MHB Uncovering Solutions for x and y in Nonlinear Equations

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$\left\{\begin{matrix}
x+y+xy=11 ----(1)& \\
x^2+xy+y^2=19---(2)&
\end{matrix}\right.
$
find all $x,y\in R$
 
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$$x+y+xy=11\quad[1]$$

$$x^2+xy+y^2=19\quad[2]$$

$$\text{Note that }x=0,y=0\text{ and }x=y=0\text{ do not give solutions.}$$

$$[1]+[2]\Rightarrow(x+y)^2+(x+y)-30=0\Rightarrow(x+y-5)(x+y+6)=0$$

$$x+y=-6\Rightarrow xy=17\quad(\text{from }[1])$$

$$x+\frac{17}{x}=-6\Rightarrow x^2+6x+17=0\Rightarrow\text{ no solutions.}$$

$$x+y=5\Rightarrow xy=6\quad(\text{from }[1])$$

$$x+\frac6x=5\Rightarrow x^2-5x+6=0\Rightarrow x\in\{2,3\}$$

$$\text{As equations }[1]\text{ and }[2]\text{ are symmetrical, we have }(x,y)=(2,3),(3,2)\text{ as solutions.}$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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