Answer
$x=\dfrac{1}{3}(v-u)$ and $y=\dfrac{1}{3}(u+2v)$
where $S=${$(u,v) | -1 \leq u \leq 1, 1\leq v \leq 3$}
Work Step by Step
Let us consider $u=y-2x$ and $v=y+x$
Then, we have $x=v-y$
Now, we have $u=y-2x$
or, $y-2(v-y)=-2v+3y=\dfrac{2v+u}{3}$
and $x=v-(\dfrac{2v+u}{3})=\dfrac{v}{3}-\dfrac{u}{3}$
Hence, $x=\dfrac{1}{3}(v-u)$ and $y=\dfrac{1}{3}(u+2v)$
where $S=${$(u,v) | -1 \leq u \leq 1, 1\leq v \leq 3$}
