How do you find the exact solutions to the system $x^{2} + y^{2} = 25$ $x^2+y^2=25$ and $2 x^{2} + 3 y^{2} = 66$ $2x^2+3y^2=66$ ?

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Answer 1 · 2017-01-01T01:58:18

$(x, y) = (\pm 4, \pm 3)$ $(x, y) = (+-4, +-3)$

These are two linear equations in $x^{2} and y^{2}$ $x^2 and y^2$ .

The second - twice the first gives $x^{2} = 16$ $x^2=16$ , giving $x = \pm 4$ $x = +-4$ .

Substituting in the first, $y^{2} = 9$ $y^2=9$ , giving $y = \pm 3$ $y =+-3$ .

The solutions $(x, y) = (\pm 4, \pm 3)$ $(x, y) = (+-4, +-3)$ give the common points of the

graphs, for the circle and the ellipse, respectively.

graph{(x^2+y^2-25)(2x^2+3y^2-66)=0x^2 [-16, 16, -8, 8]}

How do you find the exact solutions to the system x^2+y^2=25x2+y2=25x^2+y^2=25 and 2x^2+3y^2=662x2+3y2=662x^2+3y^2=66?