Answer:
Solution: [tex](-\sqrt{3},1),(\sqrt{3},1)[/tex]
Step-by-step explanation:
System of equation:
[tex]y=x^2-2\text{ and }x^2+y=4[/tex]
Using substitution method.
[tex]y=x^2-2[/tex] substitute into [tex]x^2+y=4[/tex]
[tex]x^2+x^2-2=4[/tex] (combine like term)
[tex]2x^2=4+2[/tex] (Addition property of equality)
[tex]2x^2=6[/tex] (division property of equality)
[tex]x^2=3[/tex] (Taking square root)
[tex]x=\pm \sqrt{3}[/tex]
Substitute x into [tex]y=x^2-2[/tex]
[tex]y=(\sqrt{3})^2-2[/tex]
[tex]y=1[/tex]
Solution: [tex](-\sqrt{3},1),(\sqrt{3},1)[/tex]
Now cross check the solution using graph.
Please find attachment for graph.
