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
We need to re-arrange the given equations.
we have $y-2x=-1$; $y-2x=1$, $y+x=1$ and $y+x=3$
Consider $u=y-2x$ and $v=y+x$
Here, $v=y+x$ or, $x=v-y$
and $u=y-2x=-2v+3y$
$\implies y=\dfrac{2v+u}{3}$
Therefore, $x=\dfrac{v}{3}-\dfrac{u}{3}$
Hence, $x=\dfrac{1}{3}(v-u)$ and $y=\dfrac{1}{3}(u+2v)$
where $S=${$(u,v) | -1 \leq u \leq 1, 1\leq v \leq 3$}
