Suppose that $x$ and $y$ satisfy $\frac{x-y}{x+y}=9$ and $\frac{x y}{x+y}=-60$.

The value of $(x+y)+(x-y)+x y$ is
(A) 210
(B) -150
(C) 14160
(D) -14310
(E) -50
The first equation $\frac{x-y}{x+y}=9$ gives $x-y=9 x+9 y$ and so $-8 x=10 y$ or $-4 x=5 y$.

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

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

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

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

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

ANSWER: (B)