Answer
The values of x, y are $\left( 2,1 \right),\left( 0,-1 \right)$
Work Step by Step
$ x-y=1$ (I)
${{x}^{2}}-x-y=1$ (II)
And subtract (I) from (II), to get:
$\begin{align}
& {{x}^{2}}-x-y-x+y=1-1 \\
& {{x}^{2}}-2x=0 \\
& x\left( x-2 \right)=0
\end{align}$
Therefore, $ x=2$ or $ x=0$.
And to get 'y', put the value of $ x=2$ in (I), to get:
$\begin{align}
& y=x-1 \\
& =2-1 \\
& =1
\end{align}$
Again, to get 'y', put the value of $ x=0$ in (I), to get:
$\begin{align}
& y=x-1 \\
& =0-1 \\
& =-1
\end{align}$
Therefore, the pairs of $\left( x,y \right)$ are $\left( 2,1 \right),\left( 0,-1 \right)$.
Hence, the values of x, y are $\left( 2,1 \right),\left( 0,-1 \right)$.
