# How do you graph the region 3x-4y>= -12?

Feb 10, 2015

Since $3 x - 4 y = - 12$ is linear, the easiest place to start is to draw the line for that equation by determining the x and y axii intercepts.

When $y = 0$ then $3 x - 4 y = - 12$ becomes $3 x = - 12$ or $x = - 4$

When $x = 0$ then $3 x - 4 y = - 12$ becomes $- 4 y = - 12$ or $y = 3$

We can now draw the line for the equation $3 x - 4 y = - 12$ by drawing a straight line through the two points $\left(- 4 , 0\right)$ and $\left(0 , 3\right)$

All that remains is to determine which side of that line is represented by $3 x - 4 y = - 12$

Consider the point $\left(x , y\right) = \left(0 , 0\right)$ and apply it to the given expression
$3 x - 4 y \ge - 12$ giving $3 \left(0\right) - 4 \left(0\right) \ge - 12$ or $0 \ge - 12$

Since this is obviously true $\left(x , y\right) = \left(0 , 0\right)$ must be within the area described by $3 x - 4 y \ge - 12$

The required graph is therefore the unshaded area in the diagram (plus the line for $3 x - 4 y = - 12$).