Question

How do you graph `0.25x+3y>19` 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 = 4`

`(0.25 * 4) + 3y = 19`

`1 + 3y = 19`

`1 - color(red)(1) + 3y = 19 - color(red)(1)`

`0 + 3y = 18`

`3y = 18`

`(3y)/color(red)(3) = 18/color(red)(3)`

`y = 6` or `(4, 6)`

For: `x = 16`

`(0.25 * 16) + 3y = 19`

`4 + 3y = 19`

`4 - color(red)(4) + 3y = 19 - color(red)(4)`

`0 + 3y = 15`

`3y = 15`

`(3y)/color(red)(3) = 15/color(red)(3)`

`y = 5` or `(16, 5)`

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-4)^2+(y-6)^2-0.125)((x-16)^2+(y-5)^2-0.125)(0.25x+3y-19)=0 [-30, 30, -15, 15]}

Now, we can shade the right/upper 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{(0.25x+3y-19) > 0 [-30, 30, -15, 15]}