The given system of equations is:
[tex]\begin{gathered} x^2+y-3=0 \\ x^2-y+1=0 \end{gathered}[/tex]
Add the two equations to eliminate the variable y:
[tex]\begin{gathered} 2x^2-2=0 \\ 2x^2=2 \\ x^2=1 \\ \text{ Therefore,} \\ x=\pm1 \end{gathered}[/tex]
Substite x = 1 into the first equation:
[tex]\begin{gathered} (-1)^2+y-3=0 \\ 1+y-3=0 \\ y=3-1=2 \end{gathered}[/tex]
Subtitute x=-1 into the second equation:
[tex]\begin{gathered} (1)^2+y-3=0 \\ 1+y-3=0 \\ y=3-1=2 \end{gathered}[/tex]
Therefore, the points (-1, 2) and (1, 2) are solutions of the system.
The graph of the system is:
