Answer
$x=3 \quad$ and $\quad y=-3$ or
$x=-1 \quad$ and $\quad y=-3$
Work Step by Step
We substitute -3 for $y$ in the first equation.
$$
\begin{aligned}
y & =2 x-x^2 \\
-3 & =2 x-x^2 \\
x^2-2 x-3 & =0 \\
(x-3)(x+1) & =0
\end{aligned}
$$
This gives $x=3$ or $x=-1$.
The system's solutions are:
$x=3 \quad$ and $\quad y=-3$ or
$x=-1 \quad$ and $\quad y=-3$
