# How do you graph the inequality 3x+4y<=12?

Oct 15, 2015

Graph the line of $\left(3 x + 4 y = 12\right)$ using two arbitrary solutions;
then shade in the side of the line containing the point $\left(0 , 0\right)$

#### Explanation:

$3 x + 4 y \le 12$
includes all points on the line $3 x + 4 y = 12$

The x and y intercepts provide a convenient pair of points on this line (although any other pairs would work).
$\textcolor{w h i t e}{\text{XXX}} x = 0 \rightarrow y = 4$
$\textcolor{w h i t e}{\text{XXX}} y = 0 \rightarrow x = 3$
so the pairs $\left(0 , 4\right)$ and $\left(3 , 0\right)$ are on this line.
Plot these point and draw a line through them.

Since the inequality is true for $\left(x , y\right) = \left(0 , 0\right)$
the point $\left(0 , 0\right)$ must be on the selected side of the line just drawn.
graph{3x+4y <= 12 [-10, 10, -5, 5]}