## Precalculus (6th Edition) Blitzer

$(-14,-20)$ and $(2,-4)$
Step 1. From the second equation, we have $y=x-6$ Step 2. Substitute it into the first equation; we have $2x^2-(x-6)^2=-8$, $2x^2-x^2+12x-36=-8$ and $x^2+12x-28=0$ Step 3. Factoring the equation, we have $(x-2)(x+14)=0$, which gives $x=-14$ and $x=2$ Step 4. For $x=-14$, we have $y=-14-6=-20$ Step 5. For $x=2$, we have $y=2-6=-4$ Step 6. The solutions are $(-14,-20)$ and $(2,-4)$