, 2016

The first equation $\dfrac{x-y}{x+y}=9$ gives $x-y = 9x+9y$ and so $-8x = 10y$ or $-4x = 5y$.

The second equation $\dfrac{xy}{x+y} = -60$ gives $xy = -60x - 60y$.

Multiplying this equation by 5 gives $5xy = -300x - 300y$ or $x(5y) = -300x - 60(5y)$.

Since $5y = -4x$, then $x(-4x) = -300x - 60(-4x)$ or $-4x^2 = -60x$.

Rearranging, we obtain $4x^2 - 60x = 0$ or $4x(x-15)=0$.

Therefore, $x = 0$ or $x=15$.

Since $y = -\frac{4}{5}x$, then $y = 0$ or $y = -12$.

From the first equation, it cannot be the case that $x=y=0$.

We can check that the pair $(x,y)=(15,-12)$ satisfies both equations.

Therefore, $(x+y)+(x-y)+xy = 3 + 27 + (-180) = -150$.