Respuesta :

Ryan2
[tex] \left \{ {{y=x^2-4x-5} \atop {y+x=-1}} \right. \\ \\ \left \{ {{y=x^2-4x-5} \atop {y=-1-x}} \right. \\\\ \\ x^2-4x-5=-1-x\\ \\ x^2-3x-4=0\\ \\ \Delta=(-3)^2-4.1.(-4)=9+15=25\\ \\ x=\frac{3 \pm5}{2}\\ \\ x_1=-1 \rightarrow y=-1+1=0\\ \\ x_2=4 \rightarrow y=-1-4=-5 [/tex]

We found 2 points that satisfies both equations  (4,-5) and (-1,0)

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