Given the equations y = x2 - 4x - 5 and y + x = -1, one point that satisfies both equations is

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Answer: $\left \{ {{y=x^2-4x-5} \atop {y+x=-1}} \right. \n \n \left \{ {{y=x^2-4x-5} \atop {y=-1-x}} \right. \n\n \n x^2-4x-5=-1-x\n \n x^2-3x-4=0\n \n \Delta=(-3)^2-4.1.(-4)=9+15=25\n \n x=(3 \pm5)/(2)\n \n x_1=-1 \rightarrow y=-1+1=0\n \n x_2=4 \rightarrow y=-1-4=-5$

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

Given the equations y = x2 - 4x - 5 and y + x = -1, one point that satisfies both equations is

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