/sci/ - Science & Math

forget it, someone already answered...
its pretty easy and I don't think anyone will try this.

You're right it is pretty easy lol

>>9827785
> two variables
> one equation

>>9827785
x = (1/2) + sqrt(6)/12
y = (1/2) - sqrt(6)/12

>>9827823
he asked HOW you mongoloid.

>>9827808
you can still find a solution set you dolt

x = any number

>>9827808
The top line is one equation and the bottom line is the second equation. Now I’m sure you can find the answer.

[math]\sin\theta + \cos\theta = {12 \over 7}[/math]

>>9827785
It's an algebraic variety it has infinite pairs of solutions along a curve.

>>9827785
>>9828045
This particular one is two lines in the plane can probably be simplified to something like (x-2y)*(2x-y)=0 the product of two "lines equations".

http://www.wolframalpha.com/input/?i=plot+(x-2y)*(2x-y)%3D0

>>9828005
x = 0 => 1 = 7/12

>>9827998
>guy doesn't know the simplest of algebra
>somehow he's not the mongoloid

>>9828007
I don't have a brainlet wojak suitable for this

actually the original question was asking the value of x/y+y/x.
This should make the question possible to solve.

>>9828007
I thought the same when I was answering this.
My final answer was something around 3.8, but all the alternatives were < 3.

>>9828045
>>9828051

yeah, it's wherever |(y,x)| = sqrt(7/12) * (x+y)

fix the LHS and you get a pair of lines

>>9828107
>dat division by zero
x = 0 => y = 0

it's wherever ||y,x|| = sqrt(7/12) * (x+y)

assuming y,x > 0, for a fixed norm, it's the two points where the 2-norm of y,x is equal to sqrt(7/12) times the 1-norm of y,x.

Also, OPs equation is a parabola with infinitely many pairs of <x, y> values.

>>9828333
>(x+y)

plus or minus

>>9828334

no, i think it's a pair of lines like the other anon said. c*||y,x|| = sqrt(7/12)*(c*x+c*y) so picking a different norm does not change the ratio

>>9828341
Is this bait?

>>9828354

what's wrong with it? i may have made a dumb mistake, but i'm not sure. i get stuff mixed up in my head a lot.

>>9828341

picking a value R for ||y,x|| you have two solutions (for y,x > 0) given by R=sqrt(7/12)*(y+x).

if you magnify the hypotenuse of your triangle, x, and y are also scaled by the same constant. so you have two sets of colinear solutions.

>>9828375
Never mind, sorry, I redid my calculations and you're right.

>>9828396

i'm explaining it in an incredibly stupid way. please do correct me if i'm wrong.

by the triangle inequality, you'll always have the sum of your two sides, y,x, greater than the hypotenuse. if you fix R, the 2-norm is constant, but y+x only has the right value at two points.

if you take scalar multiples cR, you can see that x and y are both scaled by c, so you have a pair of lines as your solution set

>>9828411
I tried a purely algebraic approach, and made an error forgetting to square x down the line, so I ended up with sqrt(kx) instead of sqrt(kx2), which is a line.

a bit harder than i thought it would be, fun exercise
tex preview is broke so the post might be broke too
here is the general form

[math] y=\frac{x(\sqrt{2n-1}-n)}{n-1} [/math]

in this case n is 7/12

>>9828807
figures
here's the image with the ± correction

>>9827785
x^2+y^2=7
(x+y)^2=12

2 equations, 2 variables.

>>9829008
...only for t=1 in the parametric equations
[math]x^2+y^2 =7t[/math]
[math](x+y)^2 =12t[/math]
where _t_ can be any nonzero real.

>>9828331
>division by 0
lol hope you're not serious

someone already answered this on Brainly. (x/y+y/x, but it doesn't make any sense to me, because as already said, there are infinite pairs of numbers that satisfy the equation)

This is from a national math test for high school on Brazil.

>>9829705
Just because there are an infinite number of solutions doesn't mean x/y + y/x can't be the same for all those solutions

>>9829705
[eqn] \frac{x^2 + y^2}{(x+y)^2} \\
= \frac{1}{\frac{(x+y)^2}{x^2 + y^2}} \\
= \frac{1}{1 + \frac{2xy}{x^2 + y^2}} \\
= \frac{1}{1 + \frac{2}{\frac{x^2 + y^2}{xy}}} \\
= \frac{1}{1 + \frac{2}{\frac{x}{y} + \frac{y}{x}}} [/eqn]

>>9827785
you have x!=y;
write 12(x^2+y^2)=7(x^2+2xy+y^2);
5x^2+5y^2-14xy = 0
5[x^2-14/5*xy + (7^2/5^2)y^2] - 5 ((7/5)^2-1)y^2 = 0
5[(x-7/5*y)^2]-5(24/25*y^2)=0
[x-(7+2sqrt(6))/5* y ][x-(7+2sqrt(6))/5]=0
It should be easy at this point.
method used: I relied on the Gauss algorithm for decomposition of a quadratic form into a linear combination of squares of independent linear forms.

>>9829983
>[x-(7+2sqrt(6))/5* y ][x-(7+2sqrt(6))/5]=0
It is [x-(7+2sqrt(6))/5* y ][x-(7+2sqrt(6))/5*y]=0 sorry

>>9827785
just plug in numbers until you get 7/12 its not that hard brainlets

>>9829993
tfw there is no couple of rational solutions (let alone integers)

>>9829983
>5x^2+5y^2-14xy = 0

It's useless after this step. It's a curve which has unlimited set of x and y, and that curve passes through 0,0
It's only meaningful for further decomsition if one of the variables can be elimited, which is impossible in this case.

>>9830049
Finally a smart anon

>>9830049

the quadratic form has (1,1) signature. The corresponding set is the union of two lines passing by 0.
There are no integer solutions (but the set of real couples of solution is of course infinite).

>>9829993
this

>>9830000
you don't plug in integers, stupid