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
Re-arrange the given equations as: $y-2x=-1$; $y-2x=1$, $y+x=1$ and $y+x=3$
Thus, we have $u=y-2x$ and $v=y+x$
and $v=y+x \implies x=v-y$
Now, we have $u=y-2x=y-2(v-y)=-2v+3y$
This gives: $y=\dfrac{2v+u}{3}$
Therefore, $x=v-(\dfrac{2v+u}{3})=\dfrac{v}{3}-\dfrac{u}{3}$
Hence, $x=\dfrac{1}{3}(v-u)$ and $y=\dfrac{1}{3}(u+2v)$
where $S=${$(u,v) | -1 \leq u \leq 1, 1\leq v \leq 3$}
