SOLUTION: Im REALLY stuck on this Algebra Equasion, Please Help Me!!! I have these 2 formulas 12-x^2 -2xy = 0 ...12-y^2 - 2xy = 0 ... for some reason x and y are both 2 how do

Question 309466: Im REALLY stuck on this Algebra Equasion, Please Help Me!!!
I have these 2 formulas
12-x^2 -2xy = 0
...12-y^2 - 2xy = 0
...
for some reason x and y are both 2
how do you get this
i tried and got
y= 6/x -0.5x tried to sub into second one but gets messy any way to do this on a calculator by any chance but i want to know how to do it algebraically hmmmmm........
Found 3 solutions by stanbon, Theo, richwmiller:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
12-x^2 -2xy = 0
...12-y^2 - 2xy = 0
-----------------------------
By observation you can see that -x^2 = -y^2
y^2= x^2
y^2-x^2 = 0
(y-x)(y+x) = 0
y = x or y = -x
----------------------
x = y = 0
====================
Cheers,
Stan H.
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
You have 2 equations.
They are:

12 - x^2 - 2xy = 0 and 12 - y^2 - 2xy = 0

Add 2xy to both sides of each equation to get:

2xy = 12 - x^2 and 2xy = 12 - y^2

Since these equations both equal to 2xy, then they both equal to each other and you get:

12 - x^2 = 12 - y^2

Subtract 12 from both sides of this equation to get:

-x^2 = -y^2

Multiply both sides of each equation to get:

x^2 = y^2

Take the square root of both sides of the equation to get:

x = y

Since x = y, then you can substitute x for y in each equation to get:

First equation becomes:

12 - x^2 - 2x^2 = 0 which becomes:

12 - 3x^2 = 0

Add 3x^2 to both sides of the equation to get:

3x^2 = 12

Divide both sides of the equation by 3 to get:

x^2 = 4

Take the square root of both sides of the equation to get:

x = 2

You can do the same by substituting y for x in the other equation.

That equation becomes:

12 - y^2 - 2y^2 = 0 which becomes:

12 - 3y^2 = 0

Solve for y to get:

y = 2

Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
What would happen if you set them equal to each other since they both equal zero
12-x^2 -2xy=12-y^2 - 2xy
add 2xy to both sides
12-x^2=12-y^2
subtract 12
-x^2=-y^2
multiply by -1
x^2=y^2
so we know that x=y
12-x^2-2xy=0
12-x^2-2x^2=0
12-3x^2=0
12=3x^2
4=x^2
x=+-2
solutions are
(2,2)
(-2,-2)
now check them to make sure they both work.

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