Question

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

Answer 1

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]}