How do you graph #0.25x+3y>19# on the coordinate plane?

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 = 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]}