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X = ((384x/(y+384))*y+384(384x/(y+384)))/384

  1. Dec 20, 2011 #1
    1. The problem statement, all variables and given/known data

    These are different way's I've phrased the question:

    x = ((384x/(y+384))*y+384(384x/(y+384)))/384

    x = (1280((sqrt(768^2-(((-1280)(-y))/(-y-384))^2))))/(-sqrt(768^2-((768^2-(((-1280)(-y))/(-y-384))^2)))+1280)

    x = ((sqrt(768^2-(((-1280)(-y))/(-y-384))^2)))((y+384)/(y+384-y))

    v= 384x/(y+384)
    v= (sqrt(768^2-(((-1280)(-y))/(-y-384))^2))
    y= (384x-384v)/v

    v= 768cos((pi((180t)/pi))/180)
    t= acos((sqrt(768^2-(((-1280)(-y))/(-y-384))^2))/768)


    The original version:
    x=
    (1280(768cos((pi((180acos((sqrt(768^2-(((-1280)(-y))/(-y-384))^2))/768))/pi
    ))/180)))/(-sqrt(768^2-(768cos((pi((180acos((sqrt(768^2-(((-1280)(-y))/(-y-384))^2))/768))/pi))/180))^2)+1280)


    How do I solve for y, even if there's multiple values for y?
     
    Last edited: Dec 20, 2011
  2. jcsd
  3. Dec 20, 2011 #2

    NascentOxygen

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    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    I'm having trouble figuring out what this is about. Are all of the sets of equations below supposed to have an identical solution y=f(x)?

    This reduces to 0=0

    Are you wanting to solve for y in terms of v here?

    Is acos your abbreviation for arccos?

    You know there are multiple solutions?
     
  4. Dec 20, 2011 #3
    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    Affirmative. Every equation I've written as x= gives an identical answer for a given y. If y=216 x=960.

    960 = (((384*960)/(216+384))*216+384((384*960)/(216+384)))/384

    960 = (1280((sqrt(768^2-(((-1280)(-216))/(-216-384))^2))))/(-sqrt(768^2-((768^2-(((-1280)(-216))/(-216-384))^2)))+1280)

    960 = ((sqrt(768^2-(((-1280)(-216))/(-216-384))^2)))((216+384)/(216+384-216))

    960 = (1280(768cos((pi((180acos((sqrt(768^2-(((-1280)(-216))/(-216-384))^2))/768))/pi
    ))/180)))/(-sqrt(768^2-(768cos((pi((180acos((sqrt(768^2-(((-1280)(-216))/(-216-384))^2))/768))/pi))/180))^2)+1280)


    However, I need to rewrite these to find y if the /only information i have/ is x and the rest of the equation, or rather with no knowledge of y.

    y=0, x=768:

    768 = ((sqrt(768^2-(((-1280)(-0))/(-0-384))^2)))((0+384)/(0+384-0))


    The v and t examples are just more alternative ways of writing the same equation. If I could somehow convert any given x into v or t- with no knowledge of y- I could solve this equation for y. But my goal is to solve for y so that I can find y for any given x, not just x=960 or x=768.
     
    Last edited: Dec 20, 2011
  5. Dec 20, 2011 #4

    Borek

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    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    If your equation reduces to 0=0, it is an identity, so in a way every pair x,y is a solution.
     
  6. Dec 20, 2011 #5
    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    Nascent's wrong, it isn't 0=0.

    It's ( y(384x/(y+384)) + 384(384x/(y+384)) )/384 =x

    So how do I find y=...?
     
    Last edited: Dec 20, 2011
  7. Dec 20, 2011 #6

    Borek

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  8. Dec 20, 2011 #7
    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    x-x=0, you say? I wasn't aware.

    Any value of x or y can result in a different value of x or y. I can find any x if I know y, but how can I find any y without already knowing y?
     
    Last edited: Dec 20, 2011
  9. Dec 20, 2011 #8

    Office_Shredder

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    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    The point is ANY value of y will work. It doesn't even matter what value of x you've picked. The only exception to this is if y=-384 people will complain about the denominator being 0
     
  10. Dec 20, 2011 #9

    D H

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    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    Re the first expression:
    This is almost a tautology. (It is a tautology except at y=-384, where it is undefined.)


    Re the second expression:
    This is not at all the same as your first expression. You can reduce this expression further. Those (((-1280)(-y))/(-y-384))2 terms are better written as 12802(y/(y+384))2. You also have a bunch of superfluous parentheses.

    It will help to create auxiliary variables w=x/256 and z=y/(y+384) and to write 768 as 256*3 and 1280 as 256*5. Note that common factor of 256.


    Re the third expression:
    This is the same as your second expression when y≥0. It is not the same as your second expression when y is negative. Just as you can reduce the second expression further, you can also reduce this third expression considerably. For example, y+384-y is just 384. Here it also helps to recognize that 384=256*3/2 (that common factor of 256 appears yet again).
     
  11. Dec 20, 2011 #10

    berkeman

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    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    Are you being sincere or sarcastic here? That would seem to answer your question, no?
     
  12. Dec 20, 2011 #11
    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    I get that. I just want to solve this to equal y, so that if I know what x is I can solve for y.

    If I know that y=216, I know that x=960. But what if I knew that x=961, let's say? I'd have zero clue what y=, and I'd have to guess.

    I tried to find x=y, and by guestimation the closet I came was 518.883098582601474166600894~=x=y but my calculator won't go farther.

    It's a curve, x changes with y. Like x^2=y, x= either sqrt(y) or -sqrt(y)

    y=576, x=0. y=0, x=768. y=216, x=960.

    y=-144, x=0

    It may help to know that between y=576 and y=-144 is the area I'm interested in.
     
    Last edited: Dec 20, 2011
  13. Dec 20, 2011 #12

    D H

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    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    If you know that y is non-negative, I'd go with your third expression. If y can be positive or negative, you'll need to use a simplified version of your second expression.
     
  14. Dec 20, 2011 #13
    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    If x=y, x-x=0. If x=(e^(it) + e^(-it))/2, x-x=0. If cos(x)=5, x-x=0.
    If x=0, x-x=0.
     
  15. Dec 20, 2011 #14

    SammyS

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    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    I fed these expressions to WolframAlpha.

    The one I labeled (1) above is equivalent to x = x. So, x can be any number, y can be any number except -384 .

    According to WolframAlpha, the others are all equivalent to [itex]\displaystyle x=\frac{8}{3} \sqrt{-(y-576) (y+144)}\,,[/itex] provided that y > 0. Of course this expression
    is real provided that -144 ≤ y ≤ 576 .

    Also, the first expression is the only one with x on the left hand side (as well as being on the right).
     
  16. Dec 20, 2011 #15
    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    I need to solve for y. Without already knowing y.
    All these equations depend on already knowing what y is.

    If x=2y. y=x/2.
    If x=(e^(iy) + e^(-iy))/2, y= either -ilog(x-sqrt(x^2-1)) or -ilog(x+sqrt(x^2-1))

    If x= ((384x/(y+384))*y+384(384x/(y+384)))/384

    y= ???
     
  17. Dec 20, 2011 #16

    SammyS

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    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    Now I understand what you're asking.

    It's been mentioned by some, if not all, of the responders to this thread, that any value of y (except that y≠-384, because that would give division by zero) will satisfy this equation no matter what the value of x.

    To spell it out, the right-hand side of your expression can be simplified as follows:

    [itex]\displaystyle\frac{1}{384}\left(\frac{384xy}{y+384}+\frac{384^2x}{y+384}\right)[/itex]
    [itex]\displaystyle=\left(\frac{xy}{y+384}+\frac{384x}{y+384}\right)[/itex]

    [itex]\displaystyle=\left(\frac{xy+384x}{y+384}\right)[/itex]

    [itex]\displaystyle=x\left(\frac{y+384}{y+384}\right)[/itex]
     
  18. Dec 20, 2011 #17
    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    If x= ((384x/(y+384))*y+384(384x/(y+384)))/384
    y= ???

    If y^2=x, y must either be
    y1= sqrt(x), y2= -sqrt(x)

    If ((384x/(y+384))*y+384(384x/(y+384)))/384 = x = 0

    Then y1=576 and y2=-144

    It's a curve. There are two values of y for any x except at the cusp of the curve.

    How can I take a value of x and calculate the two values of y without knowing either value of y?
     
  19. Dec 20, 2011 #18

    D H

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    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    What exactly are you trying to solve?

    Your tautology is not a curve. You made a mistake somewhere in deriving x= ((384x/(y+384))*y+384(384x/(y+384)))/384.
     
  20. Dec 20, 2011 #19
    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    ((sqrt(768^2-(((-1280)(-y))/(-y-384))^2)))((y+384)/(y+384-y)) = x

    If you swap x and y and type y=((sqrt(768^2-(((-1280)(-x))/(-x-384))^2)))((x+384)/(x+384-x)) into even the simplest graphing calculator you can see the curve.

    I'm trying to solve y=???

    Or x in the calculator-friendly version...


    I get what everyone is saying: that the equations I've provided are all identities because they are all true. But they all have y lodged in them somewhere. How do I isolate y so that I can solve for y without knowing y.

    An equation that says whatever(x)=y.

    Two equations for y, one that turns x=0 into y1=-144 and one that turns x=0 into y2=576
     
    Last edited: Dec 20, 2011
  21. Dec 20, 2011 #20

    NascentOxygen

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    Re: x=((384x/(y+384))*y+384(384x/(y+384)))/384

    This might be a good time to take a breather. :smile:

    How about describing where this equation/formula came from, for starters. Maybe you are on the wrong track all together, and even were you able to 'solve' this it might be wasted effort if you've run off-course even before reaching this point.

    You seem to have put in a lot of effort already; but let's back up a way and confirm that you are on track to solving something tangible.
     
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