How do you graph the inequality  4x+3y> -12?

Jan 23, 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: $x = 0$

$\left(4 \cdot 0\right) + 3 y = - 12$

$0 + 3 y = - 12$

$3 y = - 12$

$\frac{3 y}{\textcolor{red}{3}} = - \frac{12}{\textcolor{red}{3}}$

$y = - 4$ or $\left(0 , - 4\right)$

For: $y = 0$

$4 x + \left(3 \cdot 0\right) = - 12$

$4 x + 0 = - 12$

$4 x = - 12$

$\frac{4 x}{\textcolor{red}{4}} = - \frac{12}{\textcolor{red}{4}}$

$x = - 3$ 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.
The boundary line will be solid because the inequality operator contains an "or equal to" clause.

graph{(x^2+(y+4)^2-0.05)((x+3)^2+y^2-0.05)(4x+3y+12)=0 [-15, 15, -7.5, 7.5]}

Now, we can shade the right side of the line. We also have to make the boundary line a dashed line because the inequality operator does not contain an "or equal to" clause so the boundary line is not included in the solution set.

graph{(4x+3y+12) > 0 [-15, 15, -7.5, 7.5]}