Question 572522
ALGEBRAICALLY
{{{y=1-x^2}}} --> {{{x^2=1-y}}}
Substituting into {{{4x^2+y^2=16}}},
{{{4(1-y)+y^2=16}}} --> {{{4-4y+y^2=16}}} --> {{{y^2-4y-12=0}}}
The solutions to the above equation can be found by factoring
{{{y^2-4y-12=0}}} --> {{{(y-6)(y+2)=0}}} --> {{{y=6}}} or {{{y=-2}}}
{{{y=6}}} does not yield a real solution for x in {{{x^2=1-y}}}
{{{y=-2}}} results in {{{x^2=1-(-2)}}} --> {{{x^2=3}}} and {{{x=sqrt(3)}}} or {{{x=-sqrt(3)}}}
The solutions are ({{{-sqrt(3)}}},-2) and ({{{sqrt(3)}}},-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^2}}} graphs as a parabola with maximum at (0,1) and x-intercepts at x=-1 and x=1.
{{{4x^2+y^2=16}}} dividing both sides by 16 turns into {{{x^2/4+y^2/16=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
{{{graph(200,400,-3,3,-6,6,1-x^2,-sqrt(16-4x^2),sqrt(16-4x^2))}}} 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.