Find the ordered pair $(x,y)$ if

\begin{align*}
x+y&=(3-x)+(3-y),\\
x-y &=(x-2)+(y-2).
\end{align*}
I notice that both equations involve the sum and difference of $x$ and $y$, so I wonder if I can use some algebraic identity to simplify them. For the first equation, I can rewrite it as $x+y=6-(x+y)$ by combining the terms on the right.
[*  ] Then, adding $x+y$ to both sides, I get $2(x+y)=6$, or $x+y=3$. For the second equation, I can rewrite it as $x-y=x+y-4$ by expanding the parentheses on the right. Then, subtracting $x+y$ from both sides, I get $-2y=-4$, or $y=2$. Now that I have $y$, I can plug it into either equation to find $x$. I'll use the first one: $x+2=3$, or $x=1$. So the ordered pair is $(1,2)$. 