How do you solve the system $4 x^{2} - y^{2} - 8 x + 6 y - 9 = 0$ $4x^2-y^2-8x+6y-9=0$ and $2 x^{2} - 3 y^{2} + 4 x + 18 y - 43 = 0$ $2x^2-3y^2+4x+18y-43=0$ ?

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Answer 1 · 2016-07-25T17:28:38

No real solutions

Making $x^{2} = x_{2}$ $x^2=x_2$ and $y^{2} = y_{2}$ $y^2=y_2$ solve for $x_{2}, y_{2}$ $x_2,y_2$ the system

${ (4 x_2 - y_2 - 8 x + 6 y - 9 =0), (x_2 - 3 y_2 + 4 x + 18 y - 43 = 0) :}$

resulting in

${x_2 = 4/11 (7 x-4), y_2 = (24 x)/11 + 6 y-163/11}$

Now, solving for $x$

$x^2-28/11x+16/11=0$ we obtain

$x = 2/11 (7 pm sqrt[5])$

Now, substituting $x$ in

$y^2 = (24 x)/11 + 6 y-163/11$

and solving for $y$ we cannot obtain real solutions.

Attached a graphic of the two conics.

enter image source here

How do you solve the system 4x^2-y^2-8x+6y-9=04x2−y2−8x+6y−9=04x^2-y^2-8x+6y-9=0 and 2x^2-3y^2+4x+18y-43=02x2−3y2+4x+18y−43=02x^2-3y^2+4x+18y-43=0?