# How do you graph 2x-3y<6 on the coordinate plane?

Mar 18, 2018

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: $x = 0$

$\left(2 \cdot 0\right) - 3 y = 6$

$0 - 3 y = 6$

$- 3 y = 6$

$\frac{- 3 y}{\textcolor{red}{- 3}} = \frac{6}{\textcolor{red}{- 3}}$

$y = - 2$ or $\left(0 , - 2\right)$

For: $y = 0$

$2 x - \left(3 \cdot 0\right) = 6$

$2 x - 0 = 6$

$2 x = 6$

$\frac{2 x}{\textcolor{red}{2}} = \frac{6}{\textcolor{red}{2}}$

$x = 3$ or $\left(3 , 0\right)$

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+2)^2-0.035)((x-3)^2+y^2-0.035)(2x-3y-6)=0 [-10, 10, -5, 5]}

Now, we can shade the left side of the line.

The boundary line will be changed to a dashed line because the inequality operator does not contain an "or equal to" clause.

graph{(2x-3y-6) < 0 [-10, 10, -5, 5]}