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

1 Answer
Aug 16, 2017

See a solution process below:

Explanation:

First calculate two points on the line for this as an equation instead of an inequality to find the border of the inequality:

For #x = 0#: #(2 xx 0) + y = -3#

#0 + y = -3#

#y = -3# or #(0, -3)#

For #x = -3#: #(2 xx -3) + y = -3#

#-6 + y = -3#

#color(red)(6) - 6 + y = color(red)(6) - 3#

#0 + y = 3#

#y = 3# or #(-3, 3)#

We can now plot these two points and draw a line through them to find the border of the inequality:

graph{(x^2+(y+3)^2-0.05)((x+3)^2+(y-3)^2-0.05)(2x+y+3)=0}

Now that we have the border we can chart the inequality. Because the inequality operator contains a "or equal to" clause it will stay as a solid line. And because it has a "less than" clause we will shade to the left of the line:

graph{(2x+y+3)<=0}