Answer
$ \{ (-3, 6),(1,2) \}$
Work Step by Step
We can write the system of equations as:
$x^2=2y-3~~~(1) \\ x+ y=3 ~~~(2)$
Equation (2) can be re-written as: $y=3-x$
Plug $y=3-x$ in equation (1) to obtain:
$ x^2=6-2x-3 \\ x^2+2x-3=0 \\ (x+3)(x-1)=0 \implies x=-3,1$
When $x=-3$, then we have: $y=3+3=6$
When $x=1$, then we have: $y=3-1=2$
Therefore, a solution set is $ \{ (-3, 6),(1,2) \}$
