Given the system of equations:
[tex]\begin{cases}x^2+y^2=16 \\ y+4=x^2\end{cases}[/tex]
As shown in the figure, there are 3 points that are the solutions to the system
For each point given the value of x
So, we will substitute with (x) to find (y)
From the second equation:
[tex]y=x^2-4[/tex]
substitute with x = -2.65, 0, 2.65
[tex]\begin{gathered} x=-2.65\rightarrow y=(-2.65)^2-4=3 \\ x=0\rightarrow y=(0)^2-4=-4 \\ x=2.65\rightarrow y=(2.65)^2-4=3 \end{gathered}[/tex]
So, the answer will be:
[tex]\begin{gathered} (-2.65,3) \\ (0,-4) \\ (2.65,3) \end{gathered}[/tex]
