+501 There are two pairs $(x,y)$ of real numbers that satisfy the equation $x+y = 3xy = 4$. Given that the solutions $x$ are in the form $x = \frac{a \pm b\sqrt{c}}{d}$ where a, b, c, and d are positive integers and the expression is completely simplified, what is the value of $a + b + c + d$?
+600 $x+y=4, xy=\frac{4}{3}$.
Solutions to equation $z^2-4z+\frac{4}{3}=0$ by Vieta's.
So $\frac{4\pm \sqrt{\frac{32}{3}}}{2}=\frac{12\pm \sqrt{96}}{6}=\frac{12\pm 4\sqrt{6}}{6}=\frac{6\pm 2\sqrt{6}}{3}$. So $a+b+c+d=6+2+6+3=8+9=\boxed{17}$.
