Answer
$(0,3), (2,-1)$
Work Step by Step
We are given the system:
$\begin{cases}
x-y=3\\
(x-2)^2+(y+3)^2=4
\end{cases}$
We graph both equations on the same system of coordinates.
The graph shows the system has two solutions:
$P(0,-3)$
$Q(2,-1)$
We check is $(x,y)=(0,-3)$ is solution of the system:
$x-y=3$
$0-(-3)\stackrel{?}{=} 3$
$3=3\checkmark$
$(x-2)^2+(y+3)^2=4$
$(0-2)^2+(-3+3)^2\stackrel{?}{=}4$
$4+0\stackrel{?}{=}4$
$4=4\checkmark$
We check is $(x,y)=(2,-1)$ is solution of the system:
$x-y=3$
$2-(-1)\stackrel{?}{=} 3$
$3=3\checkmark$
$(x-2)^2+(y+3)^2=4$
$(2-2)^2+(-1+3)^2\stackrel{?}{=}4$
$0+4\stackrel{?}{=}4$
$4=4\checkmark$
Therefore $(0,3)$ and $(2,-1)$ are the system's solutions.
