How do you graph the inequality #y< =-2/3x+2#?

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Answer 1 · 2018-01-21T13:21:52

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#

#y = (-2/3 * 0) + 2#

#y = 0 + 2#

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

For: #x = 3#

#y = (-2/3 * 3) + 2#

#y = -2 + 2#

#y = 0# 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.

The boundary line will be solid because the inequality operator contains an "or equal to" clause.

graph{(x^2+(y-2)^2-0.025)((x-3)^2+y^2-0.025)(y + (2/3)x-2)=0 [-10, 10, -5, 5]}

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

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