Help

File: 1464143124471.jpg (1MB, 1412x1200px) Image search: [Google]

1MB, 1412x1200px

>y^2 + xy + x^2 + xy = (y+x)^2 = 50
>y^2 + xy - x^2 - xy = (y-x)^2 = 10
>y+x=+/-sqrt(50)
>y-x=+/-sqrt(10)

12

>>9074211
This. Just do the math.

[eqn]\left( x \,+\, y \right)^2 \,=\, 50 \qquad\text{then}\qquad x \,+\, y \,=\, \pm 5\sqrt2[/eqn]

Assuming [math]x \,+\, y \,=\, 5\sqrt2[/math],
[eqn]\begin{cases} x \,\left( x \,+\, y \right) \,=\, 5\sqrt2\, x \,=\, 20 \\ y \,\left( x \,+\, y \right) \,=\, 5\sqrt2\, y \,=\, 30 \end{cases} \qquad\text{then}\qquad \begin{cases} x \,=\, \frac{4}{\sqrt2} \\ y \,=\, \frac6{\sqrt2} \end{cases}[/eqn]
in which case [math]x\,y \,=\, 12[/math].

Same happens assuming [math]x \,+\, y \,=\, -5\sqrt2[/math].

>>9074193

You don't even need to think very hard. It's just 2 equations with 2 unknowns.

You just solve for one of them and substitute it into the other

File: milhouse is a meme.png (7KB, 206x244px) Image search: [Google]

7KB, 206x244px

>>9074193
[math](x + y)^2 = x^2 + 2xy + y^2 = 20 + 30 = 50[/math]
[math]x + y = \sqrt{50} = 5\sqrt{2}[/math]
[math]x(x+y) = x^2 + xy = 20 \implies x = 4/\sqrt{2}[/math]
[math]y(x+y) = y^2 + xy = 30 \implies y = 6/\sqrt{2}[/math]
[math]xy = 12[/math]

>>9074233
>"this is a linear system"
t. brainlet

y*ln(x)-x^2=-1
sin(y^-x)=0.5

go ahead, give me some solutions for x and y, it's only two equations with two unknowns.

>>9074193
x^2+xy+xy+y^2=50
(x+y)^2=50
x+y=sqrt50
x+y=7
3+4=7
3*4=12

>>9074268
3 and 4 do not satisfy the given equations, and that's a terrible leap in logic (1+6=7 too, but 1*6=6)

>>9074268
>>9074277
also I just realized since you said sqrt(50)=7 that i've been baited

>>9074280
does .07 really make a difference tho

[eqn]yx^2+xy^2=20y[/eqn]
[eqn]xy^2+yx^2=30x[/eqn]
[eqn]y=\frac32 x,~x=\frac23 y[/eqn]
[eqn]y^2-x^2=10[/eqn]
[eqn](\frac32 x+x)(y-\frac23 y)=10[/eqn]
[eqn]\frac52 x\cdot \frac13 y=10[/eqn]
[eqn]xy=\frac{60}5=12[/eqn]

>>9074193
you can interprete each equation as a quadratic equation with a parameter x respectively y.

with that interpretation we can first solve the first equation for x which yields a function x = x(y) and then use that knowledge to solve the second equation for all possible y values.

if there is only one solution x*y has only one result.

OP here. Thanks a lot for your answers - I get it now. It's been a while since I did my A-level exams and it seems obvious now. Thanks again.

>>9074193
12

factored as
x(x+y)=20
y(x+y)=30

you know that x,y≠0 and x+y≠0, so quotient between them defined
so x/y=2/3<->3x=2y

substitute and solve simple quadratic to get x=2√2 and y=3√2

>>9074284
>>9074483
read the fucking thread like 200 other people already solved it for OP dear lord

>>9074193
I think it's
[eqn] xy = \frac{(20)(30)}{50} [/eqn]

>>9074486
bump

File: 1486146506521.png (280KB, 534x436px) Image search: [Google]

280KB, 534x436px

>>9074261
x=1.007212
y=1.901031
Easy

Why does /sci/ love easy problems like OP's so much? I almost never see you guys tackle the more challenging ones posted.

holy shit i feel fucking retarded i havent done shit like this in a long ass time.

i eventually got it. this is the brainlet's solution

x^2 + xy = 20
y^2 + xy = 30
x(x+y) = 20
y(y+x) = 30
x = 20/(x+y)
y= 30/(y+x)
xy = 600/(x+y)^2

x^2 + 600/(x+y)^2 = 20
y^2 + 600/(x+y)^2 =30
1000/(x+y)^2 = 20
1500/(x+y)^2 = 30
(x+y)^2 = 1000/20
(x+y)^2 =1500/30 = 50

Xy = 600/50

>>9074193
Let us assume such numbers exist.
since x(x+y)=20 and y(x+y)=30, x and y are not 0 and y/x=3/2 ie y=3/2x. by replacing y by 3/2x in the forst equation we get: 20=x^2(1+3/2)=5x^2/2 and so x^2=8. But then xy = 3x^2/2 =12.

x:= sqrt(8) and y=3/2 sqrt(8) is the only solution of this system btw.

>>9074193
[math] x(x+y) = 20 \\ y(x + y) = 30 \\ (x-y)(x+y) = -10 \\ y(x-y)(x+y) = -10y \\ 30(x-y) = -10y \\ 30 x - 30y = -10y \\ 30x = 20y \\ x=\frac{20}{30}y[/math]

Take this last result and plug it into the two equations. Then a two single variable quadratic equations will appear. So solve for them.

>>9074205
>>y^2 + xy - x^2 - xy = (y-x)^2 = 10
>y^2-x^2=(y-x)^2
back to school

>>9074486
>implying >>9074284 isn't vastly superior to all previous solutions

>>9075124
wowie its exactly >>9074483
!!

>>9074193
Add and multiply the equations to get:
(x + y)^2 = 50
xy(x+y)^2 = 600
Divide the second equation by the first:
xy = 12

>>9075837
How did xy gain it value of 12?
Why is xy times 50 600?

I understand getting trolled, but still

>>9075837
You win.
/thread

File: IMG_20170731_142103.jpg (1MB, 2448x3264px) Image search: [Google]

1MB, 2448x3264px

>>9074193
did i solve right? with x you can easily find that xy is 20 - (370/20)2

>>9075941

>y=20x - x^2/x
You sure?

>>9076048
yea, i think so...
(x + y) = 20x^-1
so i pass the left most x to the right side
then i multiply both ( 20x^-1 and x) by x/x
resulting in y= (20x - x^2) / x

File: 20170731_195832.jpg (4MB, 4128x3096px) Image search: [Google]

4MB, 4128x3096px

>>9076125

I'm not sure how you're simplifying it, but you need to make the denominator common.

>>9076155
seems like you are right

>>9074261
Being this retarded

>>9074193
xy = 20 - x^2
xy = 30 - y^2
(xy)^2 = 600 - 20y^2 - 30x^2 + (xy)^2
20y^2 + 30x^2 = 600
2y^2 + 3x^2 = 60
2y^2 + 3x^2 = 3x^2 + 3xy
2y^2 = 3xy
2y = 3x
Then substitute and solve.

>>9074228
Holy fuck, this makes sense

How has no-one posted this yet?

x^2 + 2xy + y^2 = 50
(x + y)^2 = 50

x (x + y) = 20
y (x + y ) = 30
xy (x + y)^2 = 600
xy = 600 / (x+y)^2 = 600 / 50 = 12

>>9076437
damn my nigga whatchu doing on /sci/ with dat IQ

>>9074193
Add the equations together:
x*x + 2*x*y + y*y = 50
Solve for x using the quadratic formula:
x = (-2*y ± √(4y*y - 4*y*y + 200))/2
= -y ± 5√(2)
Solutions are x = -y + 5√(2) and x = -y - 5√(2).
Plug in to the second formula:
y*y - y*y ± y*5√(2) = 30
Solve for y:
y = ±3√(2)
Solve for x:
x = ±2√(2)
The two solutions are:
(x,y) = ±√(2)*(2,3)