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calebcalebhoffman

  • one year ago

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

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  1. kc_kennylau
    • one year ago
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    \[\frac{2x-3}{x+1}<1\]

  2. kc_kennylau
    • one year ago
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    I would split it into two cases

  3. calebcalebhoffman
    • one year ago
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    how would you do that

  4. myininaya
    • one year ago
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    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
    • one year ago
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    Since we are to multiply (x-1) on both sides, I would split it into x-1<0 and x-1>0

  6. myininaya
    • one year ago
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    you mean x+1?

  7. kc_kennylau
    • one year ago
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    Yes sorry

  8. calebcalebhoffman
    • one year ago
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    SO i take x+ 1 and get \[2x-3<x+1 and 2x-3>x+1\]

  9. kc_kennylau
    • one year ago
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    Yes

  10. kc_kennylau
    • one year ago
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    Well actually I believe we should do it case by case

  11. kc_kennylau
    • one year ago
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    When x+1<0: 2x-3>x+1 ... When x+1>0: 2x-3<x+1 ...

  12. calebcalebhoffman
    • one year ago
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    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
    • one year ago
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    Yep

  14. kc_kennylau
    • one year ago
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    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
    • one year ago
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    then do I divide by 2 on both of the equations or divide by 2x?

  16. kc_kennylau
    • one year ago
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    You subtract x from both sides

  17. calebcalebhoffman
    • one year ago
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    so It would be \[x-2x>4 where x+1<0\] and just the oposit sign on the otherone?

  18. kc_kennylau
    • one year ago
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    No, it should be 2x-x>4 where x+1<0

  19. calebcalebhoffman
    • one year ago
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    ok now im starting to understand so then the next step would be to hmove the 4 over

  20. kc_kennylau
    • one year ago
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    Nah, the next step is subtracting it

  21. calebcalebhoffman
    • one year ago
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    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
    • one year ago
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    I mean you can write 2x-x as x

  23. kc_kennylau
    • one year ago
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    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
    • one year ago
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    However, when x+1<0, x can't be >4, and the same goes to the second case

  25. kc_kennylau
    • one year ago
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    Therefore there is no solution

  26. kc_kennylau
    • one year ago
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    Wait

  27. aum
    • one year ago
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    The inequality is valid for x in the interval (-1, 4).

  28. kc_kennylau
    • one year ago
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    @aum tell me what is wrong in my solution, I have no time to double-check

  29. kc_kennylau
    • one year ago
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    bye

  30. calebcalebhoffman
    • one year ago
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    @aum I thought so. I see how to get the 4 but not the -1 could you help me solve this?

  31. aum
    • one year ago
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    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
    • one year ago
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    thank you so much

  33. aum
    • one year ago
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    you are welcome.

  34. aum
    • one year ago
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    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
    • one year ago
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    I incorrectly rejected case 1

  36. aum
    • one year ago
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    yes.

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