How do you find the exact solutions to the system $3 x^{2} - 20 y^{2} - 12 x + 80 y - 96 = 0$ $3x^2-20y^2-12x+80y-96=0$ and $3 x^{2} + 20 y^{2} = 80 y + 48$ $3x^2+20y^2=80y+48$ ?

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How do you find the exact solutions to the system $3 x^{2} - 20 y^{2} - 12 x + 80 y - 96 = 0$ $3x^2-20y^2-12x+80y-96=0$ and $3 x^{2} + 20 y^{2} = 80 y + 48$ $3x^2+20y^2=80y+48$ ?

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Answer 1 · 2016-12-24T03:21:53

$(x, y) = (- 4, 0), (- 4, 4), (6, 1) and (6, 3)$ $(x, y) = (-4, 0), (-4, 4), (6, 1) and (6, 3)$ . A nice Socratic graph illustrates the crossings of the hyperbola and the ellipse represented by the given equations.

Explanation:

Add and subtract.

$x^{2} - 2 x - 24 = 0$ $x^2-2x-24=0$ , giving $x - 4 and 6$ $x -4 and 6$ .

10y^2-40y+12+3x=0#, giving,

at #x = -4, y(y-4)=0 that gives common points (-4, 0) and (-4, 4) and ,

at $x = 6, y^{2} - 4 y + 3 = 0$ $x = 6, y^2-4y+3= 0$ , giving the common points (6, 1) and (6, 3).

The given equations represent a hyperbola and an ellipse,

respectively

graph{(3x^2-20y^2-12x+80y-96)(3x^2+20y^2-80y-48)=02 [-10, 10, -5, 5]}

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How do you find the exact solutions to the system $3 x^{2} - 20 y^{2} - 12 x + 80 y - 96 = 0$ $3x^2-20y^2-12x+80y-96=0$ and $3 x^{2} + 20 y^{2} = 80 y + 48$ $3x^2+20y^2=80y+48$ ?

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How do you find the exact solutions to the system 3x^2-20y^2-12x+80y-96=03x2−20y2−12x+80y−96=03x^2-20y^2-12x+80y-96=0 and 3x^2+20y^2=80y+483x2+20y2=80y+483x^2+20y^2=80y+48?

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How do you find the exact solutions to the system $3 x^{2} - 20 y^{2} - 12 x + 80 y - 96 = 0$ $3x^2-20y^2-12x+80y-96=0$ and $3 x^{2} + 20 y^{2} = 80 y + 48$ $3x^2+20y^2=80y+48$ ?