find all values of y such that $(y^2+y-6)(x^2-6y+9)-2(y^2-9) = 0$ One of the questions on my Algebra 1 homework was to find all values of $y$ such that $$(y^2+y-6)(x^2-6y+9)-2(y^2-9) = 0$$
This was listed under "Lecture 13: Quadratic Equations"
I can not figure out how to approach this problem? Should I expand it? I also noticed that one of the terms used $x$.

$$(y^2+y-6)(x^2-6y+9)-2(y^2-9) = 0 \\ (y-2)(y+3)(x^2-6y+9)-2(y-3)(y+3)=0 \\ (y-2)(y+3)(x^2-6y+9)=2(y-3)(y+3) \\ (y-2)(x^2-6y+9)=2(y-3) \\ x^2-6y+9=\frac{2(y-3)}{y-2} \\ x^2=\frac{2(y-3)}{y-2}+6y-9 \\ x^2= \frac{2(y-3)}{y-2}+\frac{6y(y-2)}{y-2}-\frac{9(y-2)}{y-2} \\x^2= \frac{2(y-3)+6y(y-2)-9(y-2)}{y-2} \\ x^2=\frac{2y-6+6y^2-12y-9y+18}{y-2}=\frac{6y^2-19y+12}{y-2} \\ x^2=\frac{6y^2-19y+12}{y-2}$$
That's the simplest form for the equation you've written above; However there isn't any  unique root (solution) for the equation; as the equation has two variables and needs two equations to find unique $(x,y)$, so solving for $x$ and $y$ can't be; because there isn't enough given. The equation above is said to be a "Relation between $x$ and $y$" Unless the $x$ in the equation become $y$.  

Graphed using: Geogebra.
