## Precalculus (6th Edition) Blitzer

The values of x, y are $\left( 2,1 \right),\left( 0,-1 \right)$
$x-y=1$ (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 $\left( 2,1 \right),\left( 0,-1 \right)$.