Please explain how.
Already tried graphing it.
actually, two lines ...
$ x^2+xy+x+3y=6$
$x^2 + (y+1)x + (3y-6) = 0$
$x = \dfrac{-(y+1) \pm \sqrt{(y+1)^2 - 4(3y-6)}}{2}$
$x = \dfrac{-(y+1) \pm (y-5))}{2}$
$x = \dfrac{-2y+4}{2} = -y+2$
$x = \dfrac{-6}{2} = -3$
If $\displaystyle x=-3$ then $\displaystyle y(x+3)=-(x+3)(x-2)$ is true for every value of $\displaystyle y$, hence the second line.
The best way to solve this is
$\displaystyle \begin{align*} y(x+3) &= -(x+3)(x-2) \\ (x+3)(y+x-2) &= 0 \end{align*}$
and so we have $\displaystyle x+3=0$ or $\displaystyle y+x-2 = 0$ (or both).