How do you solve the system $x^2+y^2+8x-20y+7=0$ and $x^2+9y^2+8x+4y+7=0$ ?

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Answer 1 · 2016-10-06T12:11:48

$(y = 0, x = -7)$ and $(y = 0, x = -1)$

Subtracting the first from the second we obtain

$8y^2+24y = 0$ or
$8(y(y+3))=0$ and the solutions are

$y = 0$ and $y = -3$ .

Now substituting those values in the first equation

1)
$y=0->x^2+8x+7=0$ giving $x=-7$ and $x=-1$
$y = -3->x^2+9+8x+60+7=0$ giving no real solutions

Now substituting in the second equation

2)
$y=0->x^2+8x+7=0$ with the already known solutions $x=-7$ and $x=-1$ .

Finally, the system has the following real solutions.

$(y = 0, x = -7)$ and $(y = 0, x = -1)$

How do you solve the system x^2+y^2+8x-20y+7=0x^2+y^2+8x-20y+7=0 and x^2+9y^2+8x+4y+7=0x^2+9y^2+8x+4y+7=0?