
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    then do I divide by 2 on both of the equations or divide by 2x?

  16. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You subtract x from both sides

  17. calebcalebhoffman
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so It would be \[x-2x>4 where x+1<0\] and just the oposit sign on the otherone?

  18. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No, it should be 2x-x>4 where x+1<0

  19. calebcalebhoffman
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok now im starting to understand so then the next step would be to hmove the 4 over

  20. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Nah, the next step is subtracting it

  21. calebcalebhoffman
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok so it would then be 2x-x-4> 0 and where x+1<0 and 2x-x-4< 0 and where x+1>0 and t

  22. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I mean you can write 2x-x as x

  23. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    When x+1<0: 2x-3>x+1 2x>x+4 2x-x>4 x>4 When x+1>0: 2x-3<x+1 2x<x+4 2x-x<4 x<4

  24. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    However, when x+1<0, x can't be >4, and the same goes to the second case

  25. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Therefore there is no solution

  26. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Wait

  27. aum
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The inequality is valid for x in the interval (-1, 4).

  28. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @aum tell me what is wrong in my solution, I have no time to double-check

  29. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    bye

  30. calebcalebhoffman
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @aum I thought so. I see how to get the 4 but not the -1 could you help me solve this?

  31. aum
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    My method is what @myininaya suggested. (2x - 3) / (x + 1) < 1 (2x - 3) / (x + 1) - 1 < 0 (2x-3 - x - 1) / (x + 1) < 0 (x - 4) / (x + 1) < 0 The x values that are of interest to us here are the ones that will make f(x) = (x-4)/(x+1) zero or undefined. f(x) = 0 when x = 4 f(x) is undefined when x = -1. So the number line is split into three intervals: (-infinity, -1); (-1, 4), (4, infinity). Pick a convenient number in each interval and see if the inequality is valid. f(x) = (x - 4) / (x + 1) < 0 when x = -2, f(x) = (-6) / (-1) = 6 which is greater than 0. So not a solution. when x = 0, f(x) = -4/1 = -4 which is less than 0. This is a solution. when x = 5, f(x) = 1 / 6 which is greater than 0. Not a solution. So the solution is x in the interval (-1, 4).

  32. calebcalebhoffman
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thank you so much

  33. aum
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you are welcome.

  34. aum
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    To follow @kc_kennylau 's method: (2x - 3) / (x + 1) < 1 Case 1: x + 1 > 0 Since (x+1) is positive, we can multiply both sides by (x+1) and the "less than" inequality will remain intact. 2x - 3 < x + 1 x < 4 Case 2: x + 1 < 0 Since (x+1) is negative, multiplying both sides by (x+1) would change the "less than" inequality to a "greater than" inequality. 2x - 3 > x + 1 x > 4. But for the second case we assumed x + 1 < 0 which implies x < -1 But we get the solution as x > 4. x cannot be less than -1 and also be greater than 4 simultaneously. Thus, case 2 yields NO solution. Therefore, there is no solution in the domain x < -1. Combining the two cases we get x should be less than 4 but greater than -1.

  35. kc_kennylau
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I incorrectly rejected case 1

  36. aum
    • 4 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes.

  37. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.