How do you graph #2x+y<=4# on the coordinate plane?

1 Answer
smendyka
Aug 25, 2017

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 = 4#

#0 + y = 4#

#y = 4# or #(0, 4)#

For #y = 0#

#2x + 0 = 4#

#2x = 4#

#(2x)/color(red)(2) = 4/color(red)(2)#

#x = 2# or #(2, 0)#

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 a "or equal to" clause.

graph{(x^2+(y-4)^2-0.075)((x-2)^2+y^2-0.075)(2x+y-4)=0 [-15, 15, -7.5, 7.5]}

Now, we can shade the left side of the line for the "less than" clause in the inequality.

graph{(2x+y-4)<=0 [-15, 15, -7.5, 7.5]}

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