SOLUTION: solve the following system by graphing and algebraically. y=1-x^2 4x^2+y^2=16

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Question 572522: solve the following system by graphing and algebraically.
y=1-x^2
4x^2+y^2=16
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
ALGEBRAICALLY
y=1-x%5E2 --> x%5E2=1-y
Substituting into 4x%5E2%2By%5E2=16,
4%281-y%29%2By%5E2=16 --> 4-4y%2By%5E2=16 --> y%5E2-4y-12=0
The solutions to the above equation can be found by factoring
y%5E2-4y-12=0 --> %28y-6%29%28y%2B2%29=0 --> y=6 or y=-2
y=6 does not yield a real solution for x in x%5E2=1-y
y=-2 results in x%5E2=1-%28-2%29 --> x%5E2=3 and x=sqrt%283%29 or x=-sqrt%283%29
The solutions are (-sqrt%283%29,-2) and (sqrt%283%29,-2).
GRAPHICALLY
I do not know if use of a graphing calculator is expected, but without it we could sketch, and then calculate to see if what looks like the intersection of the graphs really adds up.
y=1-x%5E2 graphs as a parabola with maximum at (0,1) and x-intercepts at x=-1 and x=1.
4x%5E2%2By%5E2=16 dividing both sides by 16 turns into x%5E2%2F4%2By%5E2%2F16=1.
That's the standard form of an ellipse centered at the origin, y intercepts (vertices) at y=-4 and y=4, and x-intercepts at x=-2 and x=2.
That would make us expect that the curves cross at negative values of y, with x between -2 and 2.
The graphs would look like this
If the graph suggested to you that the curves cross about at about y=-2, calculations would show you that y=-2 gives the same x value for both equations and you would have the solution.