Solving the system 12-x²-2xy=0 and 12-y²-2xy=0 for x and y

[dagg3r]

hi this is a hard algebra qestion
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 i have a Ti-83 but i want to know how to do it algebraically

thanks!

 

[VietDao29]

$$\left\{ \begin{array}{l} -x ^ 2 - 2xy + 12 = 0 \ \ (1) & -y ^ 2 - 2xy + 12 = 0 \ \ (2) \end{array} \right.$$
You should note that there are 12, and -2xy in both (1), and (2), so it's common to subtract both sides of (1) from both sides of (2), or to subtract both sides of (2) from both sides of (1).
So (1) - (2) gives:
y² - x² = 0
$$\Leftrightarrow \left[ \begin{array}{l} x = y & x = -y \end{array} \right.$$
So you have 2 cases: x = y and x = -y.
Case 1: x = y,
substitute that back to either (1) or (2), to solve for y, then for x (in this case x = y).
Case 2: x = -y,
substitute that back to either (1) or (2), to solve for y, then for x (in this case x = -y).
Can you go from here?

 

[VietDao29]

12-x^2 -2xy = 0
12-y^2 - 2xy = 0
subtract 2nd Eqn from the 1st to get
-x^2+y^2=0 so x=+/-y,
go back to the 1st eqn & complete the square to get
(x+y)^2 - y^2 = 12
substitute x=+/-y in the above to get
(+/-y+y)^2 - y^2 = 12
simplify
4y^2 - y^2 = 12 or 0 - y^2 = 12
so
y^2 = 4 or (no solution)
hence
y=+/-2 and x=+/-y= so x=y=+/-2 (the solutions are x=y=2 and x=y=-2)

No complete solution, please!
We all know how intelligent you are!
https://www.physicsforums.com/showthread.php?t=28

Read line #9!

 

[benorin]

Line #9! My bad.

 

[giodamelio]

Take a look at this page

http://www.wolframalpha.com/input/?i=12-x^2+-2xy+%3D+0

it should give you the awensers