# Thread: What is the shape of the function: x^2+xy+x+3y=6

2. ## Re: What is the shape of the function: x^2+xy+x+3y=6

$x^2 + x y + x + 3y = 6$

$y(3+x)=6-x-x^2$

$y(x+3) = -(x^2+x-6) = -(x+3)(x-2)$

$y = -(x-2) = 2-x$

$y = -x + 2$

it's a line

3. ## Re: What is the shape of the function: x^2+xy+x+3y=6

Originally Posted by romsek
$x^2 + x y + x + 3y = 6$
it's a line
Actually it is the graph of two lines. SEE HERE

4. ## Re: What is the shape of the function: x^2+xy+x+3y=6

actually, two lines ...

$x^2+xy+x+3y=6$

$x^2 + (y+1)x + (3y-6) = 0$

$x = \dfrac{-(y+1) \pm \sqrt{(y+1)^2 - 4(3y-6)}}{2}$

$x = \dfrac{-(y+1) \pm (y-5))}{2}$

$x = \dfrac{-2y+4}{2} = -y+2$

$x = \dfrac{-6}{2} = -3$

5. ## Re: What is the shape of the function: x^2+xy+x+3y=6

Originally Posted by romsek
$x^2 + x y + x + 3y = 6$

$y(3+x)=6-x-x^2$

$y(x+3) = -(x^2+x-6) = -(x+3)(x-2)$
... it's a line
If $x=-3$ then $y(x+3)=-(x+3)(x-2)$ is true for every value of $y$, hence the second line.

The best way to solve this is

\begin{align*} y(x+3) &= -(x+3)(x-2) \\ (x+3)(y+x-2) &= 0 \end{align*}

and so we have $x+3=0$ or $y+x-2 = 0$ (or both).