WMC 2018 Pursuit - Pack 1, D3

If we solve the system for $x$ and $y$ , we get:

$x=\dfrac{{{a}^{2}}+ab+{{b}^{2}}}{ab(a+b)}\hspace{0.2cm}$ and $\hspace{0.2cm}y=-\dfrac{1}{a+b}$

Therefore, $\left| \dfrac{x}{y} \right|=\dfrac{{{a}^{2}}+ab+{{b}^{2}}}{ab}=1+\dfrac{a}{b}+\dfrac{b}{a}$ . But $\dfrac{a}{b}+\dfrac{b}{a}\geq 2$ by AM-GM, so the minimum possible value for $\left| \dfrac{x}{y} \right|$ is $1+2=\boxed{3}$ .