How do you solve the system $x^{2} - 4 y^{2} - 20 x - 64 y - 172 = 0$ $x^2-4y^2-20x-64y-172=0$ and $4 x^{2} + y^{2} - 80 x + 16 y + 400 = 0$ $4x^2+y^2-80x+16y+400=0$ ?

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Answer 1 · 2016-08-11T11:45:29

$((x=6,y=-8),(x=14,y=-8))$

Calling $x2=x^2$ and $y2=y^2$ we have the system

${ (x2 - 4 y2 - 20 x - 64 y - 172 = 0), (4 x2 + y2 - 80 x + 16 y + 400 = 0) :}$

we obtain

$((x2 = 4 (5 x-21)), (y2 = -(64 +16 y)))$

Solving now

$x^2-20 x+84 = 0$ and
$y^2+16 y +64 = 0$

we obtain

$((x=6),(x=14))$

and a double root

$y = -8$
The solutions are

$((x=6,y=-8),(x=14,y=-8))$

How do you solve the system x^2-4y^2-20x-64y-172=0x2−4y2−20x−64y−172=0x^2-4y^2-20x-64y-172=0 and 4x^2+y^2-80x+16y+400=04x2+y2−80x+16y+400=04x^2+y^2-80x+16y+400=0?