**Q: Solve the system of equations (both of which are conic sections)**

1: x^{2} + y^{2} – 20x + 8y + 7 = 0

2: 9x^{2} + y^{2} + 4x + 8y + 7 = 0

A: I am going to solve this by using the elimination method (since I see that I can cancel out all of the y’s)

I am going to multiply equation 1 by (-1) and then add it to equation 2:

2: 9x^{2} + y^{2} + 4x + 8y + 7 = 0

1: -x^{2} – y^{2} + 20x – 8y – 7 = 0 [after it has been multiplied by -1]

Now, add them together to get equation 3:

3: 8x^{2} +24x = 0

Now, equation 3 only has x’s so we can solve for x (by factoring):

(8x)(x + 3) = 0

So, x = 0 or -3

But, we aren’t done… Find the solution to this system of conics means that we are finding the point(s) of intersection — and it turns out there are two points of intersection since there were two solutions for x. So, our points (solutions) are:

(0, ?) and (-3, ?)

We need to find the y-coordinate of each solution separately. We do this by plugging in the x value to either of the original equations (both will lead us to the same correct solution):

First x-value: (0, ?)

x^{2} + y^{2} – 20x + 8y + 7 = 0

(0)^{2} + y^{2} – 20(0) + 8y + 7 = 0

y^{2} + 8y + 7 = 0

Solve for y by factoring:

(y + 7)(y + 1)=0

y = -7, -1

OH! So this really means that there are more than two solutions…. When x = 0, we found two answers for y.

**(0, -7) and (0, -1)**

Second x-value: (-3, ?)

x^{2} + y^{2} – 20x + 8y + 7 = 0

(-3)^{2} + y^{2} – 20(-3) + 8y + 7 = 0

9 + y^{2} +60 + 8y + 7 = 0

y^{2} + 8y + 76 = 0

*Since this does not factor, we need to solve for y by using the quadratic formula. I did not show my work here, but we end up with a negative under the square root, so there are no real solutions for when x = -3.*

Therefore, there are two real solutions to this equation (2 points of intersection) and they are:

**(0, -7) and (0, -1)**