Question 194702
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Either method works, but the substitution method is computationally more challenging, so let's use elimination.


Rearrange the first equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -x^2 + y = 0]


Now add that to the second equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0x^2 + y^2 + y = 12 \ \ \Rightarrow\ \  y^2 + y = 12 \ \ \Rightarrow\ \ y^2 + y - 12 = 0 \ \ \Rightarrow\ \ (y - 3)(y + 4) = 0]


So, *[tex \LARGE y = 3] or *[tex \LARGE y = -4]


But


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = x^2] so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 = 3 \ \ \Rightarrow\ \ x = \pm\sqrt{3}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 = -4] which has no real solutions.


Hence, the two real solutions are the ordered pairs: *[tex \Large (\sqrt{3},3)] and *[tex \Large (-\sqrt{3},3)]


Graphically:


{{{drawing(
500, 500, -5, 5, -5, 5,
grid(1),
circle(0,0,sqrt(12)),
green(circle(-sqrt(3),3,.1)),
green(circle(sqrt(3),3,.1)),
locate(2,2.9,P1(sqrt(3),3)),
locate(-3,2.9,P2(-sqrt(3),3)),
graph(
500, 500, -5, 5, -5, 5,
x^2
))}}}


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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