In ths problem, we need to graph a linear equation in standard form: ax + by = c
The easiest way to do this is to use the x-intercept and y-intercept:
- x-intercept, where the graph crosses the x-axis when y = 0
- y-intercept, where the graph crosses the y-axis when x = 0
We are given the equation

To find the x-intercept, we will let y = 0, then solve for x. So,

Divide by -12 on both sides:

The x-intercept is at x = 1, so the point we can graph is (1,0):
Next, we can find the y-intercept by letting x = 0. That gives us:

Divide both sides by 6:

So the y-intercept is y = -2 at the point (0,-2). We can graph that point next:
Finally, we can u