How do you graph #3x + 2y<6#?

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Answer 1 · 2018-06-03T19:54:55

See a solution process below:

First, solve for two points as an equation instead of an inequality to find the boundary line for the inequality.

For: #x = 0#

#(3 * 0) + 2y = 6#

#0 + 2y = 6#

#2y = 6#

#(2y)/color(red)(2) = 6/color(red)(2)#

#y = 3# or #(0, 3)#

For: #x = 2#

#(3 * 2) + 2y = 6#

#6 + 2y = 6#

#-color(red)(6) + 6 + 2y = -color(red)(6) + 6#

#0 + 2y = 0#

#2y = 0#

#(2y)/color(red)(2) = 0/color(red)(2)#

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

We can then 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-3)^2-0.035)((x-2)^2+y^2-0.035)(3x+2y-6)=0 [-10, 10, -5, 5]}

Now, we can shade the left side of the line.

We also need to change the boundary line to a dashed line because the inequality operator does not contain an "or equal to" clause.

graph{(3x+2y-6) < 0 [-10, 10, -5, 5]}