SOLUTION: Im REALLY stuck on this Algebra Equasion, Please Help Me!!! I have these two 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

Algebra ->  Equations -> SOLUTION: Im REALLY stuck on this Algebra Equasion, Please Help Me!!! I have these two 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       Log On


   



Question 305894: Im REALLY stuck on this Algebra Equasion, Please Help Me!!!
I have these two 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 i have a Ti-83 but i want to know how to do it algebraically hmmmmm........

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
1.12-x%5E2+-2xy+=+0
2.12-y%5E2+-+2xy+=+0
Use eq. 1 to solve for y.
1.12-x%5E2+-2xy+=+0
12-x%5E2=2xy
y=%2812-x%5E2%29%2F2x
From that, then,
y%5E2=+%2812-x%5E2%29%5E2%2F%284x%5E2%29
and
2xy=12-x%5E2
Substitute into eq. 2,
2.12-y%5E2+-+2xy+=+0
12-+%2812-x%5E2%29%5E2%2F%284x%5E2%29+-+%2812-x%5E2%29+=+0
-+%2812-x%5E2%29%5E2%2F%284x%5E2%29+%2Bx%5E2+=+0
+%2812-x%5E2%29%5E2%2F%284x%5E2%29+=x%5E2
+%2812-x%5E2%29%5E2+=4x%5E4
+12-x%5E2+=0+%2B-+2x%5E2
Two solutions:
12-x%5E2=2x%5E2
3x%5E2=12
x%5E2=4
x=2 and x=-2
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12-x%5E2=-2x%5E2
-x%5E2=12
x%5E2=-12
No real solution.
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When x=-2,
y=%2812-x%5E2%29%2F2x
y=%2812-4%29%2F2%28-2%29
y=-2
When x=2,
y=%2812-x%5E2%29%2F2x
y=%2812-4%29%2F4
y=2
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(-2,-2) and (2,2) are the solutions.
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You are searching for the intersection point between two curves.
The first curve is
y=%2812-x%5E2%29%2F2x (in red below)
The second curve needs to be worked out by completing the square for the second equation,
12-y%5E2+-+2xy+=+0
12=y%5E2%2B2xy
12%2Bx%5E2=y%5E2%2B2xy%2Bx%5E2
12%2Bx%5E2=%28y%2Bx%29%5E2
0+%2B-+sqrt%2812%2Bx%5E2%29=y%2Bx
y=-x+%2B-+sqrt%2812%2Bx%5E2%29 (in blue and green below)
The curves and the two solution points are shown below.