4x^2 - 4xy + y^2 - 2x + y - 6 = 0
First can someone tell me what I'm supposed to actually do?
there are two solutions for y in terms of x ...
$\displaystyle 4x^2 - 4xy + y^2 - 2x + y - 6 = 0$
$\displaystyle (4x^2 - 4xy + y^2) - (2x - y) - 6 = 0$
$\displaystyle (2x - y)^2 - (2x - y) - 6 = 0$
$\displaystyle [(2x-y) - 3][(2x-y) + 2] = 0$
$\displaystyle 2x - y = 3$ ... $\displaystyle y = 2x-3$
$\displaystyle 2x - y = -2$ ... $\displaystyle y = 2x+2$