# How do you graph using slope and intercept of 2x+3y= -1?

See below

#### Explanation:

The first thing I'd do is to change the equation from the current standard form and put it into slope-intercept form. We do that by solving for $y$:

$2 x + 3 y = - 1$

$3 y = - 2 x - 1$

$y = - \frac{2}{3} x - \frac{1}{3}$

We're now in slope-intercept form, where

$y = m x + b , m = \text{slope" and b = y-"intercept}$

And so in our question, $m = - \frac{2}{3}$ and $b = - \frac{1}{3}$

Let's first graph the $y$-intercept. That's at $\left(0 , - \frac{1}{3}\right)$:

graph{((x-0)^2+(y+1/3)^2-.1)=0}

Now let's plot a second point.

$m = \frac{\Delta y}{\Delta x} = \text{rise"/"run}$

Our $m = - \frac{2}{3}$. For every 2 that we move up, we move 3 to the left (I'm dealing with the negative sign by having us move left - with a positive slope we'd move to the right). We can start from our first point and move in that way, and so our second point can be found by writing:

$\left(0 - 3 , - \frac{1}{3} + 2\right) = \left(- 3 , \frac{5}{3}\right)$

Let's plot that:

graph{((x-0)^2+(y+1/3)^2-.1)((x+3)^2+(y-5/3)^2-.1)=0}

And now connect the two points with a line:

graph{((x-0)^2+(y+1/3)^2-.1)((x+3)^2+(y-5/3)^2-.1)(2x+3y+1)=0}