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Answer?

The stories and information posted here are artistic works of fiction and falsehood.
Only a fool would take anything posted here as fact.

Thread replies: 117
Thread images: 25

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Answer?
>>
Not solvable
>>
x^2 + y^2 - xy = (x+y)^2 - 3xy = 4
so
x^3+y^3 = (x+y)(x^2 + y^2 - xy) = 20
>>
>>714485185
you can check this by plugging in the system in WA
http://www.wolframalpha.com/input/?i=xy+%3D+7,+x%2By+%3D+5,+find+x%5E3%2By%5E3
>>
>>714485185
I honestly expected /b/ to be too stupid to solve this
>>
250. Too easy, faggot
>>
>>714485065
x=7/y
x=5-y

5-y=7/y

5y-y^(2)=7

nigga is this diffeq

sum wierd intergration with e?
>>
>>714485752
ignore me

-y^(2)+5y-7=0
>>
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>>714485185
wow you're really good at math kek
>>
>>714485576
X + y = 5

Y = -x + 5

X = -y + 5

(-y + 5)^3 + (-x +5)^3 = 0

-y^3 + 125 - x^3 +125 = 0

125 - x^3 + 125 = y^3

125 + 125 = x^3 + y^3

250 = x^3 + y^3
>>
>>714485966
y=2.5
>>
>>714485185
Hey idiot, how about proving that the two functions relate?
>>
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>>714486073
>He distributes the exponent
>>
>>714486620
OP here, his answer is correct. Also I have no idea what you're saying. What functions? You mean x and y? They relate because I said so. That's how math works.
>>
>>714485752
Naw. The two equations are inconsistent. The only possible values for either variable according to the first are: (-1, -7), or (1, 7). According to the second, the value pairs are anything where y=5-x.
>>
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>>714486627
Well what else do you want me to do with them?
>>
>>714486759
You can relate the equation in terms of x to make a function. Did you ever learn calculus?
>>
>>714486795

(x+y)^2 != x^2 + y^2
(x+y)^2 == x^2 + 2xy +y^2
>>
>>714486795
Then you plug that in for y cubed retard
>>
>>714486783
x and y can be irrational/complex
>>714486940
????? what
>>
>>714486073
Nigga you dont know how to foil
>>
>>714486620
what the fuck does this mean you retard
>>
>>714486783
Abstract. The math is perfectly possible and there is one correct answer but it has no real world application.
>>
>>714485065

x^3+y^3=5^3
>>
>>714487122
are you saying that algebraic manipulations have no real world applciations
>>
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>>714486795
I made this for you <3
>>
>>714485065
y=7/x
y=5-x
7/x=5-x
(5-x)/x=7
(5/x)-1=7
5/x=8
[x=5/8]
y=7-(5/8)
[y=51/8]

(5/8)³+(51/8)³≈259.328125?
>>
>>714486377
my bad again

x=1.634
y=3.366

1.634+3.366=5

1.
>>
>>714486940
OP here, yes I did. None of what you are saying relates to calculus even a little. Also I think it's wrong but I'm not even 100% certain what you're trying to say.
>>
>>714485065
>x = God
>y = God

Because God is always the answer, Bless you all
>>
>>714486795
Not distribute them because that's not how math works you moron.
>>
>>714486759
>>714487118

I think he means solving simultaneously the two definitions of x and y (that is xy=7 and x+y=5)
>>
>>714487662
yes and that's exactly what
>>714485185
does except he doesn't need to explicitly calculate x and y to find x^3+y^3
>>
>>714487344
looks good better than what i came up with
>>
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>>714487344
pic

>>714487407
no
>>
>>714487605
Nigga if I foil, I get -y^3-y^2 -y+815

Let me guess...

Durrrrr factor out a multiple of y + a factor of 815, thats not how math works, retard
>>
>>714485185
Verified via WolframAlpha.
>>
>>714486620
>calls smart anon idiot
>asks idiotic question

stay mad dumbfuck
>>
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>>714485065
((((1/2)*(5 - i sqrt(3))-5)^3)+((1/2)*(5 - i sqrt(3)))
>>
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>>714487831
>imaginary numbers

Im sure that answer exists in lala land, anon
>>
>>714488009
>shouldnt need imaginary numbers for a fucking linear function
>>
>>714488009
you're a retard congratulations
>>
>>714488009
Obviously you know what imaginary numbers really are, since nowhere in that image does it define what i is. So quit trying to look stupid, you're bad at it.
>>
>>714488099
wow an even bigger retard replied to the retard
>>
>>714488182
yeah he should take lessons from you

kek
>>
>>714488099
what imaginary numbers used for?.....srs....
>>
>>714488099
>mfw xy = 7 is a linear function
>>
>>714487284
Nigga, did I ever say you distribute an exponent with two constants? You uwve a variable and a constant
>>
>>714485065
x = r.cos(θ) or ρ.sen(Ф).cos(θ)
y= r.sen(θ) or ρ.sen(Ф).sen(θ)
>>
>>714485185 >>714485185 >>714485185 >>714485185 >>714485185 >>714485185 >>714485185 >>714485185 >>714485185 >>714485185 >>714485185 >>714485185
THIS IS THE CORRECT ANSWER EVERYONE ELSE LEAVE
>>
>>714488264
>he
Don't acting like you're not him.
>>
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>>714488300
MFW Y = -x + 5

Elementary school shit, retard
>>
>>714488293
Imaginary numbers are used when you get the square root of a negative number, which isn't a real number. So you isolate the non real, or "imaginary" number as the square root of -1 as the letter i, then continue your math from there. Otherwise, you say "no real solutions" and leave it at that.
>>
>>714488293
Imaginary numbers are used to describe the roots of negative numbers, which do not have roots in the reals.

They are useful because the real numbers are not closed under multiplication, but if we add in the imaginary axis to create the complex plane we can make a group.

Unless you meant real world application in which case I know they are used a lot in electrical engineering, I don't know what else.
>>
>>714485065
7 is a prime number
>>
>>714486620
Math is self-actualized, jerk.
>>
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>>714488182
>shouldnt need an imaginary number if you dont even have a fuckubf negatice value to take the root of
>>
>>714488099
>linear
>>
>>714488472
>mfw you ignore one equation and insist that the whole system is linear
>>
>>714488616
But its where the function hits the y-axis
>>
>>714485065
No two numbers can add to 5 and multiply and make 7 so OP is a giant faggot
>>
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>>714488301
It doesn't matter. You can't distribute like that.
>>
>>714488783
The two functions are non intersecting. Ergo, there are no real solutions for the system of equations. Which means you can't find X or Y without imaginary numbers.
>>
>>714488853
Who the fuck ever said this was linear?
>>
>>714488301
(a + b)^2 = a^2 + b^2
a^2 + 2ab + b^2 = a^2 + b^2

there ya go fam, back to middle school algebra with you
>>
>>714488301
youre so dumb, its the same fucking thing you dipshit
>>
Math: No one knows all the rules and people will argue for hours and not change their minds.
>>
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>>714489092
No you retard. One is an unknown value and another is attached to it as part of the term as a whole
>>
>>714486627
Freshman's dream baby
>>
>>714488616

Correct but we don't have to stick to the set of Natural Numbers, i.e. 3.5*2 is 7, where 3.5 is an integer.

I mean the solution to the two equations isn't even in the real numbers.

>>714488902

No two natural numbers.
>>
>>714488574
You don't seem to know what you're talking about.
R is perfectly closed under multiplication. If you take two real numbers a and b, ab is always going to be a real number.
Real numbers are not algebraically closed, and C is the algebraic closure of R.
>>
>>714489310
Youre forgetting that

y = -x + 5
Y' = -1
>>
(2.5 + 0.86i)^3+(2.5 - 0.86i)^3
>>
>>714489361
C is an algebraic closure of R. There are others.
>>
>>714489517

Sorry I'm not sure what you mean by that, it's been a while since I've studied math.
>>
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>>714489012
Nigga, according to your logic, 1+2 = 3i

I think i lost a brain cell reading your stupid explanation
>>
>>714485065
Oh, complex numbers. Solution can br either 20+(2401/(10+9Ꭵsqrt(3)) or 20-(2401/(10-9Ꭵsqrt(3))).
>>
>>714486073
(5 - y)^3 = (5 - y) * (5 - y) * (5 - y)
= (25 - 10y + y^2) * (5 - y)
= 125 - 25y - 50y +10y^2 + 5y^2 -y^3
= 125 - 75y + 15y^2 - y^3

fyi
>>
y = 7/x, y = 5 - x, equate the two and you set it to 0. you get a quadratic equation which you can solve for x.

once you get x, y is simply 7/x. plug x and y into the x^3 + y^3 equation and you get 20.

but answer is already posted, thought id give it anyway for people still debating for stupid reasons
>>
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>>714485065
Not solvable. It's only for geniuses.
>>
>>714487794
Ah shit my bad
>>
>>714489655
It might be that the english definition is different from the french one.
Otherwise there are others but they're isomorphic to C so you're really just nitpicking.
>>
>>714485065
thats for reminding me i havent touched serious math in over 15 years @_@ damn....

now all i can think about is proving why sin^2(that) + cos^2(theat) = 1

GAHHHHHHHH
>>
>>714489877
Compared to other popular sites the people here are geniuses. Sad but true.
>>
>>714485065

xy=7
x+y=5

y=5-x
x(5-x)=7
-x^2+5x-7=0

x^2-5x+7=0

do the quadratic formula and boom
>>
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>>714485065
20
>>
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If you cube each side of x+y=5 you can figure this out
>>
>>714485065
my conclusion is, you can only solve that through complex numbers and im too lazy to do that.
>>
>>714490203
you are a genius confrimed
>>
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>>714490203
But wut about imaginary numbuhs, anon?
>>
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>>714490176
tell me how now lel
>>
>>714490552
oshit nigga
no real solutions
we just stepped this up to high school math
>>
>>714490459
Imaginary numbers are another way to do it. Math doesn't necessarily have only one way to the solution.
>>
>>714490968
I can do it but Id rather do what >>714490203
did.
>>
>>714490203
This one i get

Can someone explain how this>>714485185
Anon did it. He didnt explain himself at all
>>
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>>714485065
>>
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>>714491471
>>
>>714491471
He didnt find that on his own, he plugged it into wolfram alpha and was embarrassed to use real numbers
>>
>>714491471
well, you're trying to solve for x^3 + y^3
try to break x^3 + y^3 into elements that you have values for
factor out x+y because we know what x+y is
now you get (x+y)(x^2+y^2-xy) = 5*(x^2+y^2-xy)
now you need to find the value of x^2+y^2-xy
(x+y)^2=x^2+2xy+y^2
so (x+y)^2-3xy = x^2+y^2-xy
you know x+y and xy, so
x^2+y^2-xy=5^2-3*7=25-21=4
x^3 + y^3 = (x+y)(x^2+y^2-xy) = 5*4=20
>>
>>714491471
What? Yes he did. His answer was simpler and more graceful than the guy you understood. You're just too stupid to see it.
>>714492056
Wolfram doesn't show the information he posted.
>>
>>714491471
Not that anon but I believe he subtracted the first equation from the second, which gives you half of the sum of cubes formula. The other half coincidentally is the second equation, so he multiplies everything by the second equation, leaving you with the full sum of cubes.

It is a really elegant solution, even if the steps were not shown.
>>
>>714485065
x³-y³=x³+(2,5+i(3^0,5))

too lazy for the rest
>>
>>714492561
>What? Yes he did. His answer was simpler and more graceful than the guy you understood. You're just too stupid to see it.

I hope you arent so retarded that you dont see how contradictory that is
>>
>>714492715
It's not just elegant, it also generalizes to any commutative ring
>>
>>714485065
This is the kind of shit they try to make us do on school, like how the hell will this be relevant in every day life
>>
>>714492715
>>714493159
OP here, this is why I was impressed right away. Didn't expect this from /b/

>>714492888
Just because YOU don't understand it...
>>
>>714491550

x = -6
>>
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New one
>>
>>714486795
cross product term
>>
You brain so hard...you smart
>>
>>714493901
B
>>
>>714493901
C
>>
>>714493901
G

Each step things move left->right.
Also each shape switches to another.
>>
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Expanded explanation?
>>
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>>
>>714493484
Anon was looking for an explanation. While one was more elegant, the other was more clearly understood by someone not in math.
>>
>>714493484
>so elegant wow impressive yes so smart
>very common question template taught in math classes everywhere
Thread posts: 117
Thread images: 25


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