How do you graph the inequality #2x + y > 1#?

1 Answer
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 = 0#

#(2 * 0) + y = 1#

#0 + y = 1#

#y = 1# or #(0, 1)#

For: #x = 1#

#(2 * 1) + y = 1#

#2 + y = 1#

#-color(red)(2) + 2 + y = -color(red)(2) + 1#

#0 + y = -1#

#y = -1# or #(1, -1)#

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-1)^2-0.025)((x-1)^2+(y+1)^2-0.025)(2x+y-1)=0 [-10, 10, -5, 5]}

Now, we can shade the right 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{(2x+y-1) > 0 [-10, 10, -5, 5]}