Answer?

Answer?

Other urls found in this thread:

wolframalpha.com/input/?i=xy = 7, x+y = 5, find x^3+y^3
twitter.com/AnonBabble

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

you can check this by plugging in the system in WA
wolframalpha.com/input/?i=xy = 7, x+y = 5, find x^3+y^3

I honestly expected Sup Forums to be too stupid to solve this

250. Too easy, faggot

x=7/y
x=5-y

5-y=7/y

5y-y^(2)=7

nigga is this diffeq

sum wierd intergration with e?

ignore me

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

wow you're really good at math kek

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

y=2.5

Hey idiot, how about proving that the two functions relate?

>He distributes the exponent

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.

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.

Well what else do you want me to do with them?

You can relate the equation in terms of x to make a function. Did you ever learn calculus?

(x+y)^2 != x^2 + y^2
(x+y)^2 == x^2 + 2xy +y^2

Then you plug that in for y cubed retard

x and y can be irrational/complex
????? what

Nigga you dont know how to foil

what the fuck does this mean you retard

Abstract. The math is perfectly possible and there is one correct answer but it has no real world application.

x^3+y^3=5^3

are you saying that algebraic manipulations have no real world applciations

I made this for you

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?

my bad again

x=1.634
y=3.366

1.634+3.366=5

1.

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.

>x = God
>y = God

Because God is always the answer, Bless you all

Not distribute them because that's not how math works you moron.

I think he means solving simultaneously the two definitions of x and y (that is xy=7 and x+y=5)

yes and that's exactly what
does except he doesn't need to explicitly calculate x and y to find x^3+y^3

looks good better than what i came up with

pic

no

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

Verified via WolframAlpha.

>calls smart user idiot
>asks idiotic question

stay mad dumbfuck

((((1/2)*(5 - i sqrt(3))-5)^3)+((1/2)*(5 - i sqrt(3)))

>imaginary numbers

Im sure that answer exists in lala land, user

>shouldnt need imaginary numbers for a fucking linear function

you're a retard congratulations

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.

wow an even bigger retard replied to the retard

yeah he should take lessons from you

kek

what imaginary numbers used for?.....srs....

>mfw xy = 7 is a linear function

Nigga, did I ever say you distribute an exponent with two constants? You uwve a variable and a constant

x = r.cos(θ) or ρ.sen(Ф).cos(θ)
y= r.sen(θ) or ρ.sen(Ф).sen(θ)

THIS IS THE CORRECT ANSWER EVERYONE ELSE LEAVE

>he
Don't acting like you're not him.

MFW Y = -x + 5

Elementary school shit, retard

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.

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.

7 is a prime number

Math is self-actualized, jerk.

>shouldnt need an imaginary number if you dont even have a fuckubf negatice value to take the root of

>linear

>mfw you ignore one equation and insist that the whole system is linear

But its where the function hits the y-axis

No two numbers can add to 5 and multiply and make 7 so OP is a giant faggot

It doesn't matter. You can't distribute like that.

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.

Who the fuck ever said this was linear?

(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

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.

No you retard. One is an unknown value and another is attached to it as part of the term as a whole

Freshman's dream baby

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.

No two natural numbers.

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.

Youre forgetting that

y = -x + 5
Y' = -1

(2.5 + 0.86i)^3+(2.5 - 0.86i)^3

C is an algebraic closure of R. There are others.

Sorry I'm not sure what you mean by that, it's been a while since I've studied math.

Nigga, according to your logic, 1+2 = 3i

I think i lost a brain cell reading your stupid explanation

Oh, complex numbers. Solution can br either 20+(2401/(10+9Ꭵsqrt(3)) or 20-(2401/(10-9Ꭵsqrt(3))).

(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

Not solvable. It's only for geniuses.

Ah shit my bad

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.

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

Compared to other popular sites the people here are geniuses. Sad but true.

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

20

If you cube each side of x+y=5 you can figure this out

my conclusion is, you can only solve that through complex numbers and im too lazy to do that.

you are a genius confrimed

But wut about imaginary numbuhs, user?

tell me how now lel

oshit nigga
no real solutions
we just stepped this up to high school math

Imaginary numbers are another way to do it. Math doesn't necessarily have only one way to the solution.

I can do it but Id rather do what did.

This one i get

Can someone explain how thisAnon did it. He didnt explain himself at all

...

...

He didnt find that on his own, he plugged it into wolfram alpha and was embarrassed to use real numbers

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

What? Yes he did. His answer was simpler and more graceful than the guy you understood. You're just too stupid to see it.
Wolfram doesn't show the information he posted.