Factor Quadratic Equation into two First Degree Equations

Question

What is the right way to break quadratic equation of following type into two first degree equations?

Take this equation for example: $ax^2 + by^2 + cxy = d, $ where $a,b,c$ and $d$ are constants.

Answer 1

You can solve your equation for $y$: Dividing by $b$ gives $$y^2-\frac{c}{b}xy+\frac{a}{b}x^2-\frac{d}{b}=0$$ solving for $y$ gives $$y_{1,2}=-\frac{c}{2b}x\pm\sqrt{\frac{c^2}{4b^2}x^2-\frac{a}{b}x^2+\frac{d}{b}}$$

Answer 2

Slightly long answer to include basics. For an equation of type $ax^2 + bx+ c = 0$, roots are given by quadratic formula $x = (-b {\pm} \sqrt{b^2-4ac} )/2a$

To solve single equation with two variables x & y, we solve for one variable and treat other variable as constant. So,
$\mathtt ax^2 + by^2 + cxy = d$ can be mapped to standard equation

$\mathtt aX^2 + \quad bX \quad + \quad c\quad\quad = 0 (_1)$ by rewriting original equation as
$\mathtt aX^2 + (cy)X + (by^2-d)\ = 0 (_2)$
where $a_1 = a_2, b_1 = cy, c_1 = by^2-d$ which gives
$X = (-b_1 {\pm} \sqrt{b_1^2-4a_1c_1} )/2a_1$
$X = (-cy {\pm} \sqrt{c^2y^2-4a(by^2-d)} )/2a$

Resolving $\pm$ gives us two roots in terms of y - root1 & root2.
The first degree factors would be (x-root1)(x-root2) = 0 as equation equals zero when x = root1 or root2.