How do you find the point of intersection(s) for x = 2y2 + 3y + 1 and 2x + 3y2 = 0

Question

A) You cannot find points of intersections for non-functions.
B) Plug in 0 for x into both equations and solve for y. Then plug that answer back into the other equation to find the corresponding x-coordinate.
C) Solve both equations for x and set them equal to each other. This will give you the y-coordinates of the points of intersection. Then plug back into one of the equations to find the corresponding x-coordinates.
D) Solve both equations for y and set them equal to each other. This will give you the x-coordinates of the points of intersection. Then plug back into one of the equations to find the corresponding y-coordinates.

I think its C

Answer 1

You must have meant:
x = 2y^2 + 3y + 1 and 2x + 3y^2 = 0

so you are intersecting two parabolas both with a horizontal axis.

I ran it through Wolfram, and it shows that they do not intersect.
http://www.wolframalpha.com/input/?i=plot+x+%3D+2y%5E2+%2B+3y+%2B+1+and+2x+%2B+3y%5E2+%3D+0+from+-3+to+3

let's attempt to solve them anyway:
plug x from the first into the second

2(2y^2 + 3y + 1) + 3y^2 = 0
7y^2 + 6y + 2 = 0
y = (-6 ±√-20)/14 <----- not real

so there is no intersection

check your typing of the equations