# How do you graph the inequality 2x - y + 3>0?

Sep 9, 2017

See a solution process below

#### Explanation:

;First, solve for two points as an equation instead of an inequality to find the boundary line for the inequality.

For: $x = 0$

$\left(2 \cdot 0\right) - y + 3 = 0$
$- y + 3 = 0$
$\textcolor{red}{y} - y + 3 = \textcolor{red}{y}$
$0 + 3 = y$
$3 = y$
$y = 3$ or $\left(0 , 3\right)$

For: $x = 1$

$\left(2 \cdot 1\right) - y + 3 = 0$
$2 - y + 3 = 0$
$- y + 2 + 3 = 0$
$- y + 5 = 0$
$\textcolor{red}{y} - y + 5 = \textcolor{red}{y}$
$0 + 5 = y$
$5 = y$
$y = 5$ or $\left(1 , 5\right)$

We can now graph the two points on the coordinate plane and draw a line through the points to mark the boundary of the inequality.

graph{(x^2+(y-3)^2-0.125)((x-1)^2+(y-5)^2-0.125)(2x-y+3)=0 [-20, 20, -10, 10]}

To draw the inequality we will now make the boundary line dashed because the inequality operator does not contain an "or equal to" clause.

We can shade the right side of the line.

graph{(2x-y+3)>0 [-20, 20, -10, 10]}