.
[tex]\begin{gathered} y+12=x^2+x------(1) \\ x+y=3------(2) \\ \text{ From equation \lparen2\rparen:} \\ y=3-x----(3) \\ \text{ Substitute equation \lparen3\rparen into equation \lparen1\rparen:} \end{gathered}[/tex][tex]\begin{gathered} 3-x+12=x^2+x \\ x^2+2x-15=0 \\ x^2+5x-3x-15=0 \\ x(x+5)-3(x+5)=0 \\ (x-3)(x+5)=0 \end{gathered}[/tex]
Therefore,
[tex]x=3,-5[/tex]
Hence, The solution set {(-5, 8), (3, 0)} → y + 12 = x²+x; x + y =3
