# How do you graph 0.25x+3y>19 on the coordinate plane?

Feb 20, 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: $x = 4$

$\left(0.25 \cdot 4\right) + 3 y = 19$

$1 + 3 y = 19$

$1 - \textcolor{red}{1} + 3 y = 19 - \textcolor{red}{1}$

$0 + 3 y = 18$

$3 y = 18$

$\frac{3 y}{\textcolor{red}{3}} = \frac{18}{\textcolor{red}{3}}$

$y = 6$ or $\left(4 , 6\right)$

For: $x = 16$

$\left(0.25 \cdot 16\right) + 3 y = 19$

$4 + 3 y = 19$

$4 - \textcolor{red}{4} + 3 y = 19 - \textcolor{red}{4}$

$0 + 3 y = 15$

$3 y = 15$

$\frac{3 y}{\textcolor{red}{3}} = \frac{15}{\textcolor{red}{3}}$

$y = 5$ or $\left(16 , 5\right)$

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-4)^2+(y-6)^2-0.125)((x-16)^2+(y-5)^2-0.125)(0.25x+3y-19)=0 [-30, 30, -15, 15]}

Now, we can shade the right/upper side of the line. The boundary line needs to be changed to a dashed line because the inequality operator does not contain an "or equal to" clause.

graph{(0.25x+3y-19) > 0 [-30, 30, -15, 15]}