|
|
Binabik
Responsible
Legendary Hero
|
posted February 10, 2010 09:37 AM |
|
|
Step 1. Double check your arithmetic.
Otherwise it's ok.
Just continue what you are doing.
|
|
Binabik
Responsible
Legendary Hero
|
|
posted February 10, 2010 09:39 AM |
|
|
Um, we cross posted. Let me go back and check what you changed.
|
|
angelito
Honorable
Undefeatable Hero
proud father of a princess
|
|
posted February 10, 2010 09:52 AM |
|
|
Quote:
Got as far as...
2x-y+z=1
3x+y+2z=-2
=========
5x+3z=-2
Are you sure the red bolded part is correct? Shouldn't it be -1 ?
____________
Better judged by 12 than carried by 6.
|
|
Mytical
Responsible
Undefeatable Hero
Chaos seeking Harmony
|
|
posted February 10, 2010 09:53 AM |
|
|
Yes, it should. Corrected.
____________
Message received.
|
|
Binabik
Responsible
Legendary Hero
|
|
posted February 10, 2010 10:12 AM |
|
|
I think I'd better bow out of this one so I don't lead you astray. I've realized that I forgot more than I thought.
That kind of sucks because this is fairly basic stuff. I guess almost 40 years without doing it is just a little too long.
Anyway, whatever you did after you said "I believe the next step would be" certainly doesn't look right and I'm not even sure where you got that first equation.
I seem to vaguely recall that there are a couple approaches to simultaneous equations. But I'm afraid I'd just get you mixed up while I try to clear the old cobwebs and figure it out.
|
|
Ecoris
Promising
Supreme Hero
|
|
posted February 10, 2010 10:30 AM |
|
|
You have arrived at the equations
5x+3z=-1
7x+5z=3
From there you proceed as in the first problems you posted. You can continue as follows: Multiply the first equation by -5 and the second one by 3, then add them to eliminate z.
____________
|
|
Corribus 
Hero of Order
The Abyss Staring Back at You
|
|
posted February 10, 2010 03:30 PM |
|
|
@Mytical
These types of linear equations can be solved much more easily using matrices. Just FYI.
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
|
|
TheDeath
Responsible
Undefeatable Hero
with serious business
|
|
posted February 10, 2010 08:16 PM |
|
|
or a computer
____________
The above post is subject to SIRIOUSness.
No jokes were harmed during the making of this signature.
|
|
ohforfsake
Promising
Legendary Hero
Initiate
|
|
posted February 10, 2010 08:47 PM |
|
|
Hey sorry about stealing the thread (just in case you still want help with the equations mysti then I'll submit my 2 cents of what I hope is both understandable and easy (If I recall correct, you have something with a lessened attention span, so I don't know how to address you most effectively) (though I fully support Corribus in that matrices makes it a lot easier, at least in my experience, I don't know if it can be made even easier than that)).
I was considering writing an equation to solve any possible XvsX situation in heroes, like let's say 1 azure dragon vs. 75 ogre magi.
Now what I believe matters are damage given and damage taken. Considering attack and defense should be rather easy, as it is merely the difference that is important and not the number.
Only health and damage changes with the amount of units.
So I have tried to write equations for this and I end up with equations that goes back one step depending on the amount of rounds.
Something like this xn = Ayn + Bxn+1 yn = Cxn + Dyn-1
Or something like that, I honestly can't remember completely, but the method is rather simple.
Now my problem is that I don't know how to find out how many rounds a battle will be? First I tried to generalise it for n rounds and then setting the final health of one side to zero. If the other side ends up at a negative number, it should be clear it is the other side that wins and the roles can be reversed.
However solving for such an equation, I get infinite sums (as n is any number) and everytime I remove one part, it seems like I get another infinite sum (goes to n-1), now coupled with more terms, so the sums are also different (though solveable it seems).
Anyone who has an easier method to deriving such an equation?
About the 3 equations with 3 unknowns, I'd typical try to reduce it to 2 equations of 2 unknowns by multiplying with a factor so I can subtract both equations and one variable is removed.
Like this:
1) 2x-y+z=1
2) 3x+y+2z=-2
3) x-3y+2z=0
Now I'll do it like you and eliminate y, first I'll add equation 1) and 2)
that gives 5x + 3z = -1
Then I'll multiply equation 2) through with 3
9x+3y+6z=-6
Then I'll add this with equation 3)
10x+8z=-6
Now there are 2 equations of 2 unknowns
4) 5x+3z=-1
5) 10x+8z=-6
I'll solve for z first, so I'll multiply equation 4) with -2 and add it to equation 5)
2z=-4
Whereby z=-2
Then I'll use this to solve for x via equation 4) or 5), shouldn't matter, but to be sure I did both multiply and add correct I'll do it for both.
5x+3(-2)=-1 Whereby x = 1
10x+8(-2)=-6 Whereby x = 1
So that went well enough it seems, now solving for y, I check all 3 equations to be certain I made no mistakes
2*1 + y + (-2) = -1 Whereby y = -1
3*1 + y + 2(-2) = -2 Whereby y = -1
1 - 3y + 2(-2) = 0 Whereby y = -1
So it seems like the solution x=1,y=-1,z=-2 holds true for all 3 equations.
____________
Living time backwards
|
|
Ecoris
Promising
Supreme Hero
|
|
posted February 11, 2010 05:28 PM |
|
|
Quote:
I was considering writing an equation to solve any possible XvsX situation in heroes, like let's say 1 azure dragon vs. 75 ogre magi.
What exactly do you wish to know? The probability that the combat lasts n rounds? The probability that one side wins?
____________
|
|
ohforfsake
Promising
Legendary Hero
Initiate
|
|
posted February 11, 2010 06:11 PM |
|
|
Well the more information the merrier I suppose, however most of all it was simpy to find out who is most likely to win.
Especially trying to figure out how much quantity matters, so one maybe could find out stuff like how many demon is required to be raised for inferno pr. skeleton raised for necropolis given the game lasts X weeks.
That is, for Inferno to stand a chance I mean.
____________
Living time backwards
|
|
Ecoris
Promising
Supreme Hero
|
|
posted February 11, 2010 06:32 PM |
|
|
You can'y set up simple formulas to describe XvsY combats. You can get probabilistic estimates. Why not just simulate such combats on a computer? Just set up a generic XvsY simulator.
____________
|
|
Mytical
Responsible
Undefeatable Hero
Chaos seeking Harmony
|
|
posted February 17, 2010 02:45 AM |
|
|
You invest 7000 dollars at 5% and 6%, you make $520. Let x = ammount invested at 5% and y at 6%. Need all formulas and work shown.
x+y = 7000
.05x+.06y=520 seems right, but somehow y = 17000 and since x+y = 7000 that would be impossible....
____________
Message received.
|
|
Binabik
Responsible
Legendary Hero
|
|
posted February 17, 2010 03:50 AM |
|
|
Wow, if I could do that then I'd be rich. You created money out of thin air.
520/7000=7.4%
The answer must be between 5% and 6%.
Therefore?
|
|
Corribus
Hero of Order
The Abyss Staring Back at You
|
|
posted February 17, 2010 04:14 AM |
|
|
@Mytical
Quote:
You invest 7000 dollars at 5% and 6%, you make $520. Let x = ammount invested at 5% and y at 6%. Need all formulas and work shown.
Are you asking us to do your homework for you? If yes: TSK TSK TSK.
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
|
|
Mytical
Responsible
Undefeatable Hero
Chaos seeking Harmony
|
|
posted February 17, 2010 05:46 AM |
|
|
It is similar to a math homework yes, but not exact. .
____________
Message received.
|
|
Binabik
Responsible
Legendary Hero
|
|
posted February 17, 2010 06:36 AM |
|
|
So you took your homework problem and substituted different numbers? Except the new numbers you used don't have a solution.
You had the equations right, you just need to use numbers that work.
|
|
Mytical
Responsible
Undefeatable Hero
Chaos seeking Harmony
|
|
posted February 17, 2010 07:01 AM |
|
Edited by Mytical at 07:11, 17 Feb 2010.
|
It was a different problem that is not part of our homework that was similar enough that if I understood how it was reached I would understand the homework problem.
Here is the complete problem word for word.
You invested a total of 7000 dollars in two different stocks. At the end of the year one paid a profit of 5% and one a profit of 6%. Combined you made a profit of $520. Find the ammount of money invested in each stock.
Guess I am going to have to find another problem that is similar enough  .
____________
Message received.
|
|
Binabik
Responsible
Legendary Hero
|
|
posted February 17, 2010 07:17 AM |
|
|
Try making the interest around 400, then it has a solution.
____________
|
|
Mytical
Responsible
Undefeatable Hero
Chaos seeking Harmony
|
|
posted February 17, 2010 07:19 AM |
|
Edited by Mytical at 07:23, 17 Feb 2010.
|
Hehe then I am just making up the problem, which doesn't really help me. Using similar problems that my book has is the only way I can understand it. The question is..was my formulas correct, ignoring the numbers? Did I have the right formulas?
____________
Message received.
|
|
|
|
This thread is pages long: 
