Answer:
Step-by-step explanation:
Given system of equations :
[tex]y=-x^{2}-x+2;x+y=-2[/tex]
On substituting [tex]y=-2-x[/tex] from second equation in the first equation,
[tex]-2-x=-x^{2}-x+2\\x^{2}=4\\x=\pm2[/tex]
From first equation, when [tex]x=2[/tex], [tex]y=-4[/tex];
when [tex]x=-2[/tex], [tex]y=0[/tex]
∴ [tex]x=2,y=-4[/tex] and [tex]x=-2,y=0[/tex] are the solutions to the system given.
