How do you graph #2x-3y<6# on the coordinate plane?

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Answer 1 · 2018-03-18T22:11:08

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#

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

#0 - 3y = 6#

#-3y = 6#

#(-3y)/color(red)(-3) = 6/color(red)(-3)#

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

For: #y = 0#

#2x - (3 * 0) = 6#

#2x - 0 = 6#

#2x = 6#

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

#x = 3# or #(3, 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.

graph{(x^2+(y+2)^2-0.035)((x-3)^2+y^2-0.035)(2x-3y-6)=0 [-10, 10, -5, 5]}

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

The boundary line will be changed to a dashed line because the inequality operator does not contain an "or equal to" clause.

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