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Help Eupho with a semi-simple algebra problem.

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euphoricism

06-03-2006, 11:10 PM Post subject: Help Eupho with a semi-simple algebra problem.

#1 (permalink)

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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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euphoricism

06-03-2006, 11:11 PM

#2 (permalink)

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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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Nehmer

06-03-2006, 11:19 PM Post subject: Re: Help Eupho with a semi-simple algebra problem.

#3 (permalink)

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2 variables needs 2 equations to solve... If you just need any solution then, how bout y = 1, x = 4997.25

euphoricism

06-03-2006, 11:54 PM

#4 (permalink)

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But no, there is a way to do this.

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euphoricism

06-04-2006, 12:06 AM

#5 (permalink)

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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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euphoricism

06-04-2006, 12:10 AM

#6 (permalink)

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fuck me. = 30

x+y=30

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Lukie

06-04-2006, 12:11 AM

#7 (permalink)

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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.

euphoricism

06-04-2006, 12:14 AM

#8 (permalink)

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Bah I give up. thats not right either.

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euphoricism

06-04-2006, 12:17 AM

#9 (permalink)

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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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euphoricism

06-04-2006, 12:20 AM

#10 (permalink)

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ahhh but 2x+5y DOES equal 10,000. Four weeks in a month, 4*10,000

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euphoricism

06-04-2006, 12:28 AM

#11 (permalink)

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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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euphoricism

06-04-2006, 12:29 AM

#12 (permalink)

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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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euphoricism

06-04-2006, 12:32 AM

#13 (permalink)

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Nehmer wins. Its just a line.
y = 1/22 (40000-8x)

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a500lbgorilla

06-04-2006, 12:35 AM

#14 (permalink)

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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.

Smithers, use the amnesia ray.
You mean the revolver, sir?
Precisely.

euphoricism

06-04-2006, 12:36 AM

#15 (permalink)

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Now whats the easiest way to look at a whole table of values for that line...

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a500lbgorilla

06-04-2006, 12:37 AM

#16 (permalink)

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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

Smithers, use the amnesia ray.
You mean the revolver, sir?
Precisely.

euphoricism

06-04-2006, 12:39 AM

#17 (permalink)

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how would I do that with excel?

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Lukie

06-04-2006, 12:54 AM

#18 (permalink)

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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

euphoricism

06-04-2006, 12:59 AM

#19 (permalink)

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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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Lukie

06-04-2006, 01:09 AM

#20 (permalink)

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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.

euphoricism

06-04-2006, 01:15 AM

#21 (permalink)

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To an extent, you are correct. To an extent, you are far from correct.

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euphoricism

06-04-2006, 01:17 AM

#22 (permalink)

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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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Lukie

06-04-2006, 01:22 AM

#23 (permalink)

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Quote:

Originally Posted by euphoricism

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?

euphoricism

06-04-2006, 01:36 AM

#24 (permalink)

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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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Lukie

06-04-2006, 07:28 AM

#25 (permalink)

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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.

euphoricism

06-04-2006, 08:08 AM

#26 (permalink)

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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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givememyleg

06-04-2006, 09:09 AM

#27 (permalink)

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06-04-2006, 10:33 AM

#28 (permalink)

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Wouldn't you want to play longer sessions on Saturday to take advantage of the drunks?

euphoricism

06-04-2006, 07:40 PM

#29 (permalink)

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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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mb2447

06-04-2006, 08:25 PM

#30 (permalink)

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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.

a500lbgorilla

06-04-2006, 10:51 PM

#31 (permalink)

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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.

Smithers, use the amnesia ray.
You mean the revolver, sir?
Precisely.

Lukie

06-05-2006, 02:57 AM

#32 (permalink)

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Quote:

Originally Posted by a500lbgorilla

I think we already came to that conclusion there ol' buddy.

a500lbgorilla

06-05-2006, 04:14 AM

#33 (permalink)

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Quote:

Originally Posted by Lukie

Quote:

Originally Posted by a500lbgorilla

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.

Smithers, use the amnesia ray.
You mean the revolver, sir?
Precisely.

Lukie

06-05-2006, 05:08 AM

#34 (permalink)

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Quote:

Originally Posted by a500lbgorilla

Quote:

Originally Posted by Lukie

Quote:

Originally Posted by a500lbgorilla

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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Help Eupho with a semi-simple algebra problem.

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