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Help Eupho with a semi-simple algebra problem.

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  1. #1
    euphoricism's Avatar
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    Default Help Eupho with a semi-simple algebra problem.

    Math is not my strong suit..

    8x+22y=40,000
    x=? y=?
    Round to nearest whole number

    Code:
    8x+22y=40,000
          -22y                   -22y
    
    8x = 40,000 - 22y
    x = (40,000 - 22y)/8
    
    Replace back to main problem.. and I'm lost.
    
    8((40000-22y)/8) = 40,000
    -----                               -----------
     8                                         8
    
    (40000-22y) /8     = 5000
    multiply by 8 again..
    40000-22y = 400000
    -22y = 0
    y = 0
    So, thats not right. Help?
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  2. #2
    euphoricism's Avatar
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    I seem to recall it has something to do with a parabola.. maybe a circle..

    I can't remember.
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  3. #3

    Default Re: Help Eupho with a semi-simple algebra problem.

    2 variables needs 2 equations to solve... If you just need any solution then, how bout y = 1, x = 4997.25
  4. #4
    euphoricism's Avatar
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    But no, there is a way to do this.
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  5. #5
    euphoricism's Avatar
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    ooh. Youre right. I missed half the equation.

    Lets try this again:

    8x+22y=40,000
    x+y = 31


    Does that make a difference?
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  6. #6
    euphoricism's Avatar
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    fuck me. = 30

    x+y=30
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  7. #7
    Lukie's Avatar
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    8x+22y-40,000 = 0

    then factor it out. If nobody's done it by the time i'm done eating this sushi, I will.
  8. #8
    euphoricism's Avatar
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    Bah I give up. thats not right either.
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  9. #9
    euphoricism's Avatar
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    Here's what I'm trying to do, maybe you guys can help me there. Might be my methedology that is off.

    There are 30 days in june, I want to play 40,000 hands of poker. This is 1333.33 hands per day. However, I will play shorter sessions on saturday and sunday.

    Let x = hands played on weekends
    y = hands played on weekdays

    so I get 8 weekends + 22 weekdays = 40,000 total hands
    8x + 22y = 40000
    I also get that x+y = 30, but also that 2x+5y=7.
    I'm trying to find the "ideal" numbers.


    edit.. ahh x+y != 30. Thats silly. 2x+5y != 7 either.
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  10. #10
    euphoricism's Avatar
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    ahhh but 2x+5y DOES equal 10,000. Four weeks in a month, 4*10,000
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  11. #11
    euphoricism's Avatar
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    The automated solver is giving me just to play 5000 hands every weekend. Well, that makes it simple.

    Atleast I got the equation right.
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  12. #12
    euphoricism's Avatar
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    I think nehmer is close. I could just use a very long list of numbers that fit the equation and grab the one I feel is best.
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  13. #13
    euphoricism's Avatar
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    Nehmer wins. Its just a line.
    y = 1/22 (40000-8x)
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  14. #14
    a500lbgorilla's Avatar
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    If you're trying to find optimums, you were right in using just one equation.

    You would have to graph it as a line and try to find the point that you like the most.

    Now, if you thought "I think I'd like to play 2000 hands every weekday, how many do i have to hit on the weekend" then this is an easy problem.

    But if you want to know some "optimum" without any conditions like "I can hand B hrs every weekday easily playing C number of tables, what are my weekends gonna look like? If I bump that number down by an hour, what now?" then i cant help you.
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  15. #15
    euphoricism's Avatar
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    Now whats the easiest way to look at a whole table of values for that line...
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  16. #16
    a500lbgorilla's Avatar
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    Quote Originally Posted by euphoricism
    Now whats the easiest way to look at a whole table of values for that line...
    excel
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  17. #17
    euphoricism's Avatar
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    how would I do that with excel?
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  18. #18
    Lukie's Avatar
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    how about this (ballparked)

    1000 hands on weekends, 1500 hands on weekdays

    8000 + 33000 = 41,000

    subtract 1,000 for times you'd rather watch porn

    =

    40,000

    algebra problem solved
  19. #19
    euphoricism's Avatar
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    Yeah those are the numbers I got when I ballparked. 1000 on weekends, 1500 on weekdays. Its pretty close to what I want. Just wanted to know what exact numbers should be.
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  20. #20
    Lukie's Avatar
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    euph,

    to be honest, it's very difficult and certainly not efficient to always play a preset number of hands. There will be times when you are just lethargic, you don't feel like playing, playing poorly, etc., where you shouldn't play as many hands as you'd like. And there are other times when you have nothing better to do, you're running hot, you're confident, playing well, etc., where there's no reason to quit playing.
  21. #21
    euphoricism's Avatar
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    To an extent, you are correct. To an extent, you are far from correct.
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  22. #22
    euphoricism's Avatar
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    Your point is valid, however, NOT playing poker costs a lot of money. I'm not playing tonight, simply because I'm just too fucking tired. Long day, ya know?

    But its going to cost me. I'm essentially paying $250 to take the night off. Opportunity costs ain't cheap.
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  23. #23
    Lukie's Avatar
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    Quote Originally Posted by euphoricism
    Your point is valid, however, NOT playing poker costs a lot of money. I'm not playing tonight, simply because I'm just too fucking tired. Long day, ya know?

    But its going to cost me. I'm essentially paying $250 to take the night off. Opportunity costs ain't cheap.
    meh, depends. Sometimes I think it's -ev for me to play. If I'm tired, steaming, and the games aren't good, it's undoubtedly -ev for me to play. See where I'm going?
  24. #24
    euphoricism's Avatar
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    Yar. Definitely no argument that you shouldn't play when tired, stressed, steaming, as you're likely to lose money.

    But you are also losing money by not playing. Guess it's a matter of the greater of two evils.

    In any case, cheers.
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  25. #25
    Lukie's Avatar
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    Quote Originally Posted by Lukie
    8x+22y-40,000 = 0

    then factor it out. If nobody's done it by the time i'm done eating this sushi, I will.
    this is wrong, but I'll try to figure this out anyway to the best of my ability. But it's been like 2 years since I've done anything like this.

    (solve for x)

    8x+22y-40,000 = 0.

    8x+22y=40,000
    8x = -22y+40,000

    goddamnit i lost my fucking graphing calculator. windows calculator -ev.

    *x= -22/8y + 5,000*

    (solve for Y)

    8x+22y=40,000

    22y = -8x +40,000

    *y = -8/22x + 1818.18*


    (plug x value back into original equation)

    8 (-22/8y + 5,000) + 22y = 40,000
    -22y + 40,000 + 22y = 40,000

    40,000 = 40,000

    no shit asshole

    ALRIGHT LETS TRY PLUGGING Y INTO THE ORIGINAL EQUATION. GRAPHING CALCULATOR = MUCH BETTER.


    y = -8/22x + 1818.18

    8x + 22 (-8/22x + 1818.18) = 40,000

    8x + -8x + 40,000 = 40,000
    40,000 = 40,000

    omg so fucking rigg3d

    ALRIGHT LETS TRY PLUGGING THE X FORUMLA INTO THE Y FORUMLA, BOTH OF WHICH I SOLVED FOR



    x= -22/8y + 5,000
    y = -8/22x + 1818.18

    x = -22/8(-8/22x + 1818.18) + 5000
    x = x - 5000 + 5000

    x=x (as an aside, this is always, 100% true, so you can't say I'm wrong.)

    jesus christ, this was so easy for me back in the day....


    anyway I'm sure I made an error somewhere along the way and I'm gonna come here tomorrow and correct it. Seriously. Too tired now.
  26. #26
    euphoricism's Avatar
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    I don't think you made an error, those are the same results I got.

    Its a methodology thing. Since, as nehmer pointed out, there is no correct answer. It is a line, and all points on that line satisfy the equation.

    We solve for y, and graph it.

    It can be graphed using y = -8/22x + 1818.18, as you solved.

    Then its a simple matter of finding a data point within acceptable reasonable ranges. This can easily be done on any graphing calculator, but apparently nothing on the internets can do it. We can also "brute force" it.

    Say we wanted to play 1000 hands on the weekends, exactly how many hands do we need to play on weekdays?

    well, x=1000.
    8x+22y=40000
    8000+22y=40000
    y=1454.54

    so the point 1000,1454.54 satisfys the equation.

    FWIW, I think these are the exact ranges I want, but lets experiment.

    let y=1400 -- 1400 hands during weekdays
    8x+22(1400)=40000
    solve for x, x=1150 hands during weekends.

    Actually, those might be the ideal numbers. Guess I might as well try y=1300.

    8x+22(1300)=40000
    x=1425 ... but if one of our requirements be x<y -- we're playing less hands on weekends than weekdays -- this has to be ignored

    Alright one last equation, y=1500
    8x+22(1500)=40000
    x=875

    Looks like the ideal is one of these three:

    (1150,1400), (1000,1454), or (875,1500)

    Well. Guess I figured out my own answer(s). Now it doesn't matter which one of these I use, and in fact I can substitute one for the other on a weekly basis. One week I can do 1150 hands on the weekends and 1400 hands on weekdays, and one week I can do 875 hands on the weekends and 1500 hands during the week. All depends on my forseeable availability.

    Actually, thats incredibly useful. If I think I'll be very busy next weekend, I know exactly how much during the week I need to compensate.
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  27. #27
    I would rather watch beastiality than read this thread.

    But I could say that for a lot of things.
  28. #28
    Wouldn't you want to play longer sessions on Saturday to take advantage of the drunks?
  29. #29
    euphoricism's Avatar
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    Quote Originally Posted by midas06
    Wouldn't you want to play longer sessions on Saturday to take advantage of the drunks?
    Yes, in reality I'll be moving the 'weekend days' to sunday and monday, when the poker isn't usually as good. This leaves saturday as a weekday.
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  30. #30
    Quote Originally Posted by Lukie
    anyway I'm sure I made an error somewhere along the way and I'm gonna come here tomorrow and correct it. Seriously. Too tired now.
    The only error is (as said earlier) in the assumption that this equation is solvable for a unique (x,y). It is not.

    Period.
    AWOL.
  31. #31
    a500lbgorilla's Avatar
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    LOL at both of you solving one equation with two variables.

    Someone already pointed this problem out to you, then I explained that with the given data you can only make a line and have no way to glean any unique points from it without other info.
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  32. #32
    Lukie's Avatar
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    Quote Originally Posted by a500lbgorilla
    LOL at both of you solving one equation with two variables.

    Someone already pointed this problem out to you, then I explained that with the given data you can only make a line and have no way to glean any unique points from it without other info.
    I think we already came to that conclusion there ol' buddy.
  33. #33
    a500lbgorilla's Avatar
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    Quote Originally Posted by Lukie
    Quote Originally Posted by a500lbgorilla
    LOL at both of you solving one equation with two variables.

    Someone already pointed this problem out to you, then I explained that with the given data you can only make a line and have no way to glean any unique points from it without other info.
    I think we already came to that conclusion there ol' buddy.
    There's no way for you to save face after this thread, shoulda just not posted to let it die faster.
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  34. #34
    Lukie's Avatar
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    Quote Originally Posted by a500lbgorilla
    Quote Originally Posted by Lukie
    Quote Originally Posted by a500lbgorilla
    LOL at both of you solving one equation with two variables.

    Someone already pointed this problem out to you, then I explained that with the given data you can only make a line and have no way to glean any unique points from it without other info.
    I think we already came to that conclusion there ol' buddy.
    There's no way for you to save face after this thread, shoulda just not posted to let it die faster.
    rilla i have more street-cred on this block then anyone you could ever imagine, don't worry about me.

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