# System of equations - conics

What would be the best way to solve the following the following system of equations? $$x^2+y^2+6y+5=0$$ $$x^2+y^2-2x-8=0$$

This is how I did it, but I am hoping there is a simpler way.

Subtracting the two equations, I get $$6y+2x+13=0$$which simplifies to $$x=-3y-\dfrac{13}2$$I substitute it into the first equation where I get $$9y^2+39y+\dfrac{169}4+y^2+6y+5=0$$ and solving for $y$ I get $$y=-\dfrac{45\pm\sqrt{1552.5}}{20}$$which simplifies to $$y=-\dfrac{9\pm{\sqrt{62.1}}}{4}$$ Plugging this back into the equation at the beginning of this highlighted section I get $$\dfrac{-53\pm{3\sqrt{62.1}}}{2}=-2x$$or $$x=\dfrac{53\pm{3\sqrt{62.1}}}{4}$$ So my final answer is $$\big(\dfrac{53\pm{3\sqrt{62.1}}}{4},-\dfrac{9\pm{\sqrt{62.1}}}{4})$$

## 1 Answer

After the substitution, we have

$$10y^2 + 45 y + \dfrac{189}{4} \implies y_{1, 2} = -\dfrac{9}{4} \pm \dfrac{3~ \sqrt{\dfrac{3}{5}}}{4}$$

This leads to

$$x_{1, 2} = \dfrac{1}{4} \mp \dfrac{9~ \sqrt{\dfrac{3}{5}}}{4}$$