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Why counting means nothing to me anymore.



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Eikre

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Say x = y.

We can say that x - y = 0

We can also say that 2x - 2y = x - y = 0

Divide by x-y and we end up with the following:

2 = 1 = 0


If you don't get how I explained it, Let me break it down for you.

[Say x = y]

X and Y are variables. They can be ANY number, so long as the two are equal in value. (Like 1337, for example.)

[We can say that x - y = 0]

Duh. Any number subtracted by itself no longer has a value. (1337 - 1337 = 0)

[We can also say that 2x - 2y = x - y = 0]

Multiply both variables by 2. Or 3. Or any number >9000 if you want. It'll be the same number. (2*1337 - 2*1337 = 1337 - 1337 = 0)

[Divide by x-y and we end up with the following: 2 = 1 = 0]

Take the variables out. Leave everything else in. If you multiplied the variable by 2, leave the 2s behind. If you didn't multiply the variable by anything, leave a 1 behind, because Any number divided by itself is 1.

And I was taught counting was important.
 

Alaude Drenxta

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....shiiiitttttttt

I hate math.

Doesn't work though.

x =/= y

That's why they're different variables.
 

very differentiable
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[Boast mode on] With my fenomanol math skills i will show you your huge mistake [boast mode off]

But seriously the mistake is in divide by variable, if you wrote it out completely you would get this:

2x-2y=x-y=0
divide by variable gives 2-2=1-1=0

Why, well if you say x=y you basically say 2x-2y=2x-2x or 2y-2y.
2x-2x can also be written like 2(x-x) then you divide by the x and get 2(1-1)=2-2=0 and this can be used by replacing every x with a y.

It's called reduction, before you solve equations you have to write them out in their most simple form. If you try to solve 7+3log(243) you have to write the 7 as a log, in this case 3log(3^7).

Nice try, but epic failure, you can't beat maths.
 
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Let's save this until 8:35 Central Time on Monday August 25.

That's when I have Math, on the first day of school.
 

very differentiable
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That's why you have to reduce the equation, then you can solve it, see my earlier post.
 

very differentiable
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Well you told him it was a divide by zero, which would mean it's insolvable, but you can solve it hence reduction. A ten year old would say it's insolvable, i say it is solvable.

Actually, if you would divide by x-y, you would have to see the two and they x-y as tho distinct terms. That would mean you would get: (2/(x-y))*((x-y)/(x-y))= (2/(x-y))*1=2/(x-y) but in that case x≠y.
 
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Ulti

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Someone needs to think before writing a thread. If x-y is 0, then the equation implodes. So v_a is right by the method Eikre was using. Ikkuh did good. Though you should have said x(2-2) instead of 2(x-x). Got a bit confused at that part since you said divide by x and got (2-2).

My calc teach taught this lesson, but not as long. Twas fun.
 

Hollow Bastion

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It's not really as complicated as you all make it out to be. ikkuh pretty much explained it.

If x = y, then x - y = 0.

Since x = y, then x - x = 0 and y - y = 0.

If x - y = 0, then 2(x - y) = 0 which, when distributed, is 2x - 2y = 0

From this point, you can go various directions

Since x - y = 0 and 2x - 2y = 0, then x - y = 2x - 2y

Since 2x - 2y = 2(x - y), then x - y = 2(x - y).

We are not allowed to divide by (x - y), though, because of one of the givens and the piece following it: x = y and x - y = 0. We cannot divide by 0.

Substituting y with x, we get x - x = 2(x - x). Which leads to 0 = 2(0) and, simplified, 0 = 0.
 

Alaude Drenxta

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Can't be a given. They wouldn't be equal, or they'd be the same.


Besides. Maths are gay.

And I can divide by zero.

Endless Devoid / 0 = Infinite
 

Joy

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Of course it can be a given. x and y are placeholders. They mean nothing. They're only different so that we can see that there is not just one variable, but two, and they both happen to represent the same number.

There are equations where x = 0 and then we find y = 0, therefore, x = y
 

Vayne Mechanics

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It's not really as complicated as you all make it out to be. ikkuh pretty much explained it.

If x = y, then x - y = 0.

Since x = y, then x - x = 0 and y - y = 0.

If x - y = 0, then 2(x - y) = 0 which, when distributed, is 2x - 2y = 0

From this point, you can go various directions

Since x - y = 0 and 2x - 2y = 0, then x - y = 2x - 2y

Since 2x - 2y = 2(x - y), then x - y = 2(x - y).

We are not allowed to divide by (x - y), though, because of one of the givens and the piece following it: x = y and x - y = 0. We cannot divide by 0.

Substituting y with x, we get x - x = 2(x - x). Which leads to 0 = 2(0) and, simplified, 0 = 0.
:c **** you, you stole my thunder.
 
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