hi
$\displaystyle x^2 + y^2 = 12 - x - y \quad So, \quad 12 - x - y - xy = 9 \Rightarrow x + y + xy = 3 \Rightarrow x + y = 3 - xy \Rightarrow (x + y)^2 = (3 - xy)^2 \Rightarrow x^2 + 2xy + y^2 = 9 - 6xy + (xy)^2 \Rightarrow x^2 + y^2 + 8xy = 9 + (xy)^2$
You continue. At the end you will find: x = 0, y = 0, xy = 9. When x = 0 => y = +-3, when y = 0 => x = +-3, when xy = 9 => $\displaystyle x^2 + y^2 = 18 \quad and \quad x = \frac{9}{y} ....$
Hello, abualabed!
$\displaystyle \text{If }x,y\text{ are integers, solve this system:}$
. . $\displaystyle x^2 + y^2 + x + y \:=\:12\;\;[1] $
. . . . $\displaystyle x^2 + y^2 - xy \:=\:9\;\;[2]$
Subtract [1] - [2]: .$\displaystyle x + y + xy \:=\:3$
Solve for $\displaystyle y\!:\;\;y \:=\:\frac{3-x}{1+x}$
Substitute into [2]: .$\displaystyle x^2 + \left(\frac{3-x}{1+x}\right)^2 - x\left(\frac{3-x}{1+x}\right) \:=\:9$
. . which simplifies to: .$\displaystyle x^4 + 3x^3 - 9x^2 - 27x \:=\:0$
. . and which factors: .$\displaystyle x(x-3)(x+3)^2 \:=\:0$
Hence: .$\displaystyle \begin{Bmatrix}x &=& 0,3,\text{-}3 \\ y &=& 3,0,\text{-}3 \end{Bmatrix}$
Solutions: .$\displaystyle (x,y) \;=\;(3,0),\;(0,3),\;(\text{-}3,\text{-}3)$
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and
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