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mathmath333

  • one year ago

solve

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} |x-3|+|x+4|+|x-1|=14\hspace{.33em}\\~\\ \end{align}}\)

  2. alekos
    • one year ago
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    I get x = 14/3 and -14/3

  3. alekos
    • one year ago
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    Is this a test?

  4. mathmath333
    • one year ago
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    how get that

  5. mathmath333
    • one year ago
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    this is no test

  6. alekos
    • one year ago
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    well first we take all the positive values within the modulus (x-3) + (x+4) + (x-1) =14 and solve for x

  7. anonymous
    • one year ago
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    best way to do this problem is partitioning the x's

  8. alekos
    • one year ago
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    then we take all the negative values -(x-3) - (x+4) - (x-1) =14 and solve for x

  9. mathmath333
    • one year ago
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    there will be total of 6 cases but that is a long way

  10. alekos
    • one year ago
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    if you plot the graph you'll see that there are only two values

  11. mathmath333
    • one year ago
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    that can be if 4 values are not valid

  12. alekos
    • one year ago
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    that's right. the 4 values are not valid

  13. alekos
    • one year ago
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    just plot the graph and you'll see

  14. mathmath333
    • one year ago
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    i look for a pen paper method

  15. alekos
    • one year ago
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    that's fine, you can do it that way. But Desmos is easier

  16. ParthKohli
    • one year ago
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    Hey, yes, just do it casewise.

  17. mathmath333
    • one year ago
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    u mean take 6 cases ?

  18. ParthKohli
    • one year ago
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    You'd need five, so yes.

  19. mathmath333
    • one year ago
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    how is it 5 cases

  20. anonymous
    • one year ago
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    It's 4 cases

  21. ParthKohli
    • one year ago
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    Oh, four.

  22. ParthKohli
    • one year ago
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    |dw:1434037874987:dw|

  23. ParthKohli
    • one year ago
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    For some reason, I thought there were four absolute values. Oops.

  24. anonymous
    • one year ago
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    |dw:1434037949803:dw|

  25. ParthKohli
    • one year ago
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    Yes.

  26. alekos
    • one year ago
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    I'll say it again. Plot the graph and you will all see that there are only two values

  27. ParthKohli
    • one year ago
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    If we're gonna take the graphing route, might as well just solve the equation directly.

  28. alekos
    • one year ago
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    I've done it both ways. algebraically and graphically

  29. mathmath333
    • one year ago
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    how u did algebriacally

  30. alekos
    • one year ago
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    I showed you earlier on

  31. ParthKohli
    • one year ago
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    Alekos, you covered only two of the four cases...

  32. alekos
    • one year ago
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    the other cases are not valid

  33. ParthKohli
    • one year ago
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    |dw:1434038294034:dw|

  34. ParthKohli
    • one year ago
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    |dw:1434038341589:dw|

  35. ParthKohli
    • one year ago
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    |dw:1434038401841:dw|

  36. ParthKohli
    • one year ago
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    |dw:1434038477315:dw|

  37. ParthKohli
    • one year ago
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    |dw:1434038527838:dw|

  38. ParthKohli
    • one year ago
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    |dw:1434038567104:dw|

  39. ParthKohli
    • one year ago
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    Discard an answer if it doesn't fall in the given interval.

  40. mathmath333
    • one year ago
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    here how would u identify useful and nonuseful regions

  41. ParthKohli
    • one year ago
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    First thing: you have to check for all intervals. There is no shortcut. Second thing: nonuseful intervals are those which don't contain the answer obtained in that interval's case. If you're considering an interval, say, [1,10] and the answer comes out to be 20, then you don't accept 20 as an answer.

  42. mathmath333
    • one year ago
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    ok so i have to check x by putting x in original equation,ok

  43. ParthKohli
    • one year ago
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    Did you see those drawings?

  44. mathmath333
    • one year ago
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    yes

  45. ParthKohli
    • one year ago
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    In the interval \(\left[-4, 1\right]\):\[|x-3| = -(x-3)\]\[|x+4| = x+4\]\[|x-1 | = - (x-1) \]

  46. ParthKohli
    • one year ago
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    ^ do you see why those are true?

  47. mathmath333
    • one year ago
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    because \(|x|=x,\ \ if\ \ x>0\\~\\and\\~\\|x|=-x\ \ if\ \ x<0\)

  48. ParthKohli
    • one year ago
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    Exactly. So you can solve it now, yes?

  49. mathmath333
    • one year ago
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    where did u learnt this method from

  50. ParthKohli
    • one year ago
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    This is a standard method.

  51. ParthKohli
    • one year ago
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    If you go to coaching, they do teach it there.

  52. alekos
    • one year ago
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    Tedious and unnecessary. If you follow what I've done it will give you the most efficient answer

  53. ParthKohli
    • one year ago
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    What have you done?

  54. mathmath333
    • one year ago
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    @alekos u didnt defined why u choosed the two particular cases

  55. mathmath333
    • one year ago
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    *chose

  56. ParthKohli
    • one year ago
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    Some problems have neat tricks involved in them.\[|a| + |b| = |a + b |\]implies that \(a\) and \(b\) have the same sign. Similarly,\[|a| + |b| = |a-b|\]implies that \(a\) and \(b\) have the opposite sign.

  57. ParthKohli
    • one year ago
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    So for example if you want to solve\[|x+3| + |1 - x| = 4\]It reduces to the inequality\[(x+3)(1-x) \ge 0\]

  58. alekos
    • one year ago
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    That's right I didn't define it. But if you sketch the graph you'll see that there is no need. It's self apparent

  59. ParthKohli
    • one year ago
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    How would you sketch the graph?

  60. alekos
    • one year ago
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    You can use pen and paper or Desmos

  61. ParthKohli
    • one year ago
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    I'll take the pen-and-paper approach. How did you sketch the graph here?

  62. ParthKohli
    • one year ago
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    The point is... the only way you can sketch a graph is casewise. So you first consider cases, find an expression for each straight-line, and then draw straight-lines all around the graph (which is a hard task) keeping the scale in mind, then you draw y = 14 and see where it intersects the graph. Looks like it intersects in two places: the first interval and the last interval. So you go to the first and last intervals and solve their respective equations.

  63. ParthKohli
    • one year ago
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    But if you find the linear expression, why not just equate it to 14 and get done with the problem instead of sketching all of them?

  64. alekos
    • one year ago
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    I agree. Pen and paper does take a little bit of time. But, we are all using computers nowadays so we can graph using technology and come to a solution very quickly.

  65. ParthKohli
    • one year ago
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    The only reason he's been given that problem is to be able to solve it himself in spite of the existence of computers.

  66. ParthKohli
    • one year ago
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    I'm pretty sure that he knows computers exist too.

  67. alekos
    • one year ago
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    That's fine. As I said it can be solved using pen and paper . Just takes a little longer. I know what I would rather use.

  68. ParthKohli
    • one year ago
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    Nah, takes only a couple of minutes.

  69. ParthKohli
    • one year ago
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    There are four cases and each case should take you around 20-25 seconds.

  70. alekos
    • one year ago
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    OK, no problem. Either method is valid.

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