# How do you graph the system x+4y<8 and 3x-y>4?

Jun 3, 2015

First consider the boundary lines:
$\textcolor{w h i t e}{\text{XXXX}}$$x + 4 y = 8$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$this has obvious intercept points $\left(0 , 2\right)$ and $\left(8 , 0\right)$
and
$\textcolor{w h i t e}{\text{XXXX}}$$3 x - y = 4$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$with points $\left(0 , - 4\right)$ and $\left(1 , - 1\right)$

We can use the boundary line points for each of the inequalities to draw their lines and then compare $\left(x , y\right) = \left(0 , 0\right)$ to each of the inequalities to determine which side of each line is to be included.
$\textcolor{w h i t e}{\text{XXXX}}$For example $0 + 4 \left(0\right) < 8$
$\textcolor{w h i t e}{\text{XXXX}}$so $\left(0 , 0\right)$ should be on the included side of $x + 4 y < 8$